Weierstrass Function: Continuous and Bounded on $\mathbb{R}$

[mathmari]

Hey!

I am looking at the following example of a continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$ that is not differentiable at any $x\in \mathbb{R}$.

For $x\in [-1,1]$ we define $\phi (x)=|x|$ and then we extend $\phi$ to the whole $\mathbb{R}$ such that $\phi (x+2)=\phi (x)$.
Then the desired $f:\mathbb{R}\rightarrow \mathbb{R}$ is $$f(x)=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )$$

Then $f$ is continuous and bounded on $\mathbb{R}$.
For $x_0\in \mathbb{R}$ and $k\in \mathbb{N}^{\star}$ take $\delta_k=\pm \frac{1}{2}4^k$, where the sign is chosen such that there is no integer between $4^kx_0$ and $4^k(x_0+\delta_k)$ (it is $4^k\mid \delta_k$).
Define $\gamma_n=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]$.
For $n>k$ it is $\gamma_n=0$ and for $0\leq n\leq k$ it is $|\gamma_n|\leq 4^n$.
Since $|\gamma_k|=4^k$ we have that $$\left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\sum_{n=0}^k\left (\frac{3}{4}\right )^n\gamma_n\right |\geq 3^k-\sum_{n=0}^{k-1}3^n=\frac{1}{2}\left (3^k+1\right )$$ so for $k\rightarrow +\infty$, $\delta_k\rightarrow 0$ and $3^k\rightarrow +\infty$, it follows that... This is not very clear for me.

Let's start from the beginning.

Why do we define in that they the function $\phi$ ? (Wondering)

Then $f$ is continuous and bounded. Could you give me a hint how we can see that? (Wondering)

 

[I like Serena]

Hey mathmari!

The function $\phi$ is just a helper function to define the actual function $f isn't it? (Nerd)

Let's start with bounded.
The function $\phi$ has a range of $[0,1]$ doesn't it?
Isn't the series bounded above by a geometric series then? (Wondering)

Will the series be convergent for any $x$? (Wondering)

 

[mathmari]

The function $\phi$ is just a helper function to define the actual function $f isn't it? (Nerd)

Ahh ok!

Let's start with bounded.
The function $\phi$ has a range of $[0,1]$ doesn't it?

I got stuck right now. We have at the definition that the range is $[0,1]$, yes. But at the definition we have that $x\in [-1,1]$, is $x$ from the same interval at $\phi \left (4^nx\right )$ ? (Wondering)

Isn't the series bounded above by a geometric series then? (Wondering)

Will the series be convergent for any $x$? (Wondering)

So taking that the range of $\phi$ is in general $[0,1]$, then we have the following:
\begin{align*}0\leq \phi \left (4^nx\right ) \leq 1 & \Rightarrow 0\leq \left (\frac{3}{4}\right )^n \phi \left (4^nx\right ) \leq \left (\frac{3}{4}\right )^n \\ & \Rightarrow \sum_{n=0}^{\infty} 0\leq \sum_{n=0}^{\infty} \left (\frac{3}{4}\right )^n \phi \left (4^nx\right ) \leq \sum_{n=0}^{\infty} \left (\frac{3}{4}\right )^n \\ & \Rightarrow 0\leq f(x) \leq \frac{1}{1-\frac{3}{4}} \\ & \Rightarrow 0\leq f(x) \leq \frac{1}{\frac{1}{4}} \\ & \Rightarrow 0\leq f(x) \leq 4\end{align*}

Therefore $f$ is bounded! (Malthe)

 

[I like Serena]

I got stuck right now. We have at the definition that the range is $[0,1]$, yes. But at the definition we have that $x\in [-1,1]$, is $x$ from the same interval at $\phi \left (4^nx\right )$ ?

Don't forget that we "extend $ϕ$ to the whole $\mathbb R$."
Doesn't that mean that $\phi(x)$ is defined for any $x$ in $\mathbb R$? (Wondering).

So taking that the range of $\phi$ is in general $[0,1]$, then we have the following:
\begin{align*}0\leq \phi \left (4^nx\right ) \leq 1 & \Rightarrow 0\leq \left (\frac{3}{4}\right )^n \phi \left (4^nx\right ) \leq \left (\frac{3}{4}\right )^n \\ & \Rightarrow \sum_{n=0}^{\infty} 0\leq \sum_{n=0}^{\infty} \left (\frac{3}{4}\right )^n \phi \left (4^nx\right ) \leq \sum_{n=0}^{\infty} \left (\frac{3}{4}\right )^n \\ & \Rightarrow 0\leq f(x) \leq \frac{1}{1-\frac{3}{4}} \\ & \Rightarrow 0\leq f(x) \leq \frac{1}{\frac{1}{4}} \\ & \Rightarrow 0\leq f(x) \leq 4\end{align*}

Therefore $f$ is bounded!

Yep. (Nod)

Can we deduce somehow that the series converges for any $x$? (Wondering)

 

[mathmari]

Don't forget that we "extend $ϕ$ to the whole $\mathbb R$."
Doesn't that mean that $\phi(x)$ is defined for any $x$ in $\mathbb R$? (Wondering).

I got stuck right now. Does this mean that for each $x\in \mathbb{R}$ the range is $[0,1]$ ? (Wondering)

Can we deduce somehow that the series converges for any $x$? (Wondering)

From the comparison test with the series $\sum_{n=0}^{\infty} \left (\frac{3}{4}\right )^n$ do we not have that that the series converges? Or do we not look in that way if the series converges for any $x$ ? (Wondering)

 

[I like Serena]

I got stuck right now. Does this mean that for each $x\in \mathbb{R}$ the range is $[0,1]$ ?

We extend $\phi$ such that $\phi(x+2)=\phi(x)$ don't we?
So what would for instance $\phi(x)$ be for $x=1.5$? (Wondering)

From the comparison test with the series $\sum_{n=0}^{\infty} \left (\frac{3}{4}\right )^n$ do we not have that that the series converges? Or do we not look in that way if the series converges for any $x$ ?

The comparison tells us that the series has a lower bound of 0 and an upper bound of 4.
That is not sufficient for convergence though.
For instance a series like $2-1+1-1+1-1+\ldots$ is at all times within the same bounds, but it is not convergent is it? (Worried)

 

[mathmari]

We extend $\phi$ such that $\phi(x+2)=\phi(x)$ don't we?
So what would for instance $\phi(x)$ be for $x=1.5$? (Wondering)

We have that $\phi (1.5)=\phi(-0.5)=|0.5|=0.5\in [0,1]$. So we can apply that formula so many times till the result is in $[0,1]$, right? (Wondering)

The comparison tells us that the series has a lower bound of 0 and an upper bound of 4.
That is not sufficient for convergence though.
For instance a series like $2-1+1-1+1-1+\ldots$ is at all times within the same bounds, but it is not convergent is it? (Worried)

Ok... What do we have to do then? Which convergence criteria do we have to use? (Wondering)

 

[I like Serena]

We have that $\phi (1.5)=\phi(-0.5)=|0.5|=0.5\in [0,1]$. So we can apply that formula so many times till the result is in $[0,1]$, right?

Yep. (Nod)

Ok... What do we have to do then? Which convergence criteria do we have to use?

Isn't there a theorem that says that an increasing sequence that has an upper bound is convergent? (Thinking)

 

[mathmari]

Isn't there a theorem that says that an increasing sequence that has an upper bound is convergent? (Thinking)

Ohh yes (Wasntme) We have the following, or not?
\begin{align*}f(x+1)&=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x+1)\right ) =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n\right )\\ & =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n-2\right )\\ & =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n-2-2\right )=\ldots \\ & =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n-2^{2n}\right )=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )\\ & =f(x)\end{align*}

(Wondering)

 

[I like Serena]

We have the following, or not?
\begin{align*}f(x+1)&=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x+1)\right ) =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n\right )\\ & =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n-2\right )\\ & =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n-2-2\right )=\ldots \\ & =\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx+4^n-2^{2n}\right )=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )\\ & =f(x)\end{align*}

Yes. (Thinking)

 

[mathmari]

Yes. (Thinking)

So since $f(x+1)=f(x)$ it also holds that $f(x+1)\geq f(x)$, which implies that $f$ is increasing.

So since $f$ is increasing and bounded, the series converges.

Does it follow from that that $f$ is continuous? (Wondering)

 

[I like Serena]

So since $f(x+1)=f(x)$ it also holds that $f(x+1)\geq f(x)$, which implies that $f$ is increasing.

So since $f$ is increasing and bounded, the series converges.

Hold on. (Wait)

We have a series to evaluate $f(x)$ at a specific value of $x$, don't we?
I'm afraid that the value of $f(x)$ at other values of $x$ is not involved.
Moreover, we cannot tell that $f$ itself is increasing from that, and actually it is not. (Shake)

Let's examine $f$ at a specific value of $x$. Say $x_0$.
So we have:
$$f(x_0)=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )
=\phi(x_0) + \frac 34\phi(4x_0)+\left (\frac{3}{4}\right )^2\phi(4^2x_0)+\ldots
$$
We know that each partial sum has a lower bound of 0 and an upper bound of 4.
And each term is non-negative isn't it?
So the sequence of partial sums is increasing - all for a specific value of $x=x_0$ - isn't it?
Doesn't that mean that the series converges so that $f(x_0)$ is well-defined? (Wondering)

Does it follow from that that $f$ is continuous?

Let's start with continuity of $f$ at some point $x_0$.

First off, we need that $f(x_0)$ is well-defined.
That is, if we evaluate the series at $x=x_0$ it must converge to some value.

The next step is that we evaluate $\lim\limits_{x\to x_0} f(x)$.
Is it the same as $f(x_0)$? (Wondering)

 

[mathmari]

We have a series to evaluate $f(x)$ at a specific value of $x$, don't we?
I'm afraid that the value of $f(x)$ at other values of $x$ is not involved.
Moreover, we cannot tell that $f$ itself is increasing from that, and actually it is not. (Shake)

Let's examine $f$ at a specific value of $x$. Say $x_0$.
So we have:
$$f(x_0)=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )
=\phi(x_0) + \frac 34\phi(4x_0)+\left (\frac{3}{4}\right )^2\phi(4^2x_0)+\ldots
$$
We know that each partial sum has a lower bound of 0 and an upper bound of 4.
And each term is non-negative isn't it?
So the sequence of partial sums is increasing - all for a specific value of $x=x_0$ - isn't it?
Doesn't that mean that the series converges so that $f(x_0)$ is well-defined? (Wondering)

So we have that for a specific value of $x$ the series is increasing and bounded and so it converges. The resulting function must be then well-defined? (Wondering)

Let's start with continuity of $f$ at some point $x_0$.

First off, we need that $f(x_0)$ is well-defined.
That is, if we evaluate the series at $x=x_0$ it must converge to some value.

The next step is that we evaluate $\lim\limits_{x\to x_0} f(x)$.
Is it the same as $f(x_0)$? (Wondering)

We have that $$\lim\limits_{x\to x_0} f(x)=\lim\limits_{x\to x_0}\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\lim\limits_{x\to x_0}\phi \left (4^nx\right )$$
Does it hold that $\lim\limits_{x\to x_0}\phi \left (4^nx\right )=\phi \left (4^nx_0\right )$ ? (Wondering)

 

[I like Serena]

So we have that for a specific value of $x$ the series is increasing and bounded and so it converges. The resulting function must be then well-defined?

Yes. (Nod)
It means that $f(x)$ is defined for every $x\in\mathbb R$.

So we now have that $f$ is a function with domain $\mathbb R$.
And we have that $f$ is bounded. (Nerd)

We have that $$\lim\limits_{x\to x_0} f(x)=\lim\limits_{x\to x_0}\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx\right )=\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\lim\limits_{x\to x_0}\phi \left (4^nx\right )$$
Does it hold that $\lim\limits_{x\to x_0}\phi \left (4^nx\right )=\phi \left (4^nx_0\right )$ ?

Yep. (Nod)

 

[mathmari]

Yes. (Nod)
It means that $f(x)$ is defined for every $x\in\mathbb R$.

So we now have that $f$ is a function with domain $\mathbb R$.
And we have that $f$ is bounded. (Nerd)

Ahh ok! (Malthe)

Yep. (Nod)

Ok! So this means that $f$ is continuous at $x_0$.
Can we say that since this holds for any $x_0$, the function is continuous? (Wondering)

 

[I like Serena]

Ok! So this means that $f$ is continuous at $x_0$.
Can we say that since this holds for any $x_0$, the function is continuous?

Yep. (Nod)

 

[mathmari]

For $x_0\in \mathbb{R}$ and $k\in \mathbb{N}^{\star}$ take $\delta_k=\pm \frac{1}{2}4^k$, where the sign is chosen such that there is no integer between $4^kx_0$ and $4^k(x_0+\delta_k)$ (it is $4^k\mid \delta_k$).
Define $\gamma_n=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]$.

Is there a specific reason that we take $\delta_k$ in such a way? (Wondering)

For $n>k$ it is $\gamma_n=0$ and for $0\leq n\leq k$ it is $|\gamma_n|\leq 4^n$.

Let $n>k$ then $n=k+m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k+m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^k}{2}\left [\phi \left (4^{k+m}(x_0\pm \frac{1}{2}4^k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^k}{2}\left [\phi \left (4^{k+m}x_0\pm \frac{1}{2}4^{2k+m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

Then $\frac{1}{2}4^{2k+m}$ is a multiple $2$ and so using several times the property $\phi (x+2)=\phi (x)$ it holds that $\phi \left (4^{k+m}x_0\pm \frac{1}{2}4^{2k+m}\right )=\phi \left (4^nx_0\right )$ and so we get that $\gamma_n=0$.

Is this part correct? (Wondering) How do we get that $|\gamma_n|\leq 4^n$ if $0\leq n\leq k$ ? (Wondering)

 

[I like Serena]

Is there a specific reason that we take $\delta_k$ in such a way?

I'm guessing we are trying to prove that $f$ is not differentiable anywhere?
Then we want a $\delta >0$ that is 'small' and gives us some special result depending on whether $n$ is greater than $k$ or smaller.

Anyway, it says: "the sign is chosen such that there is no integer between $4^kx_0$ and $4^k(x_0+δ_k)$".
But there are always integers in between, aren't there? (Worried)
Can it be that there is a typo?
Was perhaps $\delta_k=\pm \frac{1}{2}4^{-k}$ intended? (Wondering)

How do we get that $|\gamma_n|\leq 4^n$ if $0\leq n\leq k$ ?

That would become true if we have $\delta_k=\pm \frac{1}{2}4^{-k}$ instead wouldn't it? (Thinking)

 

[mathmari]

I'm guessing we are trying to prove that $f$ is not differentiable anywhere?
Then we want a $\delta >0$ that is 'small' and gives us some special result depending on whether $n$ is greater than $k$ or smaller.

Anyway, it says: "the sign is chosen such that there is no integer between $4^kx_0$ and $4^k(x_0+δ_k)$".
But there are always integers in between, aren't there? (Worried)
Can it be that there is a typo?
Was perhaps $\delta_k=\pm \frac{1}{2}4^{-k}$ intended? (Wondering)

Ahh because with $\delta_k=\pm \frac{1}{2}4^{k}$ we would have $4^kx_0$ and $4^k(x_0+δ_k)=4^kx_0+\frac{1}{2}4^{2k}$ and between them there are integers.

If we have $\delta_k=\pm \frac{1}{2}4^{-k}$ then we get $4^kx_0$ and $4^k(x_0+δ_k)=4^kx_0+\frac{1}{2}$ which is possible that in between there are no integers.

So you are right that it must be $\delta_k=\pm \frac{1}{2}4^{-k}$ intended. (Malthe)

So, we have the following:

Let $n>k$ then $n=k+m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k+m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^{k+m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

Then $\frac{1}{2}4^{m}$ is a multiple of $2$ and so using several times the property $\phi (x+2)=\phi (x)$ it holds that $\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )=\phi \left (4^nx_0\right )$ and so we get that $\gamma_n=0$.

Is this part correct? (Wondering) For $0\leq n\leq k$ then $n=k-m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k-m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^{k-m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

How do we ontinue? (Wondering)

 

[I like Serena]

Ahh because with $\delta_k=\pm \frac{1}{2}4^{k}$ we would have $4^kx_0$ and $4^k(x_0+δ_k)=4^kx_0+\frac{1}{2}4^{2k}$ and between them there are integers.

If we have $\delta_k=\pm \frac{1}{2}4^{-k}$ then we get $4^kx_0$ and $4^k(x_0+δ_k)=4^kx_0+\frac{1}{2}$ which is possible that in between there are no integers.

So you are right that it must be $\delta_k=\pm \frac{1}{2}4^{-k}$ intended. (Malthe)

So, we have the following:

Let $n>k$ then $n=k+m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k+m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^{k+m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

Then $\frac{1}{2}4^{m}$ is a multiple of $2$ and so using several times the property $\phi (x+2)=\phi (x)$ it holds that $\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )=\phi \left (4^nx_0\right )$ and so we get that $\gamma_n=0$.

Is this part correct?

Shouldn't we have:
$$\frac{1}{\delta_k}=\frac{1}{\pm \frac 12 4^{-k}}=\pm 2\cdot 4^k$$
? (Worried)

For $0\leq n\leq k$ then $n=k-m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k-m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^{k-m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{\pm4^{-k}}{2}\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

How do we continue?

Isn't the slope of $\phi$ either $+1$ or $-1$ everywhere?
And it only flips when we pass through an integer doesn't it? (Wondering)

We picked $\delta_k$ such that there was no integer in between, so it wouldn't flip in between would it? (Thinking)

 

[mathmari]

Shouldn't we have:
$$\frac{1}{\delta_k}=\frac{1}{\pm \frac 12 4^{-k}}=\pm 2\cdot 4^k$$
? (Worried)

Ahh yes!

So, we have the following:

Let $n>k$ then $n=k+m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k+m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm2 \cdot 4^{k}\left [\phi \left (4^{k+m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm2 \cdot 4^{k}\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

Then $\frac{1}{2}4^{m}$ is a multiple of $2$ and so using several times the property $\phi (x+2)=\phi (x)$ it holds that $\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )=\phi \left (4^nx_0\right )$ and so we get that $\gamma_n=0$.

Right? (Wondering)

Isn't the slope of $\phi$ either $+1$ or $-1$ everywhere?
And it only flips when we pass through an integer doesn't it? (Wondering)

We picked $\delta_k$ such that there was no integer in between, so it wouldn't flip in between would it? (Thinking)

Do you mean that the derivative is either $1$ or $-1$ and so we have that ? (Wondering)
For $0\leq n\leq k$ then $n=k-m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k-m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm 2\cdot 4^k\left [\phi \left (4^{k-m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm 2\cdot 4^k\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

From IVT we have that $$\phi' (\xi)=\frac{\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )}{4^nx_0\pm \frac{1}{2}4^{-m}-4^nx_0}=\frac{\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )}{\pm \frac{1}{2}4^{-m}}=\pm 2\cdot 4^m\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]$$ for some $4^nx_0<\xi <4^nx_0\pm \frac{1}{2}4^{-m}$.

So we get $$\pm 2 \left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]=\frac{\phi '(\xi)}{4^m}$$

Therefore we get $$\gamma_n=4^k\cdot \frac{\phi '(\xi)}{4^m}=4^{k-m}\cdot \phi '(\xi)=4^{n}\cdot \phi '(\xi) \Rightarrow |\gamma_n|=4^n\cdot |\phi '(\xi)|\leq 4^n\cdot 1=4^n$$

 

[I like Serena]

Ahh yes!

So, we have the following:

Let $n>k$ then $n=k+m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k+m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm2 \cdot 4^{k}\left [\phi \left (4^{k+m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm2 \cdot 4^{k}\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

Then $\frac{1}{2}4^{m}$ is a multiple of $2$ and so using several times the property $\phi (x+2)=\phi (x)$ it holds that $\phi \left (4^nx_0\pm \frac{1}{2}4^{m}\right )=\phi \left (4^nx_0\right )$ and so we get that $\gamma_n=0$.

Right? (Wondering)

Do you mean that the derivative is either $1$ or $-1$ and so we have that ? (Wondering)
For $0\leq n\leq k$ then $n=k-m$ for some positive $m$.
Then we have that \begin{align*}\gamma_n&=\frac{1}{\delta_k}\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\frac{1}{\delta_k}\left [\phi \left (4^{k-m}(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm 2\cdot 4^k\left [\phi \left (4^{k-m}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^nx_0\right )\right ]\\ & =\pm 2\cdot 4^k\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]\end{align*}

From IVT we have that $$\phi' (\xi)=\frac{\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )}{4^nx_0\pm \frac{1}{2}4^{-m}-4^nx_0}=\frac{\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )}{\pm \frac{1}{2}4^{-m}}=\pm 2\cdot 4^m\left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]$$ for some $4^nx_0<\xi <4^nx_0\pm \frac{1}{2}4^{-m}$.

So we get $$\pm 2 \left [\phi \left (4^nx_0\pm \frac{1}{2}4^{-m}\right )-\phi \left (4^nx_0\right )\right ]=\frac{\phi '(\xi)}{4^m}$$

Therefore we get $$\gamma_n=4^k\cdot \frac{\phi '(\xi)}{4^m}=4^{k-m}\cdot \phi '(\xi)=4^{n}\cdot \phi '(\xi) \Rightarrow |\gamma_n|=4^n\cdot |\phi '(\xi)|\leq 4^n\cdot 1=4^n$$

Yep. All correct. (Nod)

 

[mathmari]

Yep. All correct. (Nod)

For $n=k$ we have the following:
\begin{align*}\gamma_k&=\frac{1}{\delta_k}\left [\phi \left (4^k(x_0+\delta_k)\right )-\phi \left (4^kx_0\right )\right ]\\ & =\pm 2\cdot 4^k\left [\phi \left (4^{k}(x_0\pm \frac{1}{2}4^{-k})\right )-\phi \left (4^kx_0\right )\right ]\\ & =\pm 2\cdot 4^k\left [\phi \left (4^kx_0\pm \frac{1}{2}\right )-\phi \left (4^kx_0\right )\right ]\end{align*}

From IVT we get \begin{equation*}\phi' (\xi)=\frac{\phi \left (4^kx_0\pm \frac{1}{2}\right )-\phi \left (4^kx_0\right )}{4^kx_0\pm \frac{1}{2}-4^kx_0}=\frac{\phi \left (4^kx_0\pm \frac{1}{2}\right )-\phi \left (4^kx_0\right )}{\pm \frac{1}{2}}=\pm 2\cdot \left [\phi \left (4^kx_0\pm \frac{1}{2}\right )-\phi \left (4^kx_0\right )\right ]\end{equation*}

So we get \begin{equation*}\gamma_k=4^k\cdot \phi' (\xi) \Rightarrow |\gamma_k|=|4^k\cdot \phi' (\xi)|=|4^k|\cdot |\phi' (\xi)|\end{equation*} Isn't it $|\phi '(\xi)|\leq 1$ ? How do we get $|\gamma_k|=4^k$ ? (Wondering)

 

[I like Serena]

So we get \begin{equation*}\gamma_k=4^k\cdot \phi' (\xi) \Rightarrow |\gamma_k|=|4^k\cdot \phi' (\xi)|=|4^k|\cdot |\phi' (\xi)|\end{equation*} Isn't it $|\phi '(\xi)|\leq 1$ ? How do we get $|\gamma_k|=4^k$ ?

Don't we have that $|\phi'(\xi)|=1$ everywhere? Except that it's not defined for integers? (Thinking)

 

[mathmari]

Don't we have that $|\phi'(\xi)|=1$ everywhere? Except that it's not defined for integers? (Thinking)

But at post #21 how did we get the inequality $|\gamma_n|\leq 4^n$ ? Shouldn't we get there also the equality then? I got stuck right now. (Wondering)

 

[I like Serena]

But at post #21 how did we get the inequality $|\gamma_n|\leq 4^n$ ? Shouldn't we get there also the equality then? I got stuck right now.

I believe you are right. For $n<k$ there won't be an integer in between either, so we get equality as well. (Nod)

Still, doesn't $|\gamma_n|=4^n$ imply that $|\gamma_n|\le 4^n$?
And doesn't it suffice for the purpose of the proof that we have the inequality? (Wondering)

 

[mathmari]

I believe you are right. For $n<k$ there won't be an integer in between either, so we get equality as well. (Nod)

Still, doesn't $|\gamma_n|=4^n$ imply that $|\gamma_n|\le 4^n$?
And doesn't it suffice for the purpose of the proof that we have the inequality? (Wondering)

Ok! I just wanted to understand the differnece of $|\gamma_n|$and $|\gamma_k|$, why at the one we had inequality and at the other one equality.

Since $|\gamma_k|=4^k$ we have that $$\left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\sum_{n=0}^k\left (\frac{3}{4}\right )^n\gamma_n\right |\geq 3^k-\sum_{n=0}^{k-1}3^n=\frac{1}{2}\left (3^k+1\right )$$ so for $k\rightarrow +\infty$, $\delta_k\rightarrow 0$ and $3^k\rightarrow +\infty$, it follows that...

Do we apply here the IVT for $f$ ?

If yes, then we have for $x_0<\xi <x_0+\delta_k$ that \begin{align*}f'(\xi)&=\frac{f(x_0+\delta_k)-f(x_0)}{x_0+\delta_k-x_0}=\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\\ \Rightarrow |f'(\xi)| &=\left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\frac{\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x_0+\delta_k)\right )-\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )}{\delta_k}\right | \\ & = \left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\frac{\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]}{\delta_k}\right | =\left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | \\ & = \left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\cdot 0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\end{align*}

Is it correct so far? (Wondering)

We have that $|\gamma|\leq 4^n$, so isn't the inequality symbol above at $$\left |\sum_{n=0}^k\left (\frac{3}{4}\right )^n\gamma_n\right |\geq 3^k-\sum_{n=0}^{k-1}3^n$$ wrong? (Wondering)

 

[I like Serena]

Do we apply here the IVT for $f$ ?

If yes, then we have for $x_0<\xi <x_0+\delta_k$ that \begin{align*}f'(\xi)&=\frac{f(x_0+\delta_k)-f(x_0)}{x_0+\delta_k-x_0}=\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\\ \Rightarrow |f'(\xi)| &=\left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\frac{\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x_0+\delta_k)\right )-\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )}{\delta_k}\right | \\ & = \left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\frac{\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]}{\delta_k}\right | =\left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | \\ & = \left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\cdot 0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\end{align*}

We can't because we do not know if $f'$ exists.
Worse, we are trying to prove that $f'$ does not exist, not for any value of $x$, aren't we? (Worried)

Instead I believe we should just substitute our results so far, which amounts to the same thing, except that we can't say that it is equal to $|f'(\xi)|$.

We have that $|\gamma|\leq 4^n$, so isn't the inequality symbol above at $$\left |\sum_{n=0}^k\left (\frac{3}{4}\right )^n\gamma_n\right |\geq 3^k-\sum_{n=0}^{k-1}3^n$$ wrong?

Don't we have for $n < k$:
$$\left |\left (\frac{3}{4}\right )^n\gamma_n\right | \le 3^n \implies \left (\frac{3}{4}\right )^n\gamma_n \ge -3^n$$
(Thinking)

 

[mathmari]

We can't because we do not know if $f'$ exists.
Worse, we are trying to prove that $f'$ does not exist, not for any value of $x$, aren't we? (Worried)

Instead I believe we should just substitute our results so far, which amounts to the same thing, except that we can't say that it is equal to $|f'(\xi)|$.

So do we take the right side of IVT and we show that this doesn't converge and so it cannot be equal to the derivative? (Wondering)

Don't we have for $n < k$:
$$\left |\left (\frac{3}{4}\right )^n\gamma_n\right | \le 3^n \implies \left (\frac{3}{4}\right )^n\gamma_n \ge -3^n$$
(Thinking)

We have the following:
\begin{align*}&\frac{f(x_0+\delta_k)-f(x_0)}{x_0+\delta_k-x_0}=\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\\ & \Rightarrow \left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\frac{\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x_0+\delta_k)\right )-\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )}{\delta_k}\right | \\ & = \left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\frac{\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]}{\delta_k}\right | =\left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | \\ & = \left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\cdot 0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\end{align*}

So for $0\leq n\leq k$ we have that $|\gamma_n|\leq 4^n$ so we get that \begin{align*}-4^n\leq \gamma_n\leq 4^n &\Rightarrow -\left (\frac{3}{4}\right )^n4^n\leq \left (\frac{3}{4}\right )^n\gamma_n\leq \left (\frac{3}{4}\right )^n4^n \\ & \Rightarrow -3^n\leq \left (\frac{3}{4}\right )^n\gamma_n\leq 3^n \\ & \Rightarrow \sum_{n=0}^{k}\left (-3^n\right )\leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \sum_{n=0}^{k}3^n \\ & \Rightarrow -\sum_{n=0}^{k}3^n \leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \sum_{n=0}^{k}3^n \\ & \Rightarrow -\frac{3^{k+1}-1}{2} \leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \frac{3^{k+1}-1}{2}\end{align*}

Doesn't this means that $$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\leq \frac{3^{k+1}-1}{2}$$ ? I got stuck how we get the other inequality symbol. (Wondering) Or do we take it as follows?

$$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\geq -\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n$$
(Wondering)

 

[I like Serena]

So do we take the right side of IVT and we show that this doesn't converge and so it cannot be equal to the derivative?

More specifically, we have a sub sequence and we are showing that it diverges.
Consequently that means that the limit that we need for $f'$ does not exist, and thus $f'$ does not exist. (Thinking)

We have the following:
\begin{align*}&\frac{f(x_0+\delta_k)-f(x_0)}{x_0+\delta_k-x_0}=\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\\ & \Rightarrow \left |\frac{f(x_0+\delta_k)-f(x_0)}{\delta_k}\right |=\left |\frac{\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^n(x_0+\delta_k)\right )-\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\phi \left (4^nx_0\right )}{\delta_k}\right | \\ & = \left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\frac{\left [\phi \left (4^n(x_0+\delta_k)\right )-\phi \left (4^nx_0\right )\right ]}{\delta_k}\right | =\left |\sum_{n=0}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\gamma_n\right | \\ & = \left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+\sum_{n=k+1}^{\infty}\left (\frac{3}{4}\right )^n\cdot 0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n+0\right | =\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\end{align*}

So for $0\leq n\leq k$ we have that $|\gamma_n|\leq 4^n$ so we get that \begin{align*}-4^n\leq \gamma_n\leq 4^n &\Rightarrow -\left (\frac{3}{4}\right )^n4^n\leq \left (\frac{3}{4}\right )^n\gamma_n\leq \left (\frac{3}{4}\right )^n4^n \\ & \Rightarrow -3^n\leq \left (\frac{3}{4}\right )^n\gamma_n\leq 3^n \\ & \Rightarrow \sum_{n=0}^{k}\left (-3^n\right )\leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \sum_{n=0}^{k}3^n \\ & \Rightarrow -\sum_{n=0}^{k}3^n \leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \sum_{n=0}^{k}3^n \\ & \Rightarrow -\frac{3^{k+1}-1}{2} \leq \sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\leq \frac{3^{k+1}-1}{2}\end{align*}

Doesn't this means that $$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\leq \frac{3^{k+1}-1}{2}$$ ? I got stuck how we get the other inequality symbol. (Wondering) Or do we take it as follows?

$$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |\geq -\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n$$
(Wondering)

Can't we split it up like:
$$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |
\geq \left |\left (\frac{3}{4}\right )^k\gamma_k\right | -\sum_{n=0}^{k-1}\left (\frac{3}{4}\right )^n|\gamma_n|$$
(Thinking)

 

[mathmari]

Can't we split it up like:
$$\left |\sum_{n=0}^{k}\left (\frac{3}{4}\right )^n\gamma_n\right |
\geq \left |\left (\frac{3}{4}\right )^k\gamma_k\right | -\sum_{n=0}^{k-1}\left (\frac{3}{4}\right )^n|\gamma_n|$$
(Thinking)

Ahh ok! So we split it in that way to get a lower bound that goes to infinity if $k\rightarrow \infty$ and so the limit of IVT that we need for $f'$ does not exist, and thus $f'$ does not exist and so the proof is complete, right? (Wondering)

 

[I like Serena]

Ahh ok! So we split it in that way to get a lower bound that goes to infinity if $k\rightarrow \infty$ and so the limit of IVT that we need for $f'$ does not exist, and thus $f'$ does not exist and so the proof is complete, right?

Yep. (Nod)

 

[mathmari]

Great! Thank you so much!

 

[mathmari]

For $x\in [-1,1]$ we define $\phi (x)=|x|$ and then we extend $\phi$ to the whole $\mathbb{R}$ such that $\phi (x+2)=\phi (x)$.

How does this look graphically? :unsure:

 

[I like Serena]

How does this look graphically?

\begin{tikzpicture}
\draw[help lines] (-4,-2) grid (4,2);
\draw[->] (-4.4,0) -- (4.4,0);
\draw[->] (0,-2.2) -- (0,2.2);
\draw foreach \i in {-4,...,4} { (\i,0.1) -- (\i,-0.1) node[below] {$\i$} };
\draw foreach \i in {-2,...,2} { (0.1,\i) -- (-0.1,\i) node[ left ] {$\i$} };
\draw[domain=-4.2:4.2, variable=\x, blue, ultra thick, samples=200] plot ({\x}, {abs(\x-floor((\x-1)/2)*2-2)}) node[above right] {$\phi$};
\end{tikzpicture}
:geek: