Piecewise function: differentiable but not continuously differentiable

[mathmari]

Hey! :giggle:

We have the function $$f(x,y)=\begin{cases}x^2\sin\left (\frac{1}{x}\right )+y^2\sin\left (\frac{1}{y}\right ) & \text{ if } xy\neq 0 \\ x^2\sin\left (\frac{1}{x}\right ) & \text{ if } x\neq 0, y=0 \\ y^2\sin\left (\frac{1}{y}\right ) & \text{ if } x= 0, y\neq 0 \\ 0 & \text{ if } (x,y)=(0,0)\end{cases}$$

I want to calculate the partial derivatives $\frac{\partial{f}}{\partial{x}}$ and$\frac{\partial{f}}{\partial{y}}$ for each point $(x,y)\in \mathbb{R}^2$.

For that do we have the following?
$$\frac{\partial{f}}{\partial{x}}=\begin{cases}2x\sin\left (\frac{1}{x}\right )-\cos\left (\frac{1}{x}\right ) & \text{ if } xy\neq 0 \\ 2x\sin\left (\frac{1}{x}\right )-\cos\left (\frac{1}{x}\right ) & \text{ if } x\neq 0, y=0 \\ 0 & \text{ if } x= 0, y\neq 0 \\ 0 & \text{ if } (x,y)=(0,0)\end{cases}=\begin{cases}2x\sin\left (\frac{1}{x}\right )-\cos\left (\frac{1}{x}\right ) & \text{ if } x\neq 0 \\ 0 & \text{ if } x= 0 \end{cases}$$
$$\frac{\partial{f}}{\partial{y}}=\begin{cases}2y\sin\left (\frac{1}{y}\right )-\cos\left (\frac{1}{y}\right ) & \text{ if } xy\neq 0 \\ 0 & \text{ if } x\neq 0, y=0 \\ 2y\sin\left (\frac{1}{y}\right )-\cos\left (\frac{1}{y}\right ) & \text{ if } x= 0, y\neq 0 \\ 0 & \text{ if } (x,y)=(0,0)\end{cases}=\begin{cases}2y\sin\left (\frac{1}{y}\right )-\cos\left (\frac{1}{y}\right ) & \text{ if } y\neq 0 \\ 0 & \text{ if } y= 0 \end{cases}$$
Is everything correct? Or can we not just differentiate each part of the piecewise function?

To show that $f$ is in $(0,0)$ differentiable but not continuously differentiable do we do the following?

Since we have found the partial derivatives at $(0,0)$ it follows that $f$ is differentiable in $(0,0)$.
Then we have to show that $f_x$ and $f_y$ are not continuous in $(0,0)$ which would mean that $f$ is not continuously differentiable in $(0,0)$.

Is that correct?

:unsure:

 

[I like Serena]

Is everything correct? Or can we not just differentiate each part of the piecewise function?

Hey mathmari!

Looks good to me. (Nod)

To show that $f$ is in $(0,0)$ differentiable but not continuously differentiable do we do the following?

Since we have found the partial derivatives at $(0,0)$ it follows that $f$ is differentiable in $(0,0)$.

I'm afraid that does not follow. (Shake)
Perhaps check the propositions with respect to partial derivatives and (total) differentiability?

Then we have to show that $f_x$ and $f_y$ are not continuous in $(0,0)$ which would mean that $f$ is not continuously differentiable in $(0,0)$.

If we can show the first part, then the second part does indeed follow.

 

[mathmari]

To show that the function is diferentiable in (0,0) do we have to show that for all vectors $v=(v_x, v_y)^T$ it holds that $$\lim_{v\rightarrow 0}\frac{f(v_x, v_y)-f(0,0)-\langle \nabla f(0,0), (v_x, v_y)\rangle}{\|(v_x, v_y)\|} =0$$ and for the second part, that it is not continuous diferentiable do ee calculate the limit of the partial derivatives $$\lim_{x\rightarrow 0}f_x(x,0) \text{ and } \lim_{x\rightarrow 0}f_y(0,y) $$ ? :unsure:

 

[I like Serena]

To show that the function is diferentiable in (0,0) do we have to show that for all vectors $v=(v_x, v_y)^T$ it holds that $$\lim_{v\rightarrow 0}\frac{f(v_x, v_y)-f(0,0)-\langle \nabla f(0,0), (v_x, v_y)\rangle}{\|(v_x, v_y)\|} =0$$

That is close to the definition, but let's check wiki:

A function $f$ of several real variables is said to be differentiable at a point $x_0$ if there exists a linear map $J$ such that​

$$\lim_{{h}\to {0}} \frac{\|{f}({x_0}+{h}) - {f}({x_0}) - {J}{(h)}\|}{\| {h} \|} = 0.$$​

​

The same section also lists a couple of propositions:

If a function is differentiable at $x_0$, then all of the partial derivatives exist at $x_0$, and the linear map $J$ is given by the Jacobian matrix.​

​

If all the partial derivatives of a function exist in a neighborhood of a point $x_0$ and are continuous at the point $x_0$, then the function is differentiable at that point $x_0$.​

​

However, the existence of the partial derivatives (or even of all the directional derivatives) does not in general guarantee that a function is differentiable at a point.​

So we can use the definition and show that $J$ exists at (0,0). :unsure:

Or alternatively we can show that the partial derivatives exist in a neighborhood of (0,0) and are continuous at (0,0).

and for the second part, that it is not continuous diferentiable do we calculate the limit of the partial derivatives $$\lim_{x\rightarrow 0}f_x(x,0) \text{ and } \lim_{x\rightarrow 0}f_y(0,y) $$ ?

Not quite. That is not even enough to show that $f$ is differentiable. (Shake)

Instead we can find the total derivative $Df$ of $f$ and prove that it is continuous.
That is $\lim\limits_{x\to 0} Df(x) = Df(0)$.

 

[mathmari]

Or alternatively we can show that the partial derivatives exist in a neighborhood of (0,0) and are continuous at (0,0).

For this way, do we do the following?

The set $\mathbb{R}^2\setminus \{(0,0)\}$ is open, so $f$ is continuously partially differentiable there.
Let $h \in \mathbb{R}\setminus \{0\}$ and $j \in \{1, 2\}$. Then we get \begin{equation*}\frac{f\left ((0,0)+he_j\right )-f(0,0)}{h}=\frac{f\left (he_j\right )-0}{h}=\frac{f\left (he_j\right )}{h}\end{equation*}
For $j=1$ :
\begin{equation*}\frac{f\left (he_1\right )}{h}=\frac{f\left (h(1,0)\right )}{h}=\frac{f(h,0)}{h}=\frac{h^2\sin \left (\frac{1}{h}\right )}{h}=h\sin \left (\frac{1}{h}\right )\rightarrow 0\end{equation*}
For $j=2$ :
\begin{equation*}\frac{f\left (he_2\right )}{h}=\frac{f\left (h(0,1)\right )}{h}=\frac{f(0,h)}{h}=\frac{h^2\sin \left (\frac{1}{h}\right )}{h}=h\sin \left (\frac{1}{h}\right )\rightarrow 0\end{equation*}
So $f$ is in $(0, 0)$ partially differentiable with \begin{equation*}\frac{\partial{f}}{\partial{x}}(0,0)=\frac{\partial{f}}{\partial{y}}(0,0)=0\end{equation*}
Since the partial derivatives exist in $(0,0)$ and they are continuous in that point it follows that the function is differentiable in the origin.

But $f_x$ is not continuous since $\frac{1}{n}\rightarrow 0$ but $f_x\left (\frac{1}{n}\right )=\frac{2}{n}\sin (n)-\cos (n)$ doesn't converge to $0$. So $f$ is not continuously differentiable.Is that correct? :unsure:

 

[I like Serena]

For this way, do we do the following?

The set $\mathbb{R}^2\setminus \{(0,0)\}$ is open, so $f$ is continuously partially differentiable there.

How does that follow from the fact that the particular restriction of the domain is open? :unsure:

We do have that the partial derivatives that you have calculated are well defined everywhere on that open set.

...
So $f$ is in $(0, 0)$ partially differentiable with \begin{equation*}\frac{\partial{f}}{\partial{x}}(0,0)=\frac{\partial{f}}{\partial{y}}(0,0)=0\end{equation*}

Yep. (Nod)

Since the partial derivatives exist in $(0,0)$ and they are continuous in that point it follows that the function is differentiable in the origin.

How does that follow? :unsure:

We need to verify that the partial derivatives exist in a neighborhood of $(0,0)$.
And additionally that they are continuous in (0,0).
To verify continuity of $f_x$ in $0$, we need to verify that $\lim\limits_{(x,y)\to 0} f_x(x,y)=f_x(0,0)$. And the same for $f_y$.

But $f_x$ is not continuous since $\frac{1}{n}\rightarrow 0$ but $f_x\left (\frac{1}{n}\right )=\frac{2}{n}\sin (n)-\cos (n)$ doesn't converge to $0$. So $f$ is not continuously differentiable.
Is that correct?

Yes. Moreover, $f$ is not differentiable at $0$ either.

 

[mathmari]

We need to verify that the partial derivatives exist in a neighborhood of $(0,0)$.

How do we show that? :unsure:

 

[I like Serena]

How do we show that?

You have calculated the partial derivatives everywhere in $\mathbb R^2$, which is a neighborhood of (0,0).
They are well defined on that neighborhood.
Therefore the partial derivatives exist in a neighborhood of (0,0).

 

[mathmari]

And additionally that they are continuous in (0,0).
To verify continuity of $f_x$ in $0$, we need to verify that $\lim\limits_{(x,y)\to 0} f_x(x,y)=f_x(0,0)$. And the same for $f_y$.

We have that $$\lim\limits_{(x,y)\to 0} f_x(x,y)=\lim\limits_{(x,y)\to 0} \left (2x\sin\left (\frac{1}{x}\right )-\cos\left (\frac{1}{x}\right )\right )$$ But according to Wolfram this limitis indeterminate ( Wolfram ), would that mean that it is not equal to $f_x(0,0)$ ? :unsure:

 

[I like Serena]

We have that $$\lim\limits_{(x,y)\to 0} f_x(x,y)=\lim\limits_{(x,y)\to 0} \left (2x\sin\left (\frac{1}{x}\right )-\cos\left (\frac{1}{x}\right )\right )$$ But according to Wolfram this limitis indeterminate ( Wolfram ), would that mean that it is not equal to $f_x(0,0)$ ?

Indeed.
So $f_x$ is not continuous.

 

[mathmari]

Indeed.
So $f_x$ is not continuous.

Ah since $f_x$ is not continuous, we get that $f$ is not continuously differentiable, right?

But how do we get that $f$ is differentiable? From post #5 we get that $f$ is partially differentiable in $(0,0)$. How do we get the differentiability?

:unsure:

 

[I like Serena]

Ah since $f_x$ is not continuous, we get that $f$ is not continuously differentiable, right?

But how do we get that $f$ is differentiable? From post #5 we get that $f$ is partially differentiable in $(0,0)$. How do we get the differentiability?

Indeed.
Suppose we go back to the definition of differentiability that I quoted in post #4. Can we use it to prove differentiability?
It means we have to do what you suggested in post #3.

 

[mathmari]

Suppose we go back to the definition of differentiability that I quoted in post #4. Can we use it to prove differentiability?

Do you mean the definition with the map $J$ ? :unsure:

 

[I like Serena]

Do you mean the definition with the map $J$ ?

Yes.
And if it exists, we also know that $J$ is equal to the Jacobian matrix at (0,0), which is the same as $\nabla f(0,0)=( f_x(0,0), f_y(0,0) )$.

 

[mathmari]

Yes.
And if it exists, we also know that $J$ is equal to the Jacobian matrix at (0,0), which is the same as $\nabla f(0,0)=( f_x(0,0), f_y(0,0) )$.

So do we have to check if $$\lim_{{h}\to {0}} \frac{\|{f}((0,0)+(h_1, h_2)) - {f}(0,0) - \nabla f(0,0)\|}{\| {h} \|} = 0$$ ? We have that \begin{align*}\lim_{{h}\to {0}} \frac{\|{f}(h_1, h_2) - 0 -( f_x(0,0), f_y(0,0) )\|}{\| {h} \|} &=\lim_{{h}\to {0}} \frac{\|{f}(h_1, h_2) -( 0, 0) \|}{\| {h} \|} =\lim_{{h}\to {0}} \frac{\|{f}(h_1, h_2) \|}{\| {h} \|}=\lim_{{h}\to {0}} \frac{\|h_1^2\sin\left (\frac{1}{h_1}\right )+h_2^2\sin\left (\frac{1}{h_2}\right ) \|}{\| {h} \|} \\ & =\lim_{{h}\to {0}} \frac{h_1^2\sin\left (\frac{1}{h_1}\right )+h_2^2\sin\left (\frac{1}{h_2}\right ) }{\sqrt{h_1^2+h_2^2}} \end{align*} Is that correct so far?:unsure:

 

[I like Serena]

So do we have to check if $$\lim_{{h}\to {0}} \frac{\|{f}((0,0)+(h_1, h_2)) - {f}(0,0) - \nabla f(0,0)\|}{\| {h} \|} = 0$$ ? We have that \begin{align*}\lim_{{h}\to {0}} \frac{\|{f}(h_1, h_2) - 0 -( f_x(0,0), f_y(0,0) )\|}{\| {h} \|} &=\lim_{{h}\to {0}} \frac{\|{f}(h_1, h_2) -( 0, 0) \|}{\| {h} \|} =\lim_{{h}\to {0}} \frac{\|{f}(h_1, h_2) \|}{\| {h} \|}=\lim_{{h}\to {0}} \frac{\|h_1^2\sin\left (\frac{1}{h_1}\right )+h_2^2\sin\left (\frac{1}{h_2}\right ) \|}{\| {h} \|} \\ & =\lim_{{h}\to {0}} \frac{h_1^2\sin\left (\frac{1}{h_1}\right )+h_2^2\sin\left (\frac{1}{h_2}\right ) }{\sqrt{h_1^2+h_2^2}} \end{align*} Is that correct so far?

It should be:
$$\lim_{{h}\to {0}} \frac{\|{f}((0,0)+(h_1, h_2)) - {f}(0,0) - \nabla f(0,0) \cdot (h_1, h_2)\|}{\| {h} \|} = 0$$
but otherwise it is correct since $\nabla f(0,0) = (0, 0)$.

 

[mathmari]

It should be:
$$\lim_{{h}\to {0}} \frac{\|{f}((0,0)+(h_1, h_2)) - {f}(0,0) - \nabla f(0,0) \cdot (h_1, h_2)\|}{\| {h} \|} = 0$$
but otherwise it is correct since $\nabla f(0,0) = (0, 0)$.

Ah yes (Tmi)

But how can we calculate the last limit? I got stuck right now. I thought maybe taking out of the root of the denominator the term $|h_1|$ but I am not sure if that helps. :unsure:

 

[I like Serena]

But how can we calculate the last limit? I got stuck right now. I thought maybe taking out of the root of the denominator the term $|h_1|$ but I am not sure if that helps. :unsure:

Suppose we switch to polar coordinates?
That is, we substitute $h_1=r\cos\phi$ and $h_2=r\sin\phi$.
Then we can let $r\to 0$.

 

[I like Serena]

Oh, btw, we haven't substituted $f(h_1,h_2)$ correctly.
We can't assume that $h_1h_2\ne 0$. Instead we should consider the different cases.

 

[mathmari]

Oh, btw, we haven't substituted $f(h_1,h_2)$ correctly.
We can't assume that $h_1h_2\ne 0$. Instead we should consider the different cases.

Ahh ok!

For the case $h_1h_2\neq 0$ we have then that:
\begin{align*}\lim_{{h}\to {0}} \frac{h_1^2\sin\left (\frac{1}{h_1}\right )+h_2^2\sin\left (\frac{1}{h_2}\right ) }{\sqrt{h_1^2+h_2^2}}&=\lim_{{r}\to {0}} \frac{r^2\cos^2\phi\sin\left (\frac{1}{r\cos \phi}\right )+r^2\sin^2\phi\sin\left (\frac{1}{r\sin \phi}\right ) }{\sqrt{r^2\cos^2\phi+r^2\sin^2\phi}}\\ & =\lim_{{r}\to {0}} \frac{r^2\cos^2\phi\sin\left (\frac{1}{r\cos \phi}\right )+r^2\sin^2\phi\sin\left (\frac{1}{r\sin \phi}\right ) }{\sqrt{r^2\left
(\cos^2\phi+\sin^2\phi\right )}} \\ & =\lim_{{r}\to {0}} \frac{r^2\cos^2\phi\sin\left (\frac{1}{r\cos \phi}\right )+r^2\sin^2\phi\sin\left (\frac{1}{r\sin \phi}\right ) }{\sqrt{r^2}}
\\ & =\lim_{{r}\to {0}} \frac{r^2\cos^2\phi\sin\left (\frac{1}{r\cos \phi}\right )+r^2\sin^2\phi\sin\left (\frac{1}{r\sin \phi}\right ) }{r}\\ & =\lim_{{r}\to {0}}\left (r\cos^2\phi\sin\left (\frac{1}{r\cos \phi}\right )+r\sin^2\phi\sin\left (\frac{1}{r\sin \phi}\right )\right ) \end{align*} How can we continue? :unsure:

 

[I like Serena]

The cosines and sines are bounded between -1 and +1 aren't they?

 

[I like Serena]

So we have
$$0 \le \lim_{{r}\to {0}}\left |r\cos^2\phi\sin\left (\frac{1}{r\cos \phi}\right )+r\sin^2\phi\sin\left (\frac{1}{r\sin \phi}\right )\right | \le \lim_{r\to 0} |2r| = 0\\
\implies \lim_{{r}\to {0}}\left (r\cos^2\phi\sin\left (\frac{1}{r\cos \phi}\right )+r\sin^2\phi\sin\left (\frac{1}{r\sin \phi}\right )\right ) = 0
$$


Now let's take a step back and see if we can simplify.
We can write:
$$\lim_{h\to 0} \frac{\|f(h) - f(0,0) - \nabla f(0,0)\cdot h\|}{\|h\|} = \lim_{h\to 0} \frac{\|f(h) - 0 - (0,0)\cdot h\|}{\|h\|}
=\lim_{h\to 0} \frac{\|f(h)\|}{\|h\|}$$

and we have
\begin{align*}\|f(x,y)\| &=
\left\|\begin{cases}x^2\sin\left (\frac{1}{x}\right )+y^2\sin\left (\frac{1}{y}\right ) & \text{ if } xy\neq 0 \\ x^2\sin\left (\frac{1}{x}\right ) & \text{ if } x\neq 0, y=0 \\ y^2\sin\left (\frac{1}{y}\right ) & \text{ if } x= 0, y\neq 0 \\ 0 & \text{ if } (x,y)=(0,0)\end{cases}\right\| \\
&= \begin{cases}\left|x^2\sin\left (\frac{1}{x}\right )+y^2\sin\left (\frac{1}{y}\right )\right| & \text{ if } xy\neq 0 \\
\left|x^2\sin\left (\frac{1}{x}\right )\right| & \text{ if } x\neq 0, y=0 \\
\left|y^2\sin\left (\frac{1}{y}\right )\right| & \text{ if } x= 0, y\neq 0 \\
\left|0\right| & \text{ if } (x,y)=(0,0)\end{cases} \\
&\le\begin{cases}x^2+y^2 & \text{ if } xy\neq 0 \\
x^2 & \text{ if } x\neq 0, y=0 \\
y^2 & \text{ if } x= 0, y\neq 0 \\
0 & \text{ if } (x,y)=(0,0)\end{cases}\\
&= x^2+y^2\end{align*}