Sequence of functions : pointwise & uniform convergence

[mathmari]

Hey!

Let $0<\alpha \in \mathbb{R}$ and $(f_n)_n$ be a sequence of functions defined on $[0, +\infty)$ by: \begin{equation*}f_n(x)=n^{\alpha}xe^{-nx}\end{equation*} - Show that $(f_n)$ converges pointwise on $[0,+\infty)$.

For an integer $m>a$ we have that \begin{equation*}0 \leq n^{\alpha}xe^{-nx} \leq n^{m}xe^{-nx}\end{equation*}
For $x> 0$ we have that \begin{equation*}\lim_{n\rightarrow +\infty}f_n(x)=\lim_{n\rightarrow +\infty}n^{m}xe^{-nx}=\lim_{n\rightarrow +\infty}\frac{n^{m}x}{e^{nx}}\ \overset{m\text{-times De L'Hopital}}{=}\ \lim_{n\rightarrow +\infty}\frac{m!}{x^{m-1}e^{nx}}=0\end{equation*} For $x= 0$ we have that \begin{equation*}\lim_{n\rightarrow +\infty}f_n(0)=\lim_{n\rightarrow +\infty}0=0\end{equation*} So we get \begin{equation*}\lim_{n\rightarrow +\infty}0 \leq \lim_{n\rightarrow +\infty}n^{\alpha}xe^{-nx} \leq \lim_{n\rightarrow +\infty}n^{m}xe^{-nx} \Rightarrow \lim_{n\rightarrow +\infty}0 \leq \lim_{n\rightarrow +\infty}n^{\alpha}xe^{-nx} \leq 0\end{equation*} It follows that $\displaystyle{\lim_{n\rightarrow +\infty}n^{\alpha}xe^{-nx}=0}$.

Therefore $f_n(x)$ converges pointwise to $0$.

Is everything correct? :unsure:
- Calculate $\max_{x\in [0, +\infty)}f_n(x)$ and conclude that $f_n$ converges uniformly on $[a, +\infty)$ for $a>0$.

We have that
\begin{align*}&f_n(x)=n^{\alpha}xe^{-nx}\\ &\rightarrow f_n'(x)=n^{\alpha}e^{-nx}-n^{\alpha+1}xe^{-nx}=\left (n^{\alpha}-n^{\alpha+1}x\right )e^{-nx} \\ &\rightarrow f_n'(x)=0 \Rightarrow \left (n^{\alpha}-n^{\alpha+1}x\right )e^{-nx}=0 \Rightarrow n^{\alpha}-n^{\alpha+1}x=0 \Rightarrow x=\frac{1}{n} \\ &f_n\left (\frac{1}{n}\right )=\frac{n^{\alpha-1}}{e}\end{align*}
Since we have the intervall $[\alpha, +\infty)$ we have to check also the value of the function at $x=\alpha$, right?
We get $f_n(\alpha)=n^{\alpha}\alpha e^{-n\alpha}$. But how can we compare it with $\frac{n^{\alpha-1}}{e}$ ? Or do we have to check if $f_n(x)$ is increasing or decreasing? :unsure:
- Show that $f_n$ converges uniformly on $[0, +\infty)$ iff $a<1$.

The maximum is $\frac{n^{\alpha-1}}{e}$ (since at $x=0$ we have $f_n(0)=0$ which is smaller) and for $a<1$ the limit goes to $0$, and this means that $f_n$ converges uniformly, right?
This shows that if $a<1$ then $f_n$ converges uniformly, or not? It is left to show that if $f_n$ converges uniformly then $a<1$, or not? :unsure:

 

[I like Serena]

- Show that $(f_n)$ converges pointwise on $[0,+\infty)$.

Is everything correct?

Hey mathmari!

It looks correct to me. (Nod)

- Calculate $\max_{x\in [0, +\infty)}f_n(x)$ and conclude that $f_n$ converges uniformly on $[a, +\infty)$ for $a>0$.

We have that
\begin{align*}&f_n(x)=n^{\alpha}xe^{-nx}\\ &\rightarrow f_n'(x)=n^{\alpha}e^{-nx}-n^{\alpha+1}xe^{-nx}=\left (n^{\alpha}-n^{\alpha+1}x\right )e^{-nx} \\ &\rightarrow f_n'(x)=0 \Rightarrow \left (n^{\alpha}-n^{\alpha+1}x\right )e^{-nx}=0 \Rightarrow n^{\alpha}-n^{\alpha+1}x=0 \Rightarrow x=\frac{1}{n} \\ &f_n\left (\frac{1}{n}\right )=\frac{n^{\alpha-1}}{e}\end{align*}
Since we have the intervall $[\alpha, +\infty)$ we have to check also the value of the function at $x=\alpha$, right?
We get $f_n(\alpha)=n^{\alpha}\alpha e^{-n\alpha}$. But how can we compare it with $\frac{n^{\alpha-1}}{e}$ ? Or do we have to check if $f_n(x)$ is increasing or decreasing? :unsure:

I'm a bit confused about the latin $a$ versus the greek $\alpha$.
Aren't they distinct?
Note that we can prove the statement for any latin $a>0$ independent of the value of the greek $\alpha$ that is in the exponent.

You have found that $f_n'$ has a zero at $x=\frac 1n$. We also have that $f_n'$ is decreasing.
Therefore for a given $n$ we have that $f_n$ has a maximum at $x=\frac 1n$.

Let $N=\left\lceil \frac 1 {a}\right\rceil$.
Then for any $n> N$ we have that $f_n$ is decreasing on $[a,\infty)$.
So for such $n$ the function $f_n$ has its maximum at $x=a$.
If $n$ becomes bigger, does this maximum increase or not?
If it doesn't, we have uniform convergence don't we?

- Show that $f_n$ converges uniformly on $[0, +\infty)$ iff $a<1$.

The maximum is $\frac{n^{\alpha-1}}{e}$ (since at $x=0$ we have $f_n(0)=0$ which is smaller) and for $a<1$ the limit goes to $0$, and this means that $f_n$ converges uniformly, right?
This shows that if $a<1$ then $f_n$ converges uniformly, or not? It is left to show that if $f_n$ converges uniformly then $a<1$, or not?

Indeed. Or rather, that $f_n$ does not converge uniformly when $a\ge 1$.
What happens to the maximum of $f_n$ when $n$ increases with $a\ge 1$?

 

[mathmari]

I'm a bit confused about the latin $a$ versus the greek $\alpha$.
Aren't they distinct?
Note that we can prove the statement for any latin $a>0$ independent of the value of the greek $\alpha$ that is in the exponent.

At the intervall $[a, +\infty)$ we have the latin $a$ and at the function we have the greek letter.

You have found that $f_n'$ has a zero at $x=\frac 1n$. We also have that $f_n'$ is decreasing.
Therefore for a given $n$ we have that $f_n$ has a maximum at $x=\frac 1n$.

Let $N=\left\lceil \frac 1 {a}\right\rceil$.
Then for any $n> N$ we have that $f_n$ is decreasing on $[a,\infty)$.
So for such $n$ the function $f_n$ has its maximum at $x=a$.
If $n$ becomes bigger, does this maximum increase or not?
If it doesn't, we have uniform convergence don't we?

So that means since $f_n$ is decreasing we have that for $\frac{1}{n}>a$ we have that $f_n\left (\frac{1}{n}\right )<f_n(a)$ which means that the maximum of $f_n$ on $[a, +\infty)$ is $f_n(a)$, right? :unsure:

To check the uniform convergence we have to check if $\displaystyle{\lim_{n\rightarrow +\infty}f_n(a)=0}$, right? :unsure:

Indeed. Or rather, that $f_n$ does not converge uniformly when $a\ge 1$.
What happens to the maximum of $f_n$ when $n$ increases with $a\ge 1$?

In this case the maximum is $f_n\left (\frac{1}{n}\right )=\frac{n^{\alpha-1}}{e}$ and we have that $\displaystyle{\lim_{n\rightarrow +\infty}\frac{n^{\alpha-1}}{e}=\begin{cases} +\infty& \text{ if } \alpha-1>0 \\ \frac{1}{e} & \text{ if } \alpha-1=0 \\ 0 & \text{ if } \alpha-1<0 \end{cases}}$.

Can we say then that $f_n$ converges uniformly iff $\alpha-1<0 \Rightarrow \alpha<1$ ? Or doesn't the "iff" follow in that way? :unsure:

 

[I like Serena]

So that means since $f_n$ is decreasing we have that for $\frac{1}{n}>a$ we have that $f_n\left (\frac{1}{n}\right )<f_n(a)$ which means that the maximum of $f_n$ on $[a, +\infty)$ is $f_n(a)$, right?

To check the uniform convergence we have to check if $\displaystyle{\lim_{n\rightarrow +\infty}f_n(a)=0}$, right?

Yes.
That is, it will suffice if that is the case.

Now how can we do that?

In this case the maximum is $f_n\left (\frac{1}{n}\right )=\frac{n^{\alpha-1}}{e}$ and we have that $\displaystyle{\lim_{n\rightarrow +\infty}\frac{n^{\alpha-1}}{e}=\begin{cases} +\infty& \text{ if } \alpha-1>0 \\ \frac{1}{e} & \text{ if } \alpha-1=0 \\ 0 & \text{ if } \alpha-1<0 \end{cases}}$.

Can we say then that $f_n$ converges uniformly iff $\alpha-1<0 \Rightarrow \alpha<1$ ? Or doesn't the "iff" follow in that way?

We have proven that if $a<1$ that then $f_n$ converges uniformly (to the zero function).
However, the proof that if $f_n$ converges uniformly, that then $a<1$ is still incomplete.
That is, we have the edge case $\alpha=1$ where $f_n$ might still be uniformly convergent.

So suppose $\alpha=1$ and $f_n$ is uniformly convergent.
Then we must have that there is a function $f$ such that:
$$\lim_{n\rightarrow\infty}\,\sup\{\,\left|f_n(x)-f(x)\right|: x \in [0,\infty) \,\}=0$$
We have found that in this case $f_n(x)$ is at most $\frac 1 e$.
Could there be such a function $f$ that is also at most $\frac 1 e$?

 

[mathmari]

Yes.
That is, it will suffice if that is the case.

Now how can we do that?

Just to clarify something... We have shown that $x=\frac{1}{n}$ is a critical point. We have the following:
$$f_n''(x)=-n^{\alpha+1}e^{-nx}-n\left (n^{\alpha}-n^{\alpha+1}x\right )e^{-nx}=\left (-n^{\alpha+1}-n^{\alpha+1}+n^{\alpha+2}x\right )e^{-nx}=\left (-2n^{\alpha+1}+n^{\alpha+2}x\right )e^{-nx} \\ f_n''\left(\frac{1}{n}\right )=\left (-2n^{\alpha+1}+n^{\alpha+1}\right )e^{-1}=-n^{\alpha+1}e^{-1}<0$$ That means that at $x=\frac{1}{n}$ the function $f_n(x)$ has a maximum.

So for $x<\frac{1}{n}$ the function is increasing and for $x>\frac{1}{n}$ the function is decreasing.

So if $a<\frac{1}{n}$ we have that $f_n(a)<f_n\left (\frac{1}{n}\right )$ and if $a>\frac{1}{n}$ we have that $f_n(a)<f_n\left (\frac{1}{n}\right )$ but in that case $\frac{1}{n}$ is not in the intervall $[a, +\infty)$ so the maximumis at the boundary $x=a$.

Is that correct so far? :unsure:

 

[I like Serena]

Yep. That looks correct. (Nod)

 

[mathmari]

Yep. That looks correct. (Nod)

If $a<\frac{1}{n}$ the maximum is $f_n\left(\frac{1}{n}\right )=n^{\alpha-1}e^{-1}$.
We have that \begin{equation*}\lim_{n\rightarrow +\infty}f_n\left(\frac{1}{n}\right )=\lim_{n\rightarrow +\infty}n^{\alpha-1}e^{-1}=\lim_{n\rightarrow +\infty}\frac{n^{\alpha-1}}{e}\end{equation*}

If $a>\frac{1}{n}$ the maximum is $f_n(a)=n^{\alpha}a e^{-na}$.
We have that \begin{equation*}\lim_{n\rightarrow +\infty}f_n\left(a\right )=\lim_{n\rightarrow +\infty}n^{\alpha}a e^{-na}=\lim_{n\rightarrow +\infty}\frac{n^{\alpha}a}{e^{na}}\end{equation*}

To calculate in each case the liimit do we have to distinguish cases for $\alpha$ ?

 

[I like Serena]

We can ignore latin $a<\frac 1n$, since it suffices to check for n that are 'large enough'.

So we want to know what $f_n(a)$ does for increasing n that are large enough.
I don't think it is necessary to check cases for greek $\alpha$.
How about checking the behavior of $g(n)=f_n(a)$?

 

[mathmari]

We can ignore latin $a<\frac 1n$, since it suffices to check for n that are 'large enough'.

So we want to know what $f_n(a)$ does for increasing n that are large enough.
I don't think it is necessary to check cases for greek $\alpha$.
How about checking the behavior of $g(n)=f_n(a)$?

\begin{align*}g(n)&=f_n(a)=n^{\alpha}a e^{-na}=\frac{n^{\alpha}a}{ e^{na}} \\ g'(n)&=\frac{a \alpha n^{\alpha-1}e^{na}-n^{\alpha}a^2e^{na}}{e^{2na}}=\frac{a \alpha n^{\alpha-1}-n^{\alpha}a^2}{e^{na}} \\ g''(n)&=\frac{\left (a \alpha (\alpha-1) n^{\alpha-2}-\alpha n^{\alpha-1}a^2\right )e^{na}-a\left (a \alpha n^{\alpha-1}-n^{\alpha}a^2\right )e^{na}}{e^{2na}}\\ & =\frac{a \alpha (\alpha-1) n^{\alpha-2}-\alpha n^{\alpha-1}a^2-a^2 \alpha n^{\alpha-1}+n^{\alpha}a^3}{e^{na}}\\ & =\frac{a \alpha (\alpha-1) n^{\alpha-2}-2\alpha n^{\alpha-1}a^2+n^{\alpha}a^3}{e^{na}} \\ g'(n)&=0 \Rightarrow an^{\alpha-1} \left ( \alpha -na\right )=0 \Rightarrow \alpha -na=0 \Rightarrow n=\frac{\alpha}{a} \\ g'' \left (\frac{\alpha}{a}\right )&=\frac{a \alpha (\alpha-1) \left (\frac{\alpha}{a}\right )^{\alpha-2}-2\alpha \left (\frac{\alpha}{a}\right )^{\alpha-1}a^2+\left (\frac{\alpha}{a}\right )^{\alpha}a^3}{e^{\alpha}}\\ & =\frac{a \alpha (\alpha-1) \alpha^{\alpha-2}a^{-\alpha+2}-2\alpha \alpha^{\alpha-1}a^{-\alpha+1} a^2+\alpha^{\alpha}a^{-\alpha}a^3}{e^{\alpha}}=\frac{ (\alpha-1) \alpha^{\alpha-1}a^{-\alpha+3}-2 \alpha^{\alpha}a^{-\alpha+3} +\alpha^{\alpha}a^{-\alpha+3}}{e^{\alpha}}\\ & =\frac{ (\alpha-1) \alpha^{\alpha-1}a^{-\alpha+3}- \alpha^{\alpha}a^{-\alpha+3} }{e^{\alpha}}=\frac{ \alpha \alpha^{\alpha-1}a^{-\alpha+3}-\alpha^{\alpha-1}a^{-\alpha+3}- \alpha^{\alpha}a^{-\alpha+3} }{e^{\alpha}}\\ & =\frac{ \alpha^{\alpha}a^{-\alpha+3}-\alpha^{\alpha-1}a^{-\alpha+3}- \alpha^{\alpha}a^{-\alpha+3} }{e^{\alpha}}=\frac{ -\alpha^{\alpha-1}a^{-\alpha+3}}{e^{\alpha}}<0\end{align*} So $g_n$ has a maximum at $n=\frac{\alpha}{a}$.

Is that correct? How does this help us? I got stuck right now. :unsure:

 

[I like Serena]

So $g_n$ has a maximum at $n=\frac{\alpha}{a}$.

Is that correct? How does this help us? I got stuck right now.

Looks correct.
Doesn't that mean that $g(n)$ is decreasing for sufficiently large $n$?
Its limit is $0$ isn't it?
So $\lim\limits_{n\to\infty} \sup \{|f_n(x)| : x\in [a,\infty)\} = 0$, isn't it?

 

[mathmari]

Ahh I got it! As for the last question :

We have proven that if $a<1$ that then $f_n$ converges uniformly (to the zero function).
However, the proof that if $f_n$ converges uniformly, that then $a<1$ is still incomplete.
That is, we have the edge case $\alpha=1$ where $f_n$ might still be uniformly convergent.

So suppose $\alpha=1$ and $f_n$ is uniformly convergent.
Then we must have that there is a function $f$ such that:
$$\lim_{n\rightarrow\infty}\,\sup\{\,\left|f_n(x)-f(x)\right|: x \in [0,\infty) \,\}=0$$
We have found that in this case $f_n(x)$ is at most $\frac 1 e$.
Could there be such a function $f$ that is also at most $\frac 1 e$?

In this case the maximum is $f_n\left (\frac{1}{n}\right )=\frac{n^{\alpha-1}}{e}$ and we have that $\displaystyle{\lim_{n\rightarrow +\infty}\frac{n^{\alpha-1}}{e}=\begin{cases} +\infty& \text{ if } \alpha-1>0 \\ \frac{1}{e} & \text{ if } \alpha-1=0 \\ 0 & \text{ if } \alpha-1<0 \end{cases}}$.

If $a<1$ then $f_n$ converges uniformly (to the zero function).

If $f_n$ converges uniformly, then $\displaystyle{\lim_{n\rightarrow\infty}\,\sup\{\,\left|f_n(x)-f(x)\right|: x \in [0,\infty) \,\}=0}$.
How do we continue here to take cases for $\alpha$ ? I got stuck right now. :unsure:

 

[I like Serena]

If $f_n$ converges uniformly, then $\displaystyle{\lim_{n\rightarrow\infty}\,\sup\{\,\left|f_n(x)-f(x)\right|: x \in [0,\infty) \,\}=0}$.
How do we continue here to take cases for $\alpha$ ? I got stuck right now.

We want to prove that uniform convergence of $f_n$ implies that $\alpha<1$.

Let's try a proof by contradiction.
Suppose that it doesn't. Then there must be an $\alpha \ge 1$ such that $f_n$ converges uniformly.

If $a>1$, then $\sup f_n(x)\to\infty$, so whatever we pick for $f$, we won't have that $\displaystyle{\lim_{n\rightarrow\infty}\,\sup\{\,\left|f_n(x)-f(x)\right|: x \in [0,\infty) \,\}=0}$.
Therefore $\alpha=1$.
For $\alpha=1$ we have that $f_n(x)$ has a maximum of $\frac 1e$.
So $f$ must also have a maximum of $\frac 1e$.
Suppose $f(x)=\frac 1e$ for some $x=a$.
We have seen that if $a>0$, that $f_n(a)\to 0$, so $|f_n(a)-f(a)| = |f_n(a)-\frac 1e| \to \frac 1e \ne 0$, which contradicts that $f_n$ converges uniformly.
Consequently $a=0$.
But then $|f_n(a)-f(a)| =|0-\frac 1e|=\frac 1e\ne 0$, which is again a contradiction.

Therefore $\alpha<1$, which completes the proof. :geek: