Uniform convergence of $f_{n}=\frac{e^{n^2 x^2}}{n}$ in $\mathbb{R}$

Question

Let $\displaystyle f_{n}(x)=\frac{e^{n^2 x^2}}{n}$ for $x \in \mathbb{R}$, $n\geq 1$. Show that $f_{n}\rightarrow 0$ uniformly in $\mathbb{R}$ and $f_{n}' \rightarrow 0$ pointwise in $\mathbb{R}$ but the sequence $(f_{n}')$ does not converge uniformly in a interval which contains the origin.

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My first approximation to this was with $M_{n}=\displaystyle\sup_{x\in \mathbb{R}}|f_{n}(x)-f(x)|=\sup_{x \in \mathbb{R}}\frac{e^{n^2 x^2}}{n}$ and find critical points but I only find a minimum, other approximation was with the theorem of differentiation and uniform convergence but this fail when I take $x=0$ and I don't know how I can bounded this: \begin{eqnarray*} |f_{n}'(x)-f'(x)|=2nxe^{n^2x^2} \end{eqnarray*} How can I bound this? or What way a I have to take in this problem?. Thank you. I will greatly appreciate your advice.

Answer 1

It doesn't:

$$\frac{1}{n}e^{n^2}>\frac{1}{n}\left(1+n^2\right)=\frac{1}{n}+n>n\to\infty$$

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If on the other hand

$$f_n(x)=\frac{1}{n}e^{-n^2x^2}$$

then the results follows. First, for all $x\in\mathbb{R}$ and $n\geq m\geq \epsilon^{-1}$ we have

$$|f_n-f_m|=\left|\frac{1}{n}e^{-n^2 x^2}-\frac{1}{m}e^{-m^2 x^2}\right|$$

Since $e^{-x}$ and $\frac{1}{x}$ are decreasing functions, this simplifies to

$$=\frac{1}{m}e^{-m^2 x^2}-\frac{1}{n}e^{-n^2 x^2}<\frac{1}{m}e^{-m^2 x^2}\leq\frac{1}{m}e^{0}=\frac{1}{m}\leq \epsilon$$

Thus, $f_n(x)$ converges uniformly to $0$ for all $x$. For the second portion, we have that

$$f_n^{'}(x)=-2nxe^{-n^2 x^2}$$

For any $x\in\mathbb{R}$ ($x\neq 0$) we have

$$|-2nxe^{-n^2 x^2}|=2n|x|e^{-(n|x|)^2}$$

Substituting $y=n|x|$ (note that $y\to\infty $ as $n\to\infty$ since $x\neq 0$) we see that

$$\lim_{y\to\infty}\frac{e^{y^2}}{2y}\geq \lim_{y\to\infty}\frac{1+y^2}{2y}=\lim_{y\to\infty}\left(\frac{1}{2y}+\frac{y}{2}\right)\geq \lim_{y\to\infty}\frac{y}{2}=\infty$$

Thus

$$\lim_{y\to\infty}2ye^{-y^2}=0$$

For $x=0$ it is obvious that $f_n^{'}(0)=0$. We conclude that $f_n^{'}(x)$ converges pointwise to $0$ for all $x\in\mathbb{R}$. For the final portion, consider the sequence $x_n=\frac{1}{n}$ (if the interval is only less than zero, take $x_n=-\frac{1}{n}$). This is

$$|f_n^{'}(x_n)|=2e^{-1}>\frac{1}{2}$$

Since this is constant, we have found a sequence of $x_n\to 0$ such that $f_n^{'}(x_n)>\frac{1}{2}$ for all $n\in\mathbb{N}$. We conclude $f_n^{'}(x)$ does not converge uniformly around $x=0$.