Proving that $-\frac{2nx}{e^{n^2x^2}}$ does not converge uniformly on any interval containing 0

Question

While trying to come up with a solution, I had that

$f_n(x) = n^{-1}e^{-n^2x^2}$, giving us that $f'_n(x) = -2nxe^{-n^2x^2}$.

$f'_n \rightarrow f'$ uniformly on $\mathbb{R}$ iff $\limsup_{n \rightarrow \infty} (f'_n(x)-f'(x)) =0$.

However, when $x=0$, $$ \limsup_{n \rightarrow \infty} (f'_n(0)-f'(0)) = \limsup_{n \rightarrow \infty} (0-\frac{2\cdot 0}{e^{0}}) \ = 0...??????????? $$

Haha...so I'm guessing that I'm on the wrong track..

Edit: I showed in a previous section to this that $f'_n(x) \rightarrow 0$ pointwise on $\mathbb{R}$:

$$ \lim_{n \rightarrow \infty} -\frac{2nx}{e^{n^2x^2}} = -2 \cdot \frac{x}{e^{x^2}} \lim_{n \rightarrow \infty} \frac{n}{e^{n^2}} $$ Clearly $\lim_{n \rightarrow \infty} \frac{n}{e^{n^2}} \rightarrow 0$ (according to L'Hôpital's Rule, or any suitable method). Hence it naturally follows that the pointwise limit of $f'_n$ is $0$.

Answer 1

Hint:

First prove the following result:

If $g_n$ is a sequence of continuous functions that converges uniformly to $g$ on $I$ and $x_0\in I$ then $\lim_{n\to\infty}g(x_n) = g(x_0)$ for any sequence $x_n \in I$ with $\lim_{x\to \infty} x_n = x_0$.

Proof: Use the triangle inequality

$$|g_n(x_n) - g(x_0)| \leq |g_n(x_n) - g(x_n)| + |g(x_n)-g(x_0)|$$

together with the definition of uniform convergence and the continuity of $g$ (which is implied by the uniform convergence of $g_n$).

For your problem assume that $f_n'$ converges uniformly and consider the sequence $x_n = 1/n$ (or $x_n = -1/n$). Notice that $f_n'(\pm 1/n) = \mp 2/e$ so $\lim_{n\to\infty} f_n'(\pm 1/n) = \mp 2/e$. Use the result above to derive a contradiction.