exponent of uniformly convergent sequence

Question

Is it true that if $f_n$ converges uniformly to $f$ on $\mathbb{R}$ then $\exp (-f_n^2)$ converges uniformly to $\exp (-f^2)$. I can't find any counterexample, on the other hand I can't prove that this is true.

Answer 1

I will prove later on that the function $\begin{array}{ccccc} h & : & \mathbb R^+ & \to & \mathbb R^+\\ & & x & \mapsto & \exp(-x^2) \\ \end{array}$ is Lipschitz with ratio $1$.

Now, let $\epsilon >0$ and $N$ such that $n \geq N \Rightarrow \forall x\in \mathbb R |f_n(x)-f(x)| \leq \epsilon$

Let $n \geq N$ and $x\in \mathbb R$

$|\exp (-f_n^2(x))-\exp (-f^2(x))|=|h(f_n(x)-h(f(x))| \leq |f_n(x)-f(x)^|\leq \epsilon$

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For the other proof, by the intermediate value theorem, $$\forall(a,b), \exists x \in ]a,b[, \frac{h(a)-h(b)}{a-b} = h'(x)$$

Hence $$\forall(a,b), \exists x \in ]a,b[, |h(a)-h(b)| = |a-b||2x\exp(-x^2)|\leq|a-b|\sqrt{2}\exp(-1/2)|\leq |a-b|$$

Hence $$\forall(a,b), |h(a)-h(b)| \leq |a-b|$$