Let $f(x)$ be some function defined in all of $\mathbb R$. Let $g_{n}(x)=f(\frac{x}{n})$ for any $x\in\mathbb R,n\in\mathbb N$. Suppose $g_{n}(x)$ uniformlly converges to $h(x)\equiv 0$ in $\mathbb R$. Prove that $\forall x\in\mathbb R, f(x)=0$.
My proof goes like this:
Assume that $\exists x_{0}\in\mathbb R, f(x_{0})\neq0$.
Since $g_{n}(x)$ uniformlly converges to $0$, then for $\varepsilon=|f(x_{0})|$,
$\exists N\in\mathbb N,\forall n>N \land \forall x\in\mathbb R, |g_{n}(x)|<\varepsilon$.
Then surely for $x_{1}=(N+1)x_{0}$, we get $|g_{N+1}(x_{1})|=|f(x_{0})|<\varepsilon$, which is a contradiction to our assumption.
Is this a valid proof?
Yes, your proof is correct. I would not have defined $x_1$; instead, I would just have written that$$\bigl|g_{N+1}\bigl((N+1)x_0\bigr)\bigr|=\bigl|f(x_0)\bigr|.$$But that is a matter of taste.