convergence of a succession of functions.

Question

Can you help me with this problem?

Let $f: \mathbb{R} \to \mathbb{R}$ a continuous function, and for all $n \in \mathbb{N}$ define $f_{n} : \mathbb{R} \to \mathbb{R}$ as $f_{n}(x)=f(\frac{x}{n})$, $x \in \mathbb{R}$.

a)Find $g= \lim_{n\to \infty} f_{n}$

b) Prove the convergence on a) is not necessarily uniform.

I think, $g=f(0)$ but is just my intuition.

Answer 1

Answer for b):

Suppose $f_n(x) \to f(0)$ uniformly on $\mathbb R$. Then, given $\epsilon >0$, there exist $n_0$ such that $|f(\frac x n) -f(0)| <\epsilon$ for all $x$ and all $n >n_0$. Let $t$ be any real number. Put $x=tn$ in this with $n=n_0+1$, say. You get $|f(t)-f(0)| <\epsilon$. Since $\epsilon$ is arbitrary this implies $f(t)=f(0)$. Hence $f$ is a constant. The converse also holds trivially, so the convergence is uniform iff $f$ is a constant function.

Answer 2

Your intuition is correc: since $f$ is continuous, $$\lim_{n\to\infty}f\Bigl(\frac xn\Bigr) = f\Bigl(\lim_{n\to\infty}\frac xn\Bigr)=f(0).$$