I'm trying to prove the following:
$f$ is continuous on $[0,\infty)$ and $$\lim_{x\to \infty} f(x) = f(0) = 0.$$ Define $h_n(x) = f(x/n)$ and $g_n(x)= f(nx)$. Prove that $h_ng_n$ uniformly converges on $[0,\infty)$
Can you guys help me with that please?
As a minor note: typically we use $*$ for convolution. For multiplication, I'll just write $h_ng_n$.
Since $g_n$ and $h_n$ converge to zero pointwise, the only viable candidate for convergence will be zero. We prove that $h_ng_n \to 0$ uniformly.
Note, since $f \to 0$ at $x=0$ or $x \to \infty$, we can find $N \in \mathbb N$ so that $\lvert f(x) \rvert \le 1$ when $x < 1/N$ or $x > N$ (why?). But $\lvert f \rvert$ has a maximum on $[1/N, N]$ since it is a continuous function on a compact set. Call this maximum $M_0$. Define $M = \max\{1, M_0\}$. Then $\lvert f(x) \rvert \le M$ for all $x \in [0,\infty)$. That is, $\lvert f \rvert$ is bounded.
Fix $\epsilon > 0$. Since $f\to 0$ at $\infty$, we can find $K_1 \in \mathbb N$ so that $t \ge K_1$ gives $$\lvert f(t)\rvert < \epsilon / M.$$ Likewise, since $f$ is continuous and $f(0) = 0$, we an find $K_2 \in \mathbb N$ so that $t \le 1/K_2$ gives $$\lvert f(t) \rvert < \epsilon / M.$$
Take $K = \max \{ K_1, K_2 \}$. Now take $x \in [0,\infty)$ and $n \ge K$. If $x \in [0,1]$, then $x/n \le 1/K \le 1/K_2$ and so $\lvert f(x/n) \rvert < \epsilon / M$ But then $$h_n(x) g_n(x) < (\epsilon / M) \lvert f(nx) \rvert \le (\epsilon/M)\cdot M = \epsilon.$$ Conversely, if $x \in (1,\infty)$, then $nx > n \ge K \ge K_1$ so $\lvert f(nx) \rvert< \epsilon/M$. Then $$h_n(x) g_n(x) < \lvert f(x/n) \rvert (\epsilon / M) \le M\cdot (\epsilon/M)= \epsilon.$$ Hence for $n \ge K$, we have $\lvert h_n(x) g_n(x)\rvert < \epsilon$ for all $x \in [0,\infty)$. That is, $h_n g_n \to 0$ uniformly on $[0,\infty)$.