Let $f, h \in C(\mathbb{R^d}, \mathbb{R})$. The function $f$ is bounded and uniformly continuous and $h$ is continuous, has compact support and $h(x) \geq 0$ for every $x \in \mathbb{R}$. We also know that
$$\int_{\mathbb{R}^d} h(x)dx=1 $$
For every $n \in \mathbb{N}$ we defined the functions:
$$ h_n(x) = n^d h(nx) $$ $$G_n(x)= \int_{\mathbb{R}^d} f(y) h_n(x-y) dy $$
I need to prove that $G_n \in BC(\mathbb{R}^d,\mathbb{R})$ for every $n \in \mathbb{N}$, and $G_n \rightarrow f$ unirformly when $n \rightarrow \infty$.
I already proved that $G_n$ is bounded, and I think my proof of the continuity of $G_n$ is wrong:
Proof: If $\varepsilon >0$ and $\vert x - x_0 \vert < \delta$ for any $\delta$, we know that $f$ is bounded then $f(x) < M$, $M \in \mathbb{R}^+$ then
$$\vert G_n(x) - G_n(x_0) \vert = \left\vert \int_{\mathbb{R}^d} f(y) h_n(x-y) dy - \int_{\mathbb{R}^d} f(y) h_n(x_0-y) dy \right\vert $$ $$=\left\vert \int_{\mathbb{R}^d} f(y) [h_n(x-y) - h_n(x_0-y) ]dy \right\vert \leq M \left \vert \int_{\mathbb{R}^d} n^d [h(n(x-y)) - h(n(x_0-y)) ]dy \right\vert$$
$$z=n(x-y),dz=-n dy,w=n(x_0-y),dw=-n dy$$
$$=M n^{d-1} \left\vert \int_{\mathbb{R}^d}h(z)dz - \int_{\mathbb{R}^d} h(w)dw \right\vert =0 < \varepsilon.$$
Because $\int_{\mathbb{R}^d}h(z)dz = \int_{\mathbb{R}^d} h(w)dw = 1$, then $G_n$ is continuous.
But this proof implies that $G_n$ is constant, and that is not true, what's wrong with this proof?
I also don't know how to prove that $G_n \rightarrow f$ when $n \to \infty$, I tried to prove but I obtained something similar than the proof above, I think I'm doing something wrong.
Let $\epsilon >0$ and choose $\delta>0$ such that $|f(x)-f(y)| < \epsilon$ for all $|x-y| < \delta$.
Choose $N$ such that for $n \ge N$ we have $\operatorname{supp} h_n \subset B(0,\delta)$.
We have $|G_n(x)-f(x)| \le \int |f(y)-f(x)| h_n(x-y) dy \le \epsilon$ hence $G_n$ converges to $f$ uniformly and since $f$ is bounded, we see that $G_n$ is bounded.
Furthermore, if $|x-x_0| < \delta$ \begin{eqnarray} G_n(x)-G_n(x_0) &=& \int f(y)( h_n(x-y)-h_n(x_0-y)) dy \\ &=& \int (f(x-y)-f(x_0-y)) h_n(y) dy \end{eqnarray} and so $|G_n(x)-G_n(x_0)| \le \epsilon$. Hence the $G_n$ are continuous.