Let $K: R → [0, ∞) $ be continuous, and suppose that $\int_RK(x) dx = 1$.
(1) Prove that for any bounded $f ∈ C(R)$ :
$lim_{t\rightarrow ∞} \int_Rf(x − y)tK(ty)dy = f(x)$ for all $x \in R$.
(2) Show that the convergence is uniform if $f$ is uniformly continuous.
Can anyone give me some clue on how to prove these statements? Thanks a lot in advance!
Use substitution $u = ty$. Thus $ \int f(x-y)K(ty)t\mathrm{d}y = \int f(x-\frac{u}{t}) K(u) \mathrm{d}u$. As $\int K(u)\mathrm{d}u = 1$, write : \begin{align} f(x) = f(x) \cdot 1 = f(x) \int K(u)\mathrm{d}u = \int f(x)K(u)\mathrm{d}u \end{align}
Substract the two terms : \begin{align} \int f\left(x-\frac{u}{t}\right) K(u) \mathrm{d}u - f(x) = \int \left(f\left(x-\frac{u}{t}\right) - f(x)\right) K(u) \mathrm{d}u \end{align}
$f$ is bounded by a constant $C>0$. $K$ is integrable, thus there exists $A >0$ such that $\int_{\mathbb{R}\setminus [-A,A]} K(u) \mathrm{d}u < \frac{\varepsilon}{4C}$.
$f$ is continuous at $x$ so there exists $\delta$ such that if $|x-y| < \delta$ then $|f(x) - f(y)| < \frac{\varepsilon}{2}$.
Now you can complete the proof :
\begin{align} \left|\int_{\mathbb{R}} \left(f\left(x-\frac{u}{t}\right) - f(x)\right) K(u) \mathrm{d}u \right| &\leqslant \int_{\mathbb{R}} \left|\left(f\left(x-\frac{u}{t}\right) - f(x)\right) K(u)\right| \mathrm{d}u \end{align} Divide the right hand side in two integrals over $[-A,A]$ and its complement. Use what is stated above while carefully choosing $A$ and $\delta$ and controlling $t$. Thus you have $<\varepsilon$ which is a proof of convergence.
For the uniform convergence : prove $\delta$ can be uniformly choosen so that what you did before does not depend on $x$ actually.