Is there a bounded continuous function $f$ s.t. a certain integral does not converge uniformly to zero? The following question was the reason for question Is there a bounded function $f$ with $f'$ unbounded and $f''$ bounded?, which had a negative answer.
So I'd like to post my original problem:
I'm trying to find a continuous bounded function $f:\mathbb{R}\to\mathbb{R}$
such that 
$$\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi t}}e^{-(y-x)^2/2t}f(y)dy -f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-y^2/2}(f(x+\sqrt{t}y)-f(x))dy$$
does NOT uniformly converge to $0$ for $t\to 0+$. I'm not completely sure that there is actually one. But I'm fairly certain. 
I was playing around with functions like $\text{sin}(e^{2x})$. But so far without any luck. Thanks for any comments.
 A: Here is an example.
Let me write $p_t(x) = \frac{1}{\sqrt{2\pi t}}e^{-x^2/2t}$ for short.  Let $f$ be a continuous function such that for $n \ge 2$, centered around $x=n$, there is a triangular bump of height $1$ and width $2/n^2$.  Elsewhere it is 0.  Here is a very crude picture.

Notice that the bump centered at $n$ has area $1/n^2$, so $\int_{\mathbb{R}} f(x) dx = \sum_{n=2}^\infty \frac{1}{n^2} < \infty$.  Also, since I left out $n=1$, each bump has width less than $1/2$ and so is supported in $[n-\frac{1}{2}, n+\frac{1}{2}]$.
Note $f(n) = 1$ for each $n$.  Let's estimate $\int_{\mathbb{R}} f(x) p_t(n-x)dx$.  First let's consider the part of the integral outside of  $[n-\frac{1}{2}, n+\frac{1}{2}]$, where we have $p_t(n-x) \le p_t(1/2) = \frac{1}{\sqrt{2 \pi t}} e^{-1/8t}$.  Thus we can bound this part of the integral by $p_t(1/2) \int_{\mathbb{R}} f(x) dx$.  This goes to $0$ as $t \to 0$, so take $t$ small enough that this term is less than, say, $1/4$.  (Note this part is independent of $n$.)
Now let's consider the integral over $[n-\frac{1}{2}, n+\frac{1}{2}]$ which contains a single bump.  Since $p_t(x)$ attains its maximum at $x=0$, we have $p_t(x) \le p_t(0) = (2 \pi t)^{-1/2}$.  Since the bump is actually supported in $[n-\frac{1}{n^2}, n+\frac{1}{n^2}]$ and $f \le 1$, this part of the integral is bounded by $\frac{2}{n^2} p_t(0) = \frac{2}{n^2 \sqrt{2 \pi t}}$.  We can take $n$ so large that this is also less than $1/4$.
Thus we have shown that for any sufficiently small $t$, there exists $n$ such that $\int_{\mathbb{R}} f(x) p_t(x-n)dx \le 1/2$ whereas $f(n) = 1$.  So we do not have uniform convergence.
The Wikipedia page cited by Theo Buehler is incorrect.  It cites "general facts about approximation to the identity"; however, the usual condition for $f * \psi_k \to f$ uniformly is that $f$ be uniformly continuous.  Of course, my $f$ is not uniformly continuous.  
