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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.

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I slightly edited your LaTeX to make your formula more easily readable. Consider using either $$...$$ or $\displaystyle...$ for more complicated expressions (such as the present one). – t.b. May 14 '11 at 1:49
According to the Wikipedia article on the Heat equation there is no such function (take $k = \frac{1}{2}$ in the formula linked to). – t.b. May 14 '11 at 2:10
As Nate points out in the answer below this Wiki-link is incorrect and I agree with him. Oh well, this just adds a further example showing that one shouldn't trust Wikipedia too much even if in general it is quite reliable, at least as far as mathematics is concerned. – t.b. May 14 '11 at 9:29
@Theo: OK, I will do this from now on. Thank you. – Steven May 14 '11 at 18:47
up vote 7 down vote accepted

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.

enter image description here

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.

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