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.

  • $\begingroup$ 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). $\endgroup$ – t.b. May 14 '11 at 1:49
  • $\begingroup$ According to the Wikipedia article on the Heat equation there is no such function (take $k = \frac{1}{2}$ in the formula linked to). $\endgroup$ – t.b. May 14 '11 at 2:10
  • 1
    $\begingroup$ 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. $\endgroup$ – t.b. May 14 '11 at 9:29
  • $\begingroup$ @Theo: OK, I will do this from now on. Thank you. $\endgroup$ – Steven May 14 '11 at 18:47

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.