1
$\begingroup$

I have the following problem: I'm given a linear bounded operator $P\in \mathcal{L}(L^2([a,b]))$, $a,b\in \mathbb{R}$ and I want to find a sequence of approximating linear bounded operators $(P_n)_{n\geq 1}$ satisfying the following conditions:

  1. $P_n \to P$ in $\mathcal{L}(L^2([a,b]))$ as $n\to \infty$ (i.e. in the norm topology);
  2. $P_n(H^1_0([a,b])) \subset H^1_0([a,b])$ for every $n\geq 1$;
  3. $P_n : H^1_0([a,b]) \to H^1_0([a,b])$ is bounded.

Here, $H^1_0([a,b])$ denotes the Sobolev space of order $1$ with Dirichlet boundary conditions.

Is it possible to find such an approximating sequence for a general operator $P\in \mathcal{L}(L^2([a,b]))$?

Thank you very much in advance!

$\endgroup$
2
  • 2
    $\begingroup$ I think for a counterexample you could consider something like $P(\sin nx)= \sin 2^n x$. Now an approximation of the desired type would have to map $\sin nx$ to functions $g$ with $\|g-\sin 2^n x\|_2<\epsilon$, $\|g'\|_2\le Cn$, which seems impossible (fix $\epsilon$, send $n\to\infty$). $\endgroup$ Commented Aug 11, 2021 at 14:54
  • $\begingroup$ @ChristianRemling I don't see how to derive a contradiction, because $g$ depends on $n$ here. Why can't $g$ be close to $\sin 2^n x$ while at the same time have a moderately growing $H^1_0$-norm? $\endgroup$ Commented Aug 11, 2021 at 16:07

1 Answer 1

2
$\begingroup$

Here's a more explicitly worked out version of my comment above. Consider $P(\sin nx)=\sin 2^n x$. This is an isometry, so in particular bounded in $L^2(0,\pi)$. It can not be approximated in the desired way.

Indeed, suppose we had an operator $Q$ with $\|P-Q\|<\epsilon$ that is also bounded on $H^1$. Since $\|\sin nx \|_{H^1}\simeq n$, this would have to map to functions $g=Q(\sin nx)$ with $\|g'\|_2\lesssim n$. However, we also have $\|g-\sin 2^n x\|_2<\epsilon$. This is not possible because the second inequality forces $g$ to mimic the oscillations of $\sin 2^n x$, but that will make $\|g'\|_2$ too large.

We have approximately $2^n$ intervals, of size $\simeq 2^{-n}$ each, on which $\sin 2^n x\ge 3/4$. If we had $g(x)\le 1/4$ throughout such an interval $I$, then $\|g-\sin 2^nx\|^2_{L^2(I)}\gtrsim 2^{-n}$. Thus we can have at most $\simeq \epsilon 2^n$ such intervals.

In other words, for a fraction close to one (as $\epsilon\to 0$) of the $2^n$ intervals where $|\sin 2^n x|\ge 3/4$, the function $g$ will also take a corresponding value $|g|\ge 1/4$ on these intervals. So $g$ oscillates between $\pm 1/4$ at least $\gtrsim 2^n$ times.

However, when $g$ changes from $1/4$ to $-1/4$ on $J$, then $$ \frac{1}{2} \le \int_J |g'| \le \|g'\|_{L^2(J)} |J|^{1/2} , $$ and thus in particular $\|g'\|^2_{L^2(J)}\gtrsim 1$. Since we have $\simeq 2^n$ such intervals, $\|g'\|_2$ comes out much too large.

$\endgroup$
1
  • $\begingroup$ Thank you very much for your answer. I see the point now. I overlooked that $g$ has to be continuous. $\endgroup$ Commented Aug 11, 2021 at 17:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .