3
$\begingroup$

As in title, $A$ is a differential operator with the domain $D(A)=\{f\in C^1[0,1]: \text{there is $\epsilon>0$ such that $f(x)=0$ for $x\in [0,\epsilon)$}\}$

This came up as an example of non-closed operators, but it was not explained why so - it just stated that it is an example. I do think I have a explanation:

Let $\|\cdot\|'$ denote the graph norm in $D(A)$. That is,

$$\|f\|'=\|f\|+\|f'\|$$

where $\|\cdot\|=\|\cdot\|_{\infty}$. It's enough to show that $D(A)$ is not closed with respect to this norm. For $n\ge 2$, consider following sequence of $C^1$ functions:

$$ f_n(x)=\begin{cases} 0 & \text{if $x\in [0,1/n)$} \\ \frac{1}{\left(1-\frac{1}{n}\right)^2} \left(x-\frac{1}{n}\right)^2 & x\in [1/n,1]\end{cases} $$

Clearly $f_n$ are in $D(A)$, and

$$ g_n(x):=x^2-f_n(x)=\begin{cases} x^2 & \text{if $x\in [0,1/n)$} \\ (1-c_n)x^2 +\frac{2c_n}{n}x -\frac{c_n}{n^2}& x\in [1/n,1]\end{cases} $$

where $c_n=\frac{1}{\left(1-\frac{1}{n}\right)^2}=\frac{n^2}{(n-1)^2}$. Then

$$ g_n'(x)=\begin{cases} 2x & \text{if $x\in [0,1/n)$} \\ 2(1-c_n)x +\frac{2c_n}{n} & x\in [1/n,1]\end{cases} $$

Since $c_n>1$ for all $n\ge 2$, $g_n'$ is decreasing in $[1/n , 1]$, and $g_n'=0$ when

$$x=x_0:=-\frac{c_n}{n(1-c_n)}=\frac{n}{2n-1}=1-\frac{n-1}{2n-1}\in (0.5,1) $$

Hence

$$ \|g_n'\|=\frac{2}{n}\to 0$$

and

$$ \|g_n\|=g_n(x_0)=-\frac{c_n^2}{n^2(1-c_n)}-\frac{c_n}{n^2} $$

since $c_n\to 1$ as $n\to \infty$ and as $n^2(1-c_n)=\frac{n^2-2n^3}{(n-1)^2}\to -\infty$ we also have that $\|g_n\|\to 0$

Thus $g_n\to x^2$ in $\|\cdot\|'$ norm, but $x^2\notin D(A)$ so we are done.


I think what I came up with is correct, but as one can see there is quite a bit of messy computations involved in it. And although it makes intuitive sense that $g_n$ should converge to $x^2$ I wouldn't have completely believed that this example worked without those computations.

Thus I'd like to know if there are simpler examples/ or simpler solution that does not rely on explicit construction of an example that goes wrong somehow.

  • Also, although this probably counts as another individual question, nevertheless I'd like to know if there is any connection between the graph norm and original norm (here $\|\|' \text{and} \|\|_{\infty}$), because the way I first thought up of such an example was by recognizing that the set $D(A)$ is in fact not closed with respect to $\|\|_{\infty}$ and then tried to see if the same holds for the graph norm.
$\endgroup$
  • 1
    $\begingroup$ Silly question: why did you include the prefactor $\frac1{(1-1/n)^2}$? Shouldn't everything work more cleanly without it? $\endgroup$ – s.harp Jan 23 '17 at 21:13
  • $\begingroup$ @s.harp Actually, that's an excellent question; I didn't need that factor. I could not include it and setting $c_n=1$ in subsequenct details everything works fine and much simpler! I think I just added that factor so to make sure that $f_n$ touches $x^2$ at $x=1$ and so every part of $f_n$ lies below $x^2$ hoping the uniform convergence/ and for some reason I tended to think it's "shape" (gradient) would converge to that of $x^2$ only if I added that factor. But turns out, it works just fine without it. Thanks for the input! $\endgroup$ – user160738 Jan 24 '17 at 13:21

Your Answer

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

Browse other questions tagged or ask your own question.