This question came up from definition of the sobolev space $W_0^{1,2}(0,1)$ given to me. A function $f\in L^2(0,1)$ is in $W_0^{1,2}(0,1)$ if there is $L^2(0,1)$ function $g$ such that
$$ f(t)=\int_0^t g(s)\,ds $$
and in this case $f':=g$
But it also says :equivalently $f\in W_0^{1,2}$ if $f$ is equivalent to a continuous function that vanishes at $0$ and differentiable almost everywhere with $\int_0^t f'(s)\,ds=f(t)$ and $f'\in L^2(0,1)$.
Now let's consider following function: given $g\in L^2(0,1)$, define, for $\lambda\in \mathbb{C}$
$$ f(t)=i\int_0^t g(s)e^{-\lambda i (t-s)}\,ds $$
This function is of particular interest, because it can be used to show that the spectrum of the linear operator $Af=if'$ on $D(A)=W^{1,2}_0(0,1)$ is empty (keep in mind that at this stage I'm supposed to think that $f'$ is as defined above, not (some kind of) derivative).
It's rather clear that $f$ is in $L^2(0,1)$, because it is an image of $g$ under integral operator with $L^2(0,1)^2$ kernel.
Now it is almost by definition that $f$ is a member of $D(A)$, except that, unfortunately we have a factor of $e^{-i\lambda t}$ multiplied to the integral, so we cannot really say that.
But we can instead compute that
$$i\int_0^t g(s) - \lambda f(s)\,ds=f(t)$$
which directly shows that $f\in W^{1,2}_0$ and that $if'=\lambda f - g$ or $(\lambda - A)f = g$.
As one could imagine, computations invovled are rather cumbersome, and use Fubini's theorem. Therefore, the following is very tempting:
$$f'(t)=i\left(-\lambda i \int_0^t g(s)e^{-\lambda i (t-s)}\,ds +e^{-\lambda i t} g(t)e^{\lambda i t}\right)=-i\lambda f(t) +ig(t)$$
Therefore (!) $f$ is in $W^{1,2}_0$ with $f'$ as given above. But of course, there are some problems with this:
- $f'$ is not a proper derivative, so I do not know stuff like product rule applies to them.
- Moreover, $e^{-\lambda i t}$ does not belong to $W^{1,2}_0$, so taking 'derivative of it' doesn't make sense;
However, I do know that $f$ itself vanish at $0$, and probably $f$ is equivalent to continuous function with almost everywhere derivative as given above.
So it seems that to make the above argument precise, it's sufficient to show that the function
$$\int_0^t g(s) e^{\lambda i s}\,ds$$
is continuous (equivalent to) and differentiable almost everywhere. But then this boils down to showing that any $W^{1,2}_0$ function satisfies that condition.
So long question made short: how does one show the equivalence of definitions given above? From what I decribed above I only need one direction, but it would be cool to know both ways.
I apprecite any helps/comments