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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

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    The first definition of $W^{1,2}_0$ isn't quite correct. You need $ f(1) = 0$ as well. The second definition is incorrect, since the a.e. derivative must be in $L^2$ for $f$ to be in $W^{1,2}$. In particular, functions of the form $f(t)=i\int_0^t g(s)e^{-\lambda i (t-s)}\,ds = e^{-\lambda i t} \cdot \int_0^t g(s)e^{\lambda i s}\,ds$ are indeed in $W^{1,2}$, but of course they don't generally vanish at both 0 and 1, hence they aren't in $W^{1,2}_0$. – 2017-02-04
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    @HansEngler Can you explain why $f(1)=0$ is needed? $f$ does belong to $W^{1,2}_0$ according to my definition, as $f(0)=0$. And about second definition, yes I quoted it wrong; I need $f(t)=\int_0^t f'(s)\,ds$ I think – 2017-02-04
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    Your definition is incorrect. $W^ {1,2}_0(0,1)$ is the subsetof functions in $W^{1,2}$ that vanish on the bloundary of $(0,11)$. – 2017-02-04

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