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The answer I think must be no. However I can not see what is wrong with the following reasoning.

Assume we have an estimate of the form for the adjoint of an operator $L$ in $L^{2}$ of the form:

$$||\phi||\le||L^{*}\phi||$$

for all $\phi\in C^{\infty}\bigcap H^{k}=V$ and where $||\phi||=\int\phi^{2}$ which is the usual $L^{2}$ norm and $H^{k}$ is higher order Sobolev spaces.

Now $V$ and $H^{k}$ are subspaces of $L^{2}$. Moreover, $V\subset H^{k}$

If we define an inner product space $W$ as the functions $H^{k}$ but with inner product in $L^{2}$ i.e., our Hilbert space is the vector space with elements $f,g\in H^{k}$ but with inner product $(f,g)_{L^{2}}$.

Then we can use Hahn-Banach theorem on the functional

$$k(L^{*}\omega)=\int F\omega$$

to extend the linear functional to $W$, where we are seeing $W$ as the whole space and $V$ as a subspace.

Hence we have a bounded linear functional in $W$. Hence $k\in W^{*}$.

Now using Riez representation theorem there is a $\Psi\in W$ such that

$$(\Psi,L^{*}\omega)_{W}=(\Psi,L^{*}\omega)_{L^{2}}=(F,\omega)_{L^{2}}$$

which is a weak solution, but because $\Psi\in W$ we have that $\Psi\in H^{k}$ ?!

I would appreciate a lot any comments where this went wrong.

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  • $\begingroup$ I haven't read all the details, but the space $W$ as you defined it will typically be an incomplete inner product space, not a Hilbert space. You can't use the Riesz representation theorem there. $\endgroup$ Oct 21, 2015 at 15:12
  • $\begingroup$ Thanks. I think that settles the question. $\endgroup$
    – yess
    Oct 21, 2015 at 16:37

1 Answer 1

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The space $W$ as you defined it will typically be an incomplete inner product space, not a Hilbert space. You can't use the Riesz representation theorem there.

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