2
$\begingroup$

In class, my professor went through the following construction: Let $\Omega$ be a bounded domain and define $X$ to be $H_0^1(\Omega)$ or $H^1(\Omega)$. We also define $A: X \to X^\prime$ (where $X^\prime$ denotes the dual space of $X$) by the duality pairing $$ \langle Af, g\rangle = \int_\Omega \nabla f \cdot \nabla g $$ We will also view $L^2(\Omega)$ as a subspace of $X^\prime$. Now, there exists $t_0$ such that for all $t>t_0$, $$ (\cdot, \cdot) : X\times X \to \mathbb{R}, \quad (f, g) = \langle Af, g \rangle + t\langle f, g\rangle_{L^2(\Omega)} $$ defines an inner product on $X$. For each such $t$, it follows from the Riesz representation theorem that $A+t\mathrm{I} : X\to X^\prime$ is invertible.

I am very confused as to how the Reisz representation theorem applies here and how one can deduce this. Any input is appreciated.

$\endgroup$
  • $\begingroup$ Applying the theorem to $A+tI$ you can construct the inverse as it is described in wikipedia $\endgroup$ – Javi Nov 26 '18 at 18:00
  • $\begingroup$ I think you mean $(\cdot,\cdot)$ defines an inner product on $X$, not $V$. $\endgroup$ – user10354138 Nov 26 '18 at 18:01
1
$\begingroup$

For simplicity take $t = 1$. To prove that $A+I$ is invertible you must show it is one-to-one.

The inner product on $H_0^1(\Omega)$ is given by $(f,g) = \displaystyle \int_\Omega \nabla f \cdot \nabla g \, dx$.

The operator $A+I : H_0^1(\Omega) \to H_0^1(\Omega)'$ is defined by $$\langle (A + I)f,g \rangle = \int_\Omega \nabla f \cdot \nabla g + fg \, dx.$$ The Poincare inequality gives you $$\int_\Omega f^2 \, dx \le C \int_\Omega |\nabla f|^2 \, dx$$ for all $f \in H_0^1(\Omega)$. Consequently for fixed $f$ the functional $Lg = \langle (A+I)f,g \rangle$ defines a bounded linear functional on $H_0^1(\Omega)$. The Riesz representation theorem provides you with a unique function $h \in H_0^1(\Omega)$ satisfying $$Lg = (h,g)$$ for all $g \in H_0^1(\Omega)$. Thus you can regard $A+I$ as a well-defined operator from $H_0^1(\Omega)$ to itself, with $$(A+I)f = h.$$

I think that $A+I$ is clearly linear. To prove that $A+I$ is one-to-one it suffices to show it has a trivial kernel. But if $(A+I)f = 0$ then $\langle (A+I)f,g \rangle = 0$ for all $g$. In the particular case of $f=g$ you obtain $\displaystyle \int_\Omega |\nabla f|^2 + f^2 \, dx = 0$, giving you $f=0$.

$\endgroup$
  • $\begingroup$ Thanks! I mainly got lost with the notation and it helps to see the steps written out :-) $\endgroup$ – Quoka Nov 27 '18 at 0:51

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.