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.

  • $\begingroup$ 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

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 $f=0$.

  • $\begingroup$ 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

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