# Riesz representation and inverse operator.

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.

• Applying the theorem to $A+tI$ you can construct the inverse as it is described in wikipedia – Javi Nov 26 '18 at 18:00
• I think you mean $(\cdot,\cdot)$ defines an inner product on $X$, not $V$. – 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 $$f=0$$.