Prove that solution of a variational problem exists

Question

Let $X$ is a Hilbert Spaces We define two operators $$a:X\times X\rightarrow\mathbb R$$ and $$b:X\rightarrow\mathbb R$$ where $a$ is a symmetric, bounded, strongly positive operator, and $b$ is a linear continuous functional.

By symmetric, it means that $a(u,v)=a(v,u)$ $\forall u,v\in X$, bounded means that $\exists d>0$ so that $a(u,u)\leq d\|u\|^2$ $\forall u\in X$ and strongly positive means that $\exists c>0$ so that $a(u,u)\geq c\|u\|^2$

Consider a problem to minimize the function $$2^{-1}a(u,u)-b(u).$$ I want to prove that the solution exist. In other words, there exists $u_0$ that minimize the function $$F(u):=2^{-1}a(u,u)-b(u)\quad\forall u\in X.$$ So I define $$\alpha :=\inf_{u\in X} F(u).$$ Then I'll prove that such $\alpha$ exists and then show that there is $u_0$ such that $F(u_0)=\alpha$

Now, I'm stucked on proving that $\alpha$ exists. What I now know is the consequence of the fact that $a$ is a strongly positive operator and $b$ is a functional that $$F(u)=2^{-1}a(u,u)-b(u)\geq c\|u\|^2-\|b\|\|u\| $$ Then, by definition there will be a sequence $(u_n)$ where $F(u_n)\rightarrow \alpha$. After that I'll show that such sequence must be Cauchy. Since $X$ is a Hilbert space so there must be $u_0 \in X$ such that $u_n \rightarrow u_0$. Finally, I'll show that $F(u_0)=\alpha$

Please help me on proving that such $\alpha$ exists and give me advice should you have any comment on the construction of my proof. Thankyou! :D

Answer 1

You already have the estimate \begin{equation*} F(u) \ge c \, \|u\|^2 - \|b\| \, \|u\|. \end{equation*} What is the global minimizer of $f : \mathbb{R} \to \mathbb{R}$, $f(x) = c \, x^2 - \|b\| \, x$? What is the relation of this minimizer to your infimal value $\alpha$?

I hope that these hints are helpful.