Exercise :
Let $H$ be a Hilbert space and $A \in \mathcal{L}(H)$ such that : $$\langle A(u),u \rangle \geq \langle u, u \rangle \; \forall u \in H$$ Show that $A$ is invertible.
Attempt/Thoughts :
The inequality given to hold can be transformed to $$\langle A(u),u \rangle \geq \|u\|^2$$ since $\|u\| = \sqrt{\langle u, u \rangle} $ by the definition of the inner product functional for Hilbert spaces.
Now, by the Cauchy-Schwarz inequality, one can yield : $$|\langle Au, u\rangle| \leq \|Au\|\|u\|$$
Combining the two expressions now gives us : $$\langle Au, u \rangle \geq \|u\|^2 \Rightarrow |\langle Au,u\rangle| \geq \|u\|^2 \Rightarrow \|Au\| \|u\| \geq \|u\|^2$$ $$\implies$$ $$\boxed{\|Au\| \geq \|u\|}$$ Now if I consider a sequence $\{u_n\}_{n \in \mathbb N} \subset H$ such that $Au_n \to u \in H$ then it would be : $$\|A(u_n - u_m) \| \to 0 \implies \|u_n-u_m\| \to 0 \quad \text{for} \; n,m \to \infty$$ That means that $\{u_n\}_{n \in \mathbb N}$ is Cauchy and thus $A : H \to A(H)$ is injective $("1-1")$, thus invertible ?
Is my approach correct ? Any tips, corrections and/or elaborations will be appreciated.
You have shown that the image of $A$ is closed since if $A(u_n)$ is a Cauchy sequence, so is $u_n$. Let $L$ be the adherence of $A(H)$, consider $u$ orthogonal to $L$, you have $0=\langle A(u),u\rangle\geq \langle u,u\rangle$, this implies that $u=0$, thus $L=X$. We deduce that $A$ is surjective, the open map theorem implies that $A$ is invertible.