Linear positive definite bounded operator $T:H\to H$

Question

I was trying to solve this problem:

Let $H$ be a Hilbert space and $T:H\to H$ a bounded linear operator such that $$ (Tx,x)\geq ||x||^2 \quad \forall x\in H $$ where $(\cdot,\cdot)$ and $||\cdot||$ are the scalar product and the norm on $H$.

a) Prove that the inverse operator $T^{-1}$ exists and is continuous and bounded

b) Let $K:H \to H$ be a compact linear operator. Show that if $T+K$ is injective then $T+K$ is surjective.

$T$ is clearly injective, but I don't know how to show that it is surjective (to show that the inverse exists), and if $T^{-1}$ exists then $$ ||TT^{-1}x||\cdot||T^{-1}x|| \geq (TT^{-1}x,T^{-1}x)\geq ||T^{-1}x||^2. $$ This proves that $T^{-1}$ is bounded.

I don't have any good ideas for the second point.

Answer 1

As for the second part, you can apply Fredholm's alternative theorem: for $A \in \mathcal B(H)$ compact, $$ \ker (I + A) = \{0\} \iff (I+A) [H] = H .$$ Apply this to the compact operator $A = T^{-1} K$. Since $T+K$ is injective, so is $I+ T^{-1}K$ so by the Fredhold alternative you get that $(I+T^{-1}K)[H] = H$ and thus $(T+K)[H]=H$.

Answer 2

To show that $T$ is surjective, we can show that the image is dense by showing that its orthogonal space is $\{0\}$, and closed by continuity, this it is surjective.

Assume $v\in (ImT)^\perp$, i.e. $\langle Tx,v\rangle=0$ for all $x\in H$. Choosing $x=v$ gives $\langle Tv,v\rangle=0\ge||v||^2$, thus $v=0$, which finishes.