Let $H$ be a Hilbert space and $S \in \mathcal{L}(H)$ such that $\langle Su,u \rangle \geq 0$ for every $u \in H$. I want to prove that $I+tS$ is bijective for every $t>0$.
What I was abble to do until now:
- I proved that $\text{ker}(S)=\text{Range}(S)^{\bot},$
- I proved also that $I +tS$ is injective for every $t>0.$
The hint that I saw in the book is to use Lax – Milgram Theorem to prove surjectivity. My attempt to solve the problem was the following:
Let $y \in H$, I want to show that there exists $x \in H$ such that $y=x+tS(x)$. I tried to definine the application $T(u,v)=\langle v+tS(v)-y,u \rangle$ and use Lax – Milgram to find some $v_0 \in H$ such that $\langle v_0+tS(v_0)-y,u \rangle=0$ for every $ u \in H$. The problem is that the $T$ defined as above is not even bilinear and I'm failling in construct an bilinear application to apply Lax – Milgram. Any hint will be very helpful, thank you!
Best Answer
If we define $T(u,v)=\langle v+tS(v),u \rangle$, we have that $T$ is linear, coercive and continuous. Applying Lax - Milgram, it follows that for each $y \in H$ there exists $x \in H$ such that $$\langle x+tS(x),u \rangle=\langle y,u \rangle\,\,\ \forall u \in H. $$
The result follows.