How to prove that $I + t S$ is surjective?

Question

I'm trying to solve part (2.) of below exercise

Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $S :H \to H$ be a bounded linear operator such that $\langle Su, u \rangle \ge 0$ for all $u \in H$. Let $N(S)$ be the kernel of $S$ and $R(S)$ the range of $S$. Then

1. $N(S) = R(S)^\perp$.
2. The map $I + t S$ is bijective for all $t>0$. Here $I:H \to H$ is the identity map.

Fix $y \in R(S)$. I failed to find $x \in H$ such that $(I + t S) x = y$.

Could you elaborate on how to finish the proof?

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Fix $t>0$. Consider the equation $(I + t S) u=0$. Then $S u =- \frac{1}{t} u$. We have $\langle Su, u \rangle = - \frac{1}{t} \langle u, u \rangle = - \frac{1}{t} |u|^2 \ge 0$. It follows that $u=0$. Hence $N(I+t S)=\{0\}$ and thus $I + t S$ is injective.

Let's prove that $I + t S$ is surjective. Fix $y \in H$. If $y \in N(S)$ then $(I + t S)y=y$. Let $y \in R(S)$. We need to find $x \in H$ such that $(I + t S) x = y$.

Answer 1

Positivity implies $$\|(I+tS)x\|\|x\|\ge \langle (I+tS)x,x\rangle \ge \|x\|^2$$ Thus $$\|(I+tS)x\| \ge \|x\|$$ This implies that the range of $I+tS$ is closed and that $I+tS$ is injective. The same holds for $I+S^T.$ In particular $\ker(I+tS^T)=\{0\}.$ Hence the range of $I+tS $ is equal to the entire space.

Answer 2

Below is another approach based on Lax-Milgram theorem.

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A bilinear form $[\cdot, \cdot]:H \times H \to \mathbb R$ is said to be

• continuous if there is a constant $C>0$ such that $$ |[u, v]| \le C |u| |v| \quad \forall u,v \in H. $$
• coercive if there is a constant $\alpha>0$ such that $$ [v, v] \ge \alpha |v|^2 \quad \forall v \in H. $$

Then we have

Lax-Milgram theorem Assume that $[\cdot, \cdot]:H \times H \to \mathbb R$ is a continuous coercive bilinear form. Then for each $\varphi \in H^*$, there is a unique $u\in H$ such that $$ [u, v] = \varphi (v) \quad \forall v\in H. $$ Moreover, if $[\cdot, \cdot]$ is symmetric, then $u$ is characterized as a minimizer of $$ \min_{v\in H} \bigg \{ \frac{1}{2} [v, v] - \varphi (v) \bigg \}. $$

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Let's prove that $I + t S$ is surjective. We define a bilinear form $[\cdot, \cdot] : H \times H \to \mathbb R$ by $$ [ x, z] := \langle (I + t S) x, z \rangle \quad \forall x, z \in H. $$

Then $|[ x, z]| \le \|I + t S\|_{\mathcal L(H)} |x| |z|$. Then $[\cdot, \cdot]$ is continuous. Also, $$ [ x, x] = \langle (I + t S) x, x \rangle = |x|^2 + t \langle S x, x \rangle \ge |x|^2. $$

So $[\cdot, \cdot]$ is coercive. Fix $y \in H$. We define a continuous linear functional $\varphi: H \to \mathbb R, x \mapsto \langle y, x \rangle$. By Lax-Milgram theorem, there is a unique $x \in H$ such that $$ [ x, z] = \varphi (z) \quad \forall z \in H. $$

This implies $$ \langle (I + t S) x, z \rangle = \langle y, z \rangle \quad \forall z \in H. $$

It follows that $(I + t S) x = y$. This completes the proof.