For $\langle Tx, x \rangle \geq \|x\|^2$, prove a solution exists for $Tx = y$.

Question

Edit

I've posted this a couple other times, so I now plan on deleting those, and just using this one.

Here's the original problem:

$H$ is a real Hilbert space. Let $T: H\longrightarrow H$ be a bounded linear operator on $H$ such that the inner product satisfies $$\langle Tx, x \rangle \geq \|x\|^2 \tag{A}$$ Prove that a solution for the equation $Tx = y$ exists for all $y \in H$.

I've been looking at this problem and studying orthogonal complements and the direct sum theorem for a few days now, to no avail, so I suspect there's something very fundamental I don't understand here. Below I'm posting W. Zhan's proof that I don't understand, and a couple of starting points of my own. W. Zhan's, as mentioned below, isn't working for me, so I'm starting a bounty. Thank you.

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Proof given by W. Zhan

Note that adjoint of a bounded operator always exists by Riesz Represenation Theorem. So we have $$ \langle Tx,x \rangle = \langle x, T^*x \rangle \ge ||x||^2$$ So $T^*$ is injective. We also have $im T^\perp = \ker T^*$. Hence, $im T = H$.

I'm not sure what I'm not understanding. I do understand

So $T^*$ is injective.

and also

We also have $im T^\perp = \ker T^*$

but not how those two facts yield the conclusion.

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Here's a couple starts I've had on my own

Proof (alternate start 1)

Let $Tx = 0$, then

$$0 = \langle Tx, x\rangle \geq \|x\|^2 \geq 0$$

so $\|x \| = 0$, but by definition of the norm, only if $x = 0$. So $Tx = 0 \iff x = 0$, therefore $T$ is injective. This means $T$ has an inverse linear operator $T^{-1}$.

Now, let $y \in \overline{TH}$. By definition, $y$ is a limit point of $TH$, so there exists a sequence $(x_n)$ in $H$ such that $(Tx_n) \longrightarrow y$. However, $x_n = T^{-1}Tx_n$, which implies

$$\lim_{n\to \infty}x_n = \lim_{n\to \infty} T^{-1}Tx_n = T^{-1}\lim_{n \to \infty}Tx_n = T^{-1}y$$

... here I think I'm in trouble because I don't know if I can actually show $T^{-1}y$ exits, which I think is equivalent to showint $T^{-1}$ is bounded.

Proof (alternate start 2)

Here, as W. Zhan pointed out, it's easy to see that $(TH)^\perp = \{0\}$, which implies that $TH$ is dense in $H$. In other words, $\overline{TH} = H$. So all that remains is to show that $TH$ is closed. But this is similar to the argument in my first attempt.

Discussion

I understand that $T^*$ is injective—because it is a linear operator and $Tx = 0 \implies x = 0$. In fact $T$ is as well. So I believe another option is to show surjectivity of $T$, but I can see how to do that.

Would somebody mind proving the original proposition in a detailed way?

Answer 1

Note that adjoint of a bounded operator always exists by Riesz Represenation Theorem. So we have $$ \langle Tx,x \rangle = \langle x, T^*x \rangle \ge ||x||^2$$ So $T^*x =0 \Rightarrow ||x||=0 $, hence $T^*$ is injective. We also have $im T^\perp = \ker T^*$. Thus, $im T = H$.

To see why $T$ has closed range, note $||Tx||= \sup_{||y||\le1} |\langle Tx, y \rangle| \ge \langle Tx, x \rangle /||x|| \ge ||x||$. Then result follows from characterization of closed image. The direction you need is actually already on MSE.

EDIT: Here's an easier way to see why $T$ has closed range. Suppose $Tx_n \rightarrow y$, then applying $x=x_n-x_m$ implies $x_n$ is Cauchy. Continuity implies $Tx = y$.

Answer 2

Here's an alternative to the existing answers which won't require us to show that $T$ has closed range along the way (obviously we'll get that indirectly from $T$ being surjective). Fix $y \in H$. Note that $Tx = y$ if and only if $$\alpha y - \alpha Tx + x = x$$ for $\alpha > 0$. So you're trying to find a fixed point of the operator $\Lambda:x \mapsto \alpha y - \alpha Tx + x$. We can compute \begin{align} \|\Lambda x_1 - \Lambda x_2\|^2 =& \| x_1 - x_2 - \alpha T(x_1 - x_2)\|^2 \\ = & \|x_1 - x_2\|^2 + \alpha^2 \|T(x_1 - x_2)\|^2 - 2 \alpha \langle T(x_1 - x_2), x_1 - x_2 \rangle \\ \leq& \|x_1 - x_2\|^2 + \alpha^2 \|T\|^2 \|x_1 - x_2\|^2 - 2 \alpha \|x_1 - x_2\|^2 \end{align} If we pick $\alpha = \|T\|^{-2}$, we see that $\lambda$ is a contraction and so the desired $x$ exists by the contraction mapping theorem.

Answer 3

You can show surjectivity of $T$ by using Theorem 0.1 of this paper. What you need to show is

$$ \sup_{x\in H\setminus\{0\}} \langle Tx, y\rangle/\lVert x\rVert \geq c\lVert y\rVert$$

for all $y\in H$ (for some $c>0$). So, for a given $y\in H$ you set $x$ to $y$ and get

$$ \sup_{x\in H\setminus\{0\}} \langle Tx, y\rangle/\lVert x\rVert \geq \langle Ty, y\rangle/\lVert y\rVert \geq \lVert y\rVert$$

which is equivalent to surjectivity of $T$.

Answer 4

We can prove that $T$ is not singular, and from that, the problem is solved.

Assume $T$ is singular. Then there exists a vector $v \not= 0$ such that $Tv = 0$. But then $\langle Tv, v\rangle = 0 < \|v\|^2$ which contradicts our assumption that $\langle Tv, v\rangle \ge \|v\|^2$.

Okay, so now that $T$ is not singular, it has an inverse, implying that $x = T^{-1} y$.