This is part of an old prelim exam in Analysis that I'm studying to prepare for my own prelim.

We're given that $H$ is a Hilbert space over $\mathbb R$ and $T\in\mathcal B(H,H)$ is a bounded linear operator such that $$ \langle Tx, x\rangle \geq \lVert x\rVert^2\quad \forall x\in H $$ The problem actually asks us to show that $Tx=y$ has a unique solution $\forall y\in H$ and satisfies $\lVert x\rVert \leq \lVert y\rVert$, but I got stuck just trying to show $T$ is onto. I don't really have any ideas for strategies here. I tried making something of the fact that the above implies $\langle Ty-y,y\rangle\geq 0\ \forall y$, but that didn't seem to go anywhere.

  • $\begingroup$ This is Lax-Milgram lemma in disguise. $\endgroup$ – daw Aug 11 '16 at 12:23
  • $\begingroup$ Reading up on that, I see that you're right. Proving that lemma would be way beyond the scope of the exam and that classes it is based on, though. (We've only done real, not complex analysis; inner product is the only bilinear form we've encountered; etc, etc). Anyway, the pages I find online about Lax-Milgram only explain it, they don't offer proofs, so I'm guessing that proving it would be a doozy. Do you have any suggestion for how I could establish that $T$ is surjective without proving Lax-Milgram (or citing it without proof, since we never proved it in our classes)? $\endgroup$ – Ceph Aug 11 '16 at 12:43

Due to the assumption, $T$ is injective.

Let $R(T)$ denote the range of $T$. Then $T^{-1}$ exists as a linear mapping from $R(T)$ to $H$. Due to the inequality, we have $\|T^{-1}y\|\le \|y\|$ for all $y\in H$.

Now argue $R(T)$ is closed: take a sequence $(y_k)$ in $R(T)$ converging to $y$ in $H$. Then by the estimate on $T^{-1}$ the sequence $(T^{-1}y_k)$ is Cauchy, thus converging to some $x$ with $Tx=y$. Hence, $R(T)$ is closed.

It remains to prove that $R(T)=H$. Take $x\in R(T)^\perp$. Then $\langle x, Ty\rangle=0$ for all $y$. Thus, $\langle x,Tx\rangle =0$ implying $x=0$.

So we have shown that $R(T)$ is closed, and its orthogonal complement is zero, this proves $R(T)=H$.

  • $\begingroup$ Thanks. How do you have, though, that $T$ is injective? And how do you have that $\lVert T^{-1}y\rVert\leq \lVert y\rVert$? $\endgroup$ – Ceph Aug 11 '16 at 13:43
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    $\begingroup$ @Ceph For injectivity plug $Tx=0$ in the inequality, for the estimate on $T^{-1}$ use $x:=T^{-1}y$. $\endgroup$ – daw Aug 11 '16 at 13:46
  • $\begingroup$ Ah, right, I see the injectivity. On the other issue, though, using $x:=T^{-1}y$ just gives me $$\langle T(T^{-1}y),T^{-1}y\rangle \geq \lVert T^{-1}y\rVert^2 \iff \langle y,x\rangle \geq \lVert x\rVert ^2$$ which still doesn't get me $\lVert T^{-1}y\rVert\leq \lVert y\rVert$. Of course, it's immediate from the other inequality, but that other inequality is to be proven. $\endgroup$ – Ceph Aug 11 '16 at 13:57
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    $\begingroup$ Use Cauchy-Schwartz to get $\| T^{-1} y \|^2 \leq \| y \| \| T^{-1} y \|.$ $\endgroup$ – Justthisguy Aug 11 '16 at 14:54
  • $\begingroup$ Ah, I got it now. I had been trying to apply C-S to a different part of the problem but hadn't thought to use it for this. Thanks. $\endgroup$ – Ceph Aug 11 '16 at 15:44

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