Welcome to mathimatikoi.org forum; Enjoy your visit here.

Lemma

Functional Analysis
Post Reply
Papapetros Vaggelis
Community Team
Community Team
Articles: 0
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Lemma

#1

Post by Papapetros Vaggelis » Thu Mar 02, 2017 5:52 pm

Let \(\displaystyle{\left(H,\langle{,\rangle}\right)}\) be a Hilbert space. If \(\displaystyle{U:H\to H}\)

is a \(\displaystyle{\mathbb{C}}\) - linear and bounded operator such that \(\displaystyle{||U||\leq 1}\), then prove that

\(\displaystyle{\left(\forall\,h\in H\right)\,\,\left(U(h)=h\iff U^{\star}(h)=h\right)}\).
r9m
Articles: 0
Posts: 59
Joined: Thu Dec 10, 2015 1:58 pm
Location: India
Contact:

Re: Lemma

#2

Post by r9m » Sat May 27, 2017 5:48 pm

Since, $U : H \to H$ satisfies $\lVert U \rVert \le 1$, then $$\left<(I-U)h,h\right> = \lVert h \rVert^2 - \left< Uh,h \right> \ge \lVert h \rVert^2 (1 - \lVert U \rVert) \ge 0 \text{ for all } h \in H$$
We claim that, $N(I-U) = R(I-U)^{\perp}$

If, $h \in N(I-U)$ then, $\left<(I-U)(h - th'), h - th'\right> \ge 0 \implies t^2\left<(I-U)h' , h'\right> - t\left<(I-U)h' , h\right> \ge 0$ for all $h' \in H$ and $t \in \mathbb{R}$.

It follows that $\left<(I-U)h', h\right> = 0$ for all $h' \in H$, i.e., $N(I-U) \subseteq R(I-U)^{\perp}$.

To see the other way inclusion, similarly if $h \in R(I-U)^{\perp}$, then $\left<(I-U)(th' - h),th' - h\right> \ge 0$ for all $h' \in H$ and $t \in \mathbb{R}$,

Implies $t^2\left<(I-U)h' ,h'\right> - t\left<(I-U) h,h'\right> \ge 0$ for all $t \in \mathbb{R}$.

Hence, $\left<(I-U) h,h'\right> = 0$ for all $h' \in H$. That is $h \in N(I-U)$.

Now, since $N((I-U)^{*}) = R(I-U)^{\perp}$ (this is standard result for adjoint of an operator between Banach spaces), it follows $N(I-U) = N(I - U^{*})$, i.e., $U(h) = h \iff U^{*}(h) = h$.
Papapetros Vaggelis
Community Team
Community Team
Articles: 0
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Lemma

#3

Post by Papapetros Vaggelis » Sat May 27, 2017 7:38 pm

Hi r9m.

Here is another solution.

Since \(\displaystyle{||U||\leq 1}\), we get \(\displaystyle{||U(x)||\leq ||x||\,,\forall\,x\in H}\).

Let \(\displaystyle{h\in H}\). Suppose that \(\displaystyle{U(h)=h}\). Then,

\(\displaystyle{\begin{aligned}||U^{\star}(h)-h||^2&=\langle{U^{\star}(h)-h,U^{\star}(h)-h\rangle}\\&=||U^{\star}(h)||^2-\langle{U^{\star}(h),h\rangle}-\langle{h,U^{\star}(h)\rangle}+||h||^2\\&\leq ||U(h)||^2+||h||^2-\langle{h,U(h)\rangle}-\langle{h,U(h)\rangle}\\&=2\,||h||^2-2\,||h||^2\\&=0 \end{aligned}}\)

so, \(\displaystyle{U^{\star}(h)=h}\).

Now, suppose that \(\displaystyle{U^{\star}(h)=h}\). Then,

\(\displaystyle{||U(h)-h||^2=...=0\implies U(h)=h}\).
Post Reply