Exercise 4 Conway, a course in functional analysis.

Question

I am trying to test Conway problem 4 page 49. I'm having trouble proving an implication.

Corollary 5.4. If $T$ is a compact self-adjoint operator, then there is a sequence $\left\{\mu_n\right\}$ of real numbers and an orthonormal basis $\left\{e_n\right\}$ for $(ker(T))^{\perp}$ such that for all $h$, $Th=\sum_{n=1}^{\infty} \mu_n \langle h,e_n\rangle e_n$.

Exercise 4. If $T$ is a compact self-adjoint operator and $\left\{e_n\right\}$ and $\left\{\mu\right\}$ are as in (5.4) and if $h$ is a given vector in $\mathcal{H}$, show that there is a vector $f\in \mathcal{H}$ such that $Tf=h$ if and only if $h\perp ker(T)$ and $\sum_{n}\mu_n^{-1}|\langle h,e_n\rangle|^2<\infty.$ Find the form of the general vector $f$ such that $Tf=h$.

I have this:

$(\Rightarrow)$ Let $f\in\mathcal{H}:Tf=h$. Let $g\in ker(T)$ then $T(g)=0$. \begin{eqnarray} \langle h,g\rangle&=&\langle T(f),g\rangle\\ &=&\langle f,T^*(g)\rangle\\ &=&\langle f,T(g)\rangle\\ &=&\langle f,0\rangle\\ &=&0 \end{eqnarray} Therefore $h\perp ker(T)$.

On the other hand, \begin{eqnarray} \langle h,e_n\rangle&=&\langle T(f),e_n\rangle\\ &=&\langle \sum_{n=1}^{\infty} \mu_n \langle f,e_n\rangle e_n,e_n\rangle\\ &=&\sum_{n=1}^{\infty}\mu_m \langle f,e_m\rangle \langle e_m,e_n\rangle\\ &=&\mu_n\langle f,e_n\rangle\\ \end{eqnarray} Therefore $\displaystyle \frac{\langle h,e_n\rangle}{\mu_n}=\langle f,e_n\rangle$ Now, $\displaystyle \sum_{n=1}^{\infty} |\frac{\langle h,e_n\rangle}{\mu_n}|^2=\sum_{n=1}^{\infty} |\langle f,e_n\rangle |^2\leq \|f\|^2$ (Bessel Inequality) Therefore $\displaystyle \frac{ |\langle h,e_n\rangle|}{\mu_n}\in l^2$

($\Leftarrow$) (I don't know how to prove this implication)

Let $h\in ker(T)$ and $\displaystyle \frac{ |\langle h,e_n\rangle|}{\mu_n}\in l^2(\mathbb{N})$

Formally, I think the $f$ to find is $\displaystyle f=\sum_{n=1}^{\infty}\frac{\langle h,e_n\rangle}{\mu_n}e_n$ because $T(f)=\sum_{n=1}^{\infty} \langle h,e_n\rangle e_n$ and $h=\sum_{n=1}^{\infty} \langle h,e_n\rangle e_n$ but I'm not sure if this is correct. How proves this implicance?

Answer 1

Observe that, as a direct result of the corollary, $Te_n=\mu_ne_n$. Since $h\in Ker(T)^\perp,$ you can expand it in the basis: $$h=\sum\limits_{n=1}^\infty\langle h,e_n\rangle e_n.$$ You also know that $$Th= \sum\limits_{n=1}^\infty\mu_n\langle h,e_n\rangle e_n.$$ As you recommended, let's call $$f=\sum\limits_{n=1}^\infty\frac{1}{\mu_n}\langle h,e_n\rangle e_n.$$ This makes sense due to the fact that $\mu_n^{-1}|\langle h,e_n\rangle|\in\ell^2.$ Then, applying $T$ gives $$Tf=\sum\limits_{n=1}^\infty \frac{1}{\mu_n}\langle h,e_n\rangle Te_n=\sum\limits_{n=1}^\infty \frac{1}{\mu_n}\langle h,e_n\rangle \mu_n e_n=\sum\limits_{n=1}^\infty\langle h,e_n\rangle e_n=h.$$ We were able to pull the infinite sum out as a consequence of linearity and continuity.

EDIT: Let's justify the first equality. Call $a_n=\mu_n^{-1}\langle h,e_n\rangle$. The missing work is showing that $$T\left(\sum\limits_{n=1}^\infty a_n e_n\right)=\sum\limits_{n=1}^\infty T(a_n e_n).$$ If we show this, then we just use linearity to conclude that it equals $\sum\limits_{n=1}^\infty a_n Te_n.$

First, since $T$ is linear, $$T\left(\sum\limits_{n=1}^m a_n e_n\right)=\sum\limits_{n=1}^m T(a_n e_n).$$ Taking the limit as $m\rightarrow\infty$, $$\lim_{m\rightarrow\infty}T\left(\sum\limits_{n=1}^m a_n e_n\right)=\sum\limits_{n=1}^\infty T(a_n e_n).$$ The convergence of these limits should be understood as occurring in the $\mathcal{H}$-norm.

We're done if we can justify moving the limit inside of $T$. The follows from continuity: Since $T$ is continuous, it sends convergent sequences to convergence sequences. In particular, if we call $b_n=\sum\limits_{n=1}^m a_n e_n$, then $b_n\rightarrow b:=\sum\limits_{n=1}^\infty a_n e_n$ in $\mathcal{H}$, so $T(b_n)\rightarrow T(b)$ in $\mathcal{H}.$ That is, $$\lim_{m\rightarrow\infty}T\left(\sum\limits_{n=1}^m a_n e_n\right)=T\left(\sum\limits_{n=1}^\infty a_n e_n\right).$$