Two exercises in functional analysis.

Question

I have met two questions which, after some attempts have I have yet been able to solve nor find any available solutions online. Could anyone please offer me some insights?

1. Let $ H $ be a Hilbert space over $\mathbb{C}$ and $T \in B(H,H)$ an unitary operator. For $ n \in \mathbb{N} $ set $$ S_n : = \frac {1}{n}( I + T + ... + T^{n-1} ) .$$ Show that $S_{n}v \longrightarrow P_{M}v $ as $ n \rightarrow \infty $ where $ P_{M} $ is the orthogonal projection onto the subspace $ \text{Ker}(I - T) $.

2. Let $E$ be a normed vector space over $\mathbb{K}$, let $M \subseteq E $ be a subset and suppose that $ \sup_{ v \in M} |f(v)| < \infty $ for a all functional $f \in B(E, \mathbb{K})$. Show that $M$ is bounded.

From what I have heard the second question is not meant to be difficult. However as $E$ is not complete there is not much theorem I could work with. I thought maybe by working with the Banach space $ B(E,\mathbb{K}) $ but came to no result. I'm still a noob in FA so any help would be much appreciated!

Thanks!

Answer 1

The second question is an application of uniform boundedness. Let $j:E\to E^{**}$ be the natural embedding and consider $j(M)\subset B(E^*,\Bbb K)$. For all $f\in E^*$ you have $\{ j(m)\,(f)=f(m) \mid m\in M\}$ is bounded. By uniform boundedness the set $j(M)$ is bounded in $E^{**}$. But $j$ is an isometric embedding, so $M$ must be bounded in $E$.

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The first question can be done either with the spectral theorem for unitary operators or by explicitly showing $S_nv\to 0$ if $v\in \ker(I-T)^\perp$ and noting $S_nv=v$ if $v\in \ker(I-T)$. From this follows that $S_n$ converges pointwise to the projection onto $\ker(I-T)$.

Now the second part is trivial, for the first note $\|S_n(I-T)\| = \|\frac1n(I-T^n)\|≤\frac2n$. Now $\langle(I-T)v,w\rangle=\langle v,(I-T^*)w\rangle$ from which $$\mathrm{im}(I-T)^\perp = \ker(I-T^*)=\ker(I-T).$$ This means that $\mathrm{im}(I-T)$ is dense in $\ker(I-T)^\perp$. So for any $\epsilon>0, v\in \ker(I-T)^\perp$ we find a $w$ with $\|v-(I-T)w\|<\epsilon$. It follows $$\|S_nv\|≤\|S_n\|\,\|v-(I-T)w\|+\|S_n(I-T)w\|≤\epsilon+\frac2n\|w\|.$$ As $n$ goes to infinity you find that $\limsup_n\|S_nv\|≤\epsilon$, which was arbitrary. It follows $S_nv\to 0$ for $v\in \ker(I-T)^\perp$.

Answer 2

$(I-T)(S_n(v))={1\over n}(I-T)(I+...+T^{n-1})(v)={1\over n}(v-T^n(v))$.

This implies that $\langle (I-T)(S_n(v)),(I-T)(S_n(v))\rangle={1\over n^2}(\|v\|^2+\|T^n(v)\|^2-\langle v,T^n(v)\rangle-\langle T^n(v),v\rangle)$.

Since $\|T^n(n)\|=\|v\|$ and $|\|\langle v,T^n(v)\||\leq ||v||\|T^n(v)||=\|v\|^2$.We deduce that $lim_n(I-T)(S_n(v))=0$,

$\langle S_n(v)-v,S_n(v)\rangle=0$, to see this, remark that

$\langle S_n(v),S_n(v)\rangle= {1\over n}\sum_{i=0}^{n-1}\langle T^i(v),S_n(v)\rangle$ and $\langle T^i(v),S_n(v)\rangle=\langle v,S_n(v)\rangle$ since $\langle T^p(v),T^q(v)\rangle =\langle v,T^{q-p}(v)\rangle, q\geq p$. We deduce that $\langle v,S_n(v)\rangle=\langle v,S_n(v)\rangle$ and $\langle S_n(v)-v,S_n(v)\rangle=\langle S_n(v),S_n(v)\rangle-\langle v,S_n(v)\rangle=0$. This implies that $\langle lim_nS_n(v)-v,lim_nS_n(v)\rangle=0$.