I have found an example that strong convergence does not imply convergence in norm.
Let us take $T_{n} = P_{span\{e_{1},…,e_{n}\}}$ be a projection on $span\{e_{1},…,e_{n}\}$ in Hilbert's space with a ortonormal base $(e_{n})_{n=1}^\infty$. Let
$$T_{n}x := \sum^n_{j=1} <x,e_{j}>e_{j}$$
where <.,.> is a scalar product. And now it is said, that from theorem about Fourier series:
$$ ||T_{n}x – Tx|| = ||T_{n}x – x|| \to 0 $$
for $n \to \infty$.
Can someone explain to me, from which theorem does it comes that it tends to zero, and why Tx = x here?
Regards
Best Answer
So I'm not entirely sure due to some lack of context but I suspect this is an example of the following. Given how the sequence $(T_{n})$ is set up one would expect that $(T_{n})$ converges to the identity operator $I$. And indeed for all $x\in H$ where $H$ is the Hilbert space we find that $$\lim_{n\rightarrow\infty}\|T_{n}x-Ix\|=\lim_{n\rightarrow\infty}\|T_{n}x-x\|=0.$$ This follows from the fact that $(e_{n})$ is an orthonormal basis, so $$x=\sum^{\infty}_{j=1}\langle x,e_{j}\rangle e_{j}$$ i.e. $$\lim_{n\rightarrow\infty}\sum^{n}_{j=1}\langle x,e_{j}\rangle e_{j}\rightarrow x.$$ Therefore $$\lim_{n\rightarrow\infty}\|T_{n}x-x\|=\lim_{n\rightarrow\infty}\|x-\sum^{n}_{j=1}\langle x,e_{j}\rangle e_{j}\|=0.$$
However if we consider the limit in operator norm we find $$\|I-T_{n}\|=\sup\{\|(I-T_{n})x\|:x\in H,\|x\|\leq 1\}=\|(I-T_{n})e_{n+1}\|=1$$ So even though for all $x$ $(T_{n}x)$ strongly converges to $Ix$, $(T_{n})$ does not converge in the operator norm.
I hope this answers your questions, let me know if I need to elaborate on anything.