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Let $f:[a,b]\to H$ for an arbitrary real-valued Hilbert space $H$ be continuous on the real interval $[a,b]$. We are given that $B=\{b_k:k\in \mathbb{N}\}$ is an orthonormal basis for $H$. I am trying to show that $$\sum_{k=1}^{n}\langle f(t),b_k\rangle^2$$ converges uniformly to $\|f(t)\|^2$ on the interval.

I'm easily able to show pointwise convergence just using the fact that there is a sequence of scalars $(\alpha_k)_t$ such that for any $t$ in the interval $$f(t)=\sum_{k=1}^\infty\alpha_kb_k.$$ I know I need to bring in the continuity of $f$ to gain uniform convergence but I'm not sure how to do that in my present argument. Any tips would be appreciated.

K.Power
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Let's denote $$ f_n(t) := \sum_{k=1}^n \langle f(t), b_k \rangle^2. $$ Then, the $f_n$ are continuous, increasing and converge pointwise to the continuous function $g(t) := \lVert f(t) \rVert^2$. The fact that the $f_n$ are increasing implies uniform convergence.

See for example Sequence of monotone functions converging to a continuous limit, is the convergence uniform?.

Community
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  • Thanks so much. I didn't realise that there was such a result. It certainly wasn't mentioned in my undergrad analysis courses, which is strange considering how useful it seems. – K.Power Jun 02 '16 at 20:18
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    You are welcome. Indeed, for some reason it is not well known (as it was for me, after I needed it to prove a $C^*$-algebraic result !). –  Jun 02 '16 at 20:21