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My teacher said that if $P_n f = \sum_{j=0}^n(f,w_j)w_j$, where $w_j$ is orthonormal basis of $L^2$, then $|P_n f- f|_{L^2} \to 0$ for $f \in L^2$. How do I prove this?

I thought $$|P_nf - f| = |\sum_{j=0}^n(f,w_j)w_j - \sum_{j=0}^\infty(f,w_j)w_j| = |\sum_{j={n+1}}^\infty(f,w_j)w_j| \leq \sum_{j={n+1}}^\infty|f|$$ where the last equality is by Cauchy Schwarz, but this equals infinity.

How to show it correctly using the definition??

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up vote 3 down vote accepted

The last inequality is not tight enough. However, we can write $$\left\lVert\sum_{j=n+1}^{+\infty}(f,w_j)w_j\right\rVert^2\leqslant \sum_{j=n+1}^{+\infty}|(f,w_j)|^2,$$ as $\{w_j\}$ is an orthonormal subset. By Bessel's inequality, the series $\sum_{j=1}^{+\infty}|(f,w_j)|^2$ is convergent. Conclude.

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(If you don't know Bessel's inequality, then simply apply Cauchy-Schwarz to each term in the sum on the RHS.) – Zach L. Feb 9 '13 at 16:23
Thanks. I assume you mean use the DCT. Can one do this for arbitrary Hilbert spaces (not L^2)? – michael_faber Feb 11 '13 at 12:48
@michael_faber There is no need to use the DCT, as we know that the remainder of a convergent series converges to $0$. – Davide Giraudo Feb 11 '13 at 13:09

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