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Let $(\varphi_k)$ be an orthonormal basis of a Hilbert space $\mathbf{H}$, and consider the sum $\sum_{k}\left < x, \varphi_k\right > \overline{\left < y, \varphi_k \right >}$ for some $x,y\in \mathbf{H}$. It converges to $\left < x, y\right >$. Does it necessarily absolutbly converge? I'm pretty sure it does in $\ell_2$ and even in $L_2[a,b]$. But does it hold in the general case, and how does one prove it?

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  • $\begingroup$ Could you share the $l^2$ case? I am guessing Holder inequality in place of the special case Cauchy Schwartz inequality would work $\endgroup$ – Medo Apr 16 '18 at 18:50
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Since $$\sum_k | \langle x, \phi_k \rangle |^2 = \sum_k \langle x,\phi_k \rangle \overline{ \langle x,\phi_k \rangle} = \langle x,x \rangle = \|x\|^2$$ you can use e.g. Cauchy-Schwarz to find that $$ \sum_k | \langle x,\phi_k \rangle \overline{ \langle y,\phi_k \rangle}| = \sum_k | \langle x,\phi_k \rangle | \cdot |\overline{ \langle y,\phi_k \rangle}| \le \|x\| \|y\|$$ so yes, it converges absolutely.

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  • $\begingroup$ I tried that, but doesn't Cauchy-Schwartz give me only $\sum_k | \langle x,\phi_k \rangle | \cdot |\overline{ \langle y,\phi_k \rangle}| \le \sum_k \|x\| \|y\| = \infty$ ? $\endgroup$ – Bary12 Apr 16 '18 at 18:28
  • $\begingroup$ No, there is no sum on the right-hand side. $\endgroup$ – Umberto P. Apr 16 '18 at 18:29
  • $\begingroup$ Oh, I thought of using Cauchy-Schawtz as $\left|\left\langle x,\varphi_{k}\right\rangle \right|\leq\left\Vert x\right\Vert \left\Vert \varphi_{k}\right\Vert =\left\Vert x\right\Vert $ but now I understand it should be used on the numbers themselves. $\endgroup$ – Bary12 Apr 16 '18 at 18:34
  • $\begingroup$ One last thing, how did you get from $\sum_k | \langle x, \phi_k \rangle |^2 = \|x\|^2$ to $\sum_k | \langle x, \phi_k \rangle | \leq \|x\|$? $\endgroup$ – Bary12 Apr 16 '18 at 19:26
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Yes, it converges absolutely because $$ 2|\langle x,\varphi_k\rangle\langle \varphi_k,y\rangle| \le |\langle x,\varphi_k\rangle|^2+|\langle y,\varphi_k\rangle|^2, $$ and because the sums of the terms on the right converge by Bessel's inequality.

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