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Exercise 3.6-4 in Kreyszig asks to show that $\langle x,y \rangle = \sum_k \langle x,e_k \rangle \overline{\langle y,e_k \rangle}$ using the "Parseval relation": $\sum_k |\langle x, e_k \rangle |^2 = ||x||^2$, for all $x \in X$, where the $(e_k)$ form an orthonormal set.

I'm a bit stumped here. I see how the relation on $||x||^2$ would follow from the first, but not the other way around.

Fequish
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Suppose you know Parseval's equality: $$ \|x\|^2 = \sum_{k}|\langle x,e_k\rangle|^2 $$ Then the following converges absolutely $$ [x,y] = \sum_{k} \langle x,e_k\rangle \langle e_k,y\rangle $$ And, $[x,y]$ has the properties of an inner product, with $[x,x]=\|x\|^2$. By the polarization identity, $[x,y]=\langle x,y\rangle$ must hold. For example, in the complex case, $$ [x,y] = \frac{1}{4}\sum_{n=0}^{3}i^{n}[x+i^n y,x+i^{n}y] = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|x+i^n y\|^2 = \langle x,y\rangle. $$ In the real case $$ [x,y] = \frac{1}{4}\left([x+y,x+y]-[x-y,x-y]\right) = \frac{1}{4}\left(\|x+y\|^2-\|x-y\|^2\right) = \langle x,y\rangle. $$

Disintegrating By Parts
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  • I'm still missing a few steps. I see that polarization and parallelogram imply $\langle x,y \rangle = \frac{1}{2} [ ||x||^2 + ||y||^2] = \frac{1}{2} [\sum_k |\langle x,e_k \rangle |^2 + \sum_j |\langle y,e_j \rangle |^2]$, but why is this equal to $\sum_k \langle x,e_k \rangle \overline{\langle y,e_j \rangle}$? – Fequish Feb 14 '16 at 20:05
  • @Fequish : $\langle x,y \rangle = \overline{\langle y,x\rangle}$ – Disintegrating By Parts Feb 14 '16 at 20:17
  • I still don't see it. Also, I forgot to mention my comment was for the real case. – Fequish Feb 14 '16 at 20:21
  • @Fequish : Stay simple: all you're showing is two inner products with the same norm are the same. You only have to verify that $[x,y]$ and $\langle x,y\rangle$ are two such inner products. They have the same associated norm by Parseval's equality. To prove this all you have to know is that given an inner product, you can use the norm to get back the original inner product. In the real case, $\langle x,y\rangle = \frac{1}{4}(|x+y|^2-|x-y|^2)$. – Disintegrating By Parts Feb 14 '16 at 20:24
  • Ah, thank you, I see it now. The only remaining part that I don't understand is what you mentioned about absolute convergence. I think before doing anything, you needed to justify that $\sum_k \langle x,e_k \rangle \langle e_k y \rangle$ converges, so that the new norm $[\cdot, \cdot]$ is well-defined. Right? I don't understand why Parseval allows you to conclude convergence. – Fequish Feb 14 '16 at 20:55
  • @Fequish $2|ab| \le |a|^2+|b|^2$ holds for any real or complex $a$, $b$. – Disintegrating By Parts Feb 14 '16 at 22:32