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Hi: I'm reading some introductory notes on hilbert spaces and there is a step in a proof that I don't follow. I will put the exact statement below. If someone could explain how it is obtained, it's appreciated. Note that commans between two terms when they have < and > around them denotes the innner product. Also, $e_{n}$ for $n = 1,2,3,\ldots$ is a complete orthonormal sequence in a Hilbert space $H$ and $x$ is in $H$.

Proof: Observe that

\begin{eqnarray*} 0 <= || x - \sum_{n=1}^{m} <x,e_{n}>e_{n}) ||^2 & = & \left< x - \sum_{n=1}^{m} <x,e_{n}>e_{n}, x - \sum_{n=1}^{m} <x,e_{n}>e_{n} \right> \\ & = & \left< x, x - \sum_{n=1}^{m} <x,e_{n}>e_{n} \right> - \sum_{n=1}^{m} <x, e_{n}> \left < e_{n}, x - \sum_{n=1}^{m} <x,e_{n}>e_{n} \right > \\ & = & ||x||^2 - \sum_{n=1}^{m} |<x, e_{n}>|^2 \end{eqnarray*}

I understand the first two lines of above. My question is how one goes from the second to the last line to the last line. Thanks for your help.

mark leeds
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2 Answers2

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Note that the first inner product in the second-last line is: $$ \left\langle x, x - \sum_{n=1}^m \langle x,e_n\rangle e_n\right\rangle = \langle x,x\rangle - \sum_{n=1}^m \langle x,e_n\rangle \langle x,e_n\rangle = \|x\|^2 - \sum_{n=1}^m |\langle x,e_n\rangle|^2 $$ so we would hope that the second is equal to zero. We have $$ \sum_{n=1}^m \langle x,e_n\rangle\left\langle e_n, x - \sum_{n=1}^m \langle x,e_n\rangle e_n\right\rangle = \sum_{n=1}^m \langle x,e_n\rangle\langle e_n,x\rangle - \sum_{n=1}^m \langle x,e_n\rangle^2 \sum_{k=1}^m \langle e_n,e_k\rangle $$ Luckily, $\langle e_n,e_k\rangle = 1$ if $n=k$ and $0$ otherwise, so the two terms in the difference are equal.

ptrsinclair
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For clarity's sake, I've rewritten this second to last line as

$$\left< x, x - \sum_{n=1}^{m}\left[ <x,e_{n}>e_{n}\right] \right> - \sum_{n=1}^{m}\left[ <x, e_{n}> \left < e_{n}, x - \sum_{i=1}^{m} <x,e_{i}>e_{i} \right >\right].$$ By linearity of the inner-product, $$\left< x, x - \sum_{n=1}^{m}\left[ <x,e_{n}>e_{n}\right] \right> =||x||^2-\sum_{n=1}^m \left[<x,e_n>^2\right].$$ Since $<e_n,e_m>=0$ whenever $m \neq n$ and $1$ otherwise, whenever $1 \leq n\leq m$, we find that $$\left < e_{n}, x - \sum_{i=1}^{m} <x,e_{i}>e_{i} \right >=<e_n,x>-<x,e_n>=0,$$ since inner-products are symmetric. Hence the second term in the original expression is identically zero. Hope that helps! Let me know if anything here is unclear.

Archaick
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  • I will print out, read carefully, make sure I understand it and then check it. Thank you very much for your thorough explanation. It's really appreciated. – mark leeds Aug 01 '15 at 14:18
  • both of you have answered so what am I supposed to do if I understand both explanations. which one do I check ? thanks. – mark leeds Aug 01 '15 at 14:20
  • @markleeds Don't mention it, glad to help. You can go ahead and give ptrsinclair the correct answer check, it doesn't matter to me. – Archaick Aug 01 '15 at 21:20
  • I followed both of the explanations. Thanks again to both of you. – mark leeds Aug 02 '15 at 04:30