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I can't seem to find a way of asking a sub-question in relation to does linearity of inner product hold for infinite sum, which is in itself too generic a question for my purposes. Could someone please confirm or deny my understanding that the following:

$$\left\langle \sum_{n=1}^\infty x_n, y\right\rangle =\sum_{n=1}^\infty \langle x_n, y\rangle $$ is always true in a Hilbert space equipped with the norm induced by its inner product, where the convergence on the LHS is convergence in the said norm and the convergence on the RHS is convergence in the norm of the scalar field.

Thank you very much.

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    possible duplicate of [does linearity of inner product hold for infinite sum?](http://math.stackexchange.com/questions/24137/does-linearity-of-inner-product-hold-for-infinite-sum)2011-04-26
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    @Jonas Jonas, I have explicitly linked my question to the page you cite to underline the connection. However, the page in question does not answer my query.2011-04-26
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    Josef: Zev Chonoles' answer applies. Note that the inner product *is* continuous because $|\langle x,y \rangle| \leq \|x\|\,\|y\|$.2011-04-26
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    @Josef: I think it does answer the question. Hilbert spaces are topological vector spaces, so your query follows from the answer there by taking all the $y_n=y$2011-04-26
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    @Ross: Ah, I didn't notice this unfortunate typo in Zev's answer. I fixed it.2011-04-26
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    Thanks. I did have a good look at Zev's answer but it talks about 'any norm' on an inner product space (slightly confusing as I thought that, if in an inner product space, you would generally work with the norm induced by the inner product - hence my doubts).2011-04-26
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    @Josef: I'm sorry, I didn't check your link! I just remembered that Zev had answered a question like this, I found it, and it looked like an exact duplicate. Your question was more specific and clear, but the mathematical content is virtually identical and Zev's answer applies, which is why I voted to close. Now I understand that you are basically asking for clarification on the previous question. (When you get 50 points (very soon) you will be able to comment on all posts, if that's what you meant by "sub-question".)2011-04-26
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    @Jonas No worries, I appreciate the intent to maximise the use of existing explanations and keep things lean. I'm very cautious as I'm always worried that I'm neglecting something, so sometimes feel the need to chase an explicit confirmation of simple conclusions! Was reluctant to ask a new question so will probably delete it soon. Thank you.2011-04-26
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    @Josef: Whether or not your question ends up getting closed, there's no need to delete it. There is added value in keeping it because of the clarity and specificity of the question as well as the clarification in the comments of how Zev's answer applies, and it is easy to get between the two questions because they are linked. (Also, for what it's worth, if you delete it you'll lose the points from upvotes after a rep recalc.)2011-04-26
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    Thanks to whoever voted up this question! By pushing me over 50 it enables me to comment on existing questions directly rather than create similar new ones! So I guess it's praise by anticipation.2011-04-26

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There is an implicit question about the convergence of the sums, which this answer does not address.

If $\sum_{n=1}^\infty x_n$ converges in the norm of the Hilbert space $H$, then for any $y \in H$, $\sum_{n=1}^\infty \langle x_n, y \rangle$ converges in the base field ($\mathbb{R}$ or $\mathbb{C}$), and $$\langle \sum_{n=1}^\infty x_n , y \rangle = \sum_{n=1}^\infty \langle x_n, y \rangle.$$ As mentioned, this is because the inner product is linear and continuous with respect to the $H$ norm topology (essentially, by Cauchy-Schwarz).

However, if the sum on the right side converges for some $y \in H$, or even for all $y \in H$, it does not follow that $\sum_{n=1}^\infty x_n$ converges in $H$, so the left side may be meaningless. For an example, let $\{e_i\}_{i=1}^\infty$ be an orthonormal set in $H$ (assuming it is infinite dimensional). Let $x_1 = e_1$, and $x_n = e_n - e_{n-1}$ for $n \ge 2$. Then $\sum_{n=1}^\infty x_n = \lim_{i \to \infty} e_i$ does not converge in norm. However, $\sum_{n=1}^\infty \langle x_n, y \rangle = \lim_{i \to \infty} \langle e_i, y \rangle = 0$ for all $y$, by Bessel's inequality.

Effectively, having the right side converge for all $y$ only implies that $\sum_{n=1}^\infty x_n$ converges in the weak topology.

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    Thank you for this. In retrospect, just slapping an 'equals' sign there was not the best thing to do. In my mind I had the inner product space convergence as an assumption, but an unstated one.2011-04-26