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.

  • $\begingroup$ possible duplicate of does linearity of inner product hold for infinite sum? $\endgroup$ – Jonas Meyer Apr 26 '11 at 1:55
  • $\begingroup$ @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. $\endgroup$ – Josef K. Apr 26 '11 at 1:58
  • $\begingroup$ Josef: Zev Chonoles' answer applies. Note that the inner product is continuous because $|\langle x,y \rangle| \leq \|x\|\,\|y\|$. $\endgroup$ – t.b. Apr 26 '11 at 2:03
  • $\begingroup$ @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 $y_n=y$ $\endgroup$ – Ross Millikan Apr 26 '11 at 2:05
  • $\begingroup$ @Ross: Ah, I didn't notice this unfortunate typo in Zev's answer. I fixed it. $\endgroup$ – t.b. Apr 26 '11 at 2:07

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.

  • $\begingroup$ 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. $\endgroup$ – Josef K. Apr 26 '11 at 3:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.