Is there anything wrong with the following:

If $\{f_n\}$, $\{g_n\}$ are two sequences of functions in a Hilbert space $H$, then

$$\begin{align*} \sqrt{\sum_n |\langle f,f_n \rangle|^2} - \sqrt{\sum_n |\langle f,f_n - g_n \rangle|^2}&\leq \sqrt{\sum_n |\langle f,g_n \rangle|^2}\\ &\leq \sqrt{\sum_n |\langle f,f_n \rangle|^2} + \sqrt{\sum_n |\langle f,f_n - g_n \rangle|^2} \end{align*}$$

for all $f\in H$. Thanks!


This works fine provided that you know all the sums are finite.

An easy way to see it is to let $a(n) = \langle f, f_n \rangle$, $b(n) = \langle f, f_n - g_n \rangle$. Then your inequalities read $$||a|| - ||b|| \le ||a-b|| \le ||a|| + ||b||$$ where $||\cdot||$ is the $\ell^2$ norm, and this is a simple consequence of the triangle inequality for the $\ell^2$ norm.

  • $\begingroup$ What if $||a||\leq ||b||$? $\endgroup$ – Cody Jan 12 '12 at 15:50
  • $\begingroup$ Is it should be $\big|||a||-||b||\big|\leq ||a-b|| \leq ||a||+||b||$! So in case of $||a||< ||b||$ we can change the places for $a$ and $b$ as follow: $$||b||-||a||\leq ||b-a||\leq ||b||+||a||$$. Please correct me if I'm wrong! $\endgroup$ – Cody Jan 12 '12 at 17:02
  • $\begingroup$ @Cody: Sure, you could say that. Or, observe that if $||a|| \le ||b||$ then $||a||-||b|| \le 0$, while $||a-b|| \ge 0$ by definition of the norm, so the inequality is trivially satisfied. $\endgroup$ – Nate Eldredge Jan 13 '12 at 2:56

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.