# Triangle inequality in a Hilbert space

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!

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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.
What if $||a||\leq ||b||$? – Cody Jan 12 '12 at 15:50
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! – Cody Jan 12 '12 at 17:02
@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. – Nate Eldredge Jan 13 '12 at 2:56