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For a norm $\|\cdot\|$ by triangle inequality we have $\|a+b\| \le \|a\| + \|b\|$ and by Cauchy–Schwarz inequality $\|ab\| \le \|a\|\|b\|$, but I am not sure if the following inequality holds always true: $\|a+b\|^2 \le \|a\|^2 + \|b\|^2$.

Thanks for the answers!

Then, let's say $\|a+b\|^2 \lesssim \|a\|^2 + \|b\|^2$, is that true?

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    $\begingroup$ Take $a=b$ non zero. $\endgroup$
    – Kelenner
    Apr 19 '18 at 13:57
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    $\begingroup$ Since $\|a+b\|^2=\|a\|^2+2a\cdot b\|b\|^2$, that inequality hinges on if $a\cdot b$ is positive or negative. $\endgroup$
    – anon
    Apr 19 '18 at 13:58
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    $\begingroup$ $\| a+b \|^2\leq 2(\|a\|^2+\|b\|^2)$ $\endgroup$ Apr 19 '18 at 14:19
  • $\begingroup$ Thanks! I think I should use "$\lesssim$" instead of "$\le$" as in the textbook did $\endgroup$ Apr 19 '18 at 14:31
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If we have an inner product $\langle . | . \rangle$, then: $$ ||a+b||^2 = \langle a+b | a+b \rangle = \langle a | a \rangle +2\langle a | b \rangle + \langle b | b \rangle = ||a||^2+||b||^2+2\langle a | b \rangle $$ Thus your assertion does not hold for say: $a=\alpha b$ with $\alpha>0$.

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  • $\begingroup$ So $\|a+b\|^2 \le 2(\|a\|^2 + \|b\|^2)$ is true? $\endgroup$ Apr 19 '18 at 14:25
  • $\begingroup$ Yes because then you would have $2 \langle a|b \rangle \leq \langle a|a \rangle + \langle b|b \rangle$ i.e. $0\leq \langle b-a|b-a \rangle = ||b-a||^2$ which holds $\endgroup$ Apr 19 '18 at 14:38
  • $\begingroup$ Thank you very much! By the way, can we say $\|ab\|\le\|a\|\|b\|$ holds by Cauchy–Schwarz inequality? I saw Cauchy–Schwarz inequality in textbook is always an inner product form $|<a,b>|$ on LHS. $\endgroup$ Apr 19 '18 at 16:10
  • $\begingroup$ No it is not possible, that would be a normed algebra. $|\langle a,b \rangle|$ and $||ab||$ are nothing alike and actually, in a vector space, $ab$ makes no sense at all since there is no multiplication in vector spaces. $\endgroup$ Apr 19 '18 at 16:14
  • $\begingroup$ Well. Assume $a$, $b$ are two functions integrable over the domain $\Omega$ for the norm, then can we say $\|a b\|_{\Omega} \le \|a\|_{\Omega} \|b\|_{\Omega}$, where $ab$ means the product of the two functions? $\endgroup$ Apr 20 '18 at 11:48
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No, take $a=b=1$. Then you get $\|a+b\|^2=4$ and $\|a\|^2=\|b\|^2=1$, so $4\le 1+1$ does not hold.

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If $a=b\neq 0$ then $||a+b\|^2=4\|a\|^2\neq \|a\|^2+\|a\|^2=2\|a\|^2$.

But you could prove that $x\to \|x\|^2$ is a convex function, i.e. $\|sa+tb\|^2\leq s\|a\|^2+t\|b\|^2$ whenever $s+t=1$ and $ s,t\geq 0$.

So in particular if $s=t=\frac 12$ then $\|a+b\|^2\leq 2(\|a\|^2+\|b\|^2)$

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