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Let $V$ be an inner product space and let $u$, $v$ $\in$ $V$ . Prove that $\|u+v\|=\|u\|+\|v\|$ iff either $u=\alpha v$ or $v=\alpha u$ for some $\alpha\in\Bbb C$.

I got stuck with $\|u+v\|=\|u\|+\|v\|$ if either $u=\alpha v$ or $v=\alpha u$ for some $\alpha\in\Bbb C$.

I opened $\|u+v\|=\|u\|+\|v\|$ and got: $$\langle u,v\rangle +\overline{\langle u,v\rangle}=\langle u,v\rangle +\langle v,u\rangle =2\sqrt{\langle u,u\rangle\langle v,v\rangle}=2\|u\|\|v\|$$

But I don't see how to continue.

Noa Even
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Mr.O
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3 Answers3

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The statement is not quite correct: if $\|u+v\| = \|u\|+\|v\|$ then $\dim(\text{span}_{\mathbb C}\{u,v\}) = 1$, but this is not sufficient, one needs $v\in\mathbb R_{\geq 0}.u$ and $u \in \mathbb R_{\geq 0}.v$. The most trivial case where $u \in \mathbb C.v$ but $\|u+v\| \neq \|u\|+ \|v\|$ is where $v =-u$.

If either of $u$ or $v$ is zero, the statement is trivially true. Thus we assume that both are nonzero, in which case the condition that $u\in \mathbb C.v$ is equivalent to the condition $v \in \mathbb C$. Now suppose that $\|u+v\| = \|u\|+\|v\|$. Squaring and expanding out, using the fact that $\|u\|^2= \langle u,u\rangle$, we find:

$$ 2\|u\|\|v\| = 2 \langle u,v \rangle. $$

Now by the case of equality in the Cauchy-Schwarz inequality it follows that $u =\mathbb C.v$. If $u =\lambda.v$ where $u\in \mathbb C^\times$, then $\|\lambda u+ u\| =\|(\lambda+1)u\| = |\lambda+1|\|u\|$ and $\|u\|+\|v\| = (|\lambda|+1)\|u\|$. Thus we must have $|\lambda+1| = |\lambda|+1$, which (by Cauchy-Schwarz for $\mathbb C =\mathbb R^2$) is the case if and only if $\lambda \in \mathbb R_{>0}$ (since we are assuming $u \neq 0$).

krm2233
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I don't follow how you got to that line of reasoning. But, what you'll want to do is follow the proof of triangle inequality (which is usually proven by Cauchy-Schwarz inequality), and trace the consequences when you put equality instead of an inequality.

You should find that, if $\|u + v\| = \|u\| + \|v\|$, then $\langle u, v \rangle = \|u\| \|v\|$. From this, you can follow the proof here, to show that $v = 0$ or $u = \frac{\langle u, v \rangle}{\|v\|^2}v$.

Theo Bendit
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  • can you explain how you got $\langle u, v \rangle = |u| |v|$? – Mr.O Oct 27 '17 at 06:34
  • Expand $|u + v|^2 = (|u| + |v|)^2$. The right side is a standard expansion. For the left side, you can write $|u + v|^2 = \langle u + v, u + v \rangle$ and expand. Once you cancel $|u|^2 + |v|^2$ from both sides, you get $\langle u, v \rangle + \langle v, u \rangle = 2|u||v|$. In the real case, we're done. In the complex case, we get $$2|u||v| = 2 \Re \langle u, v \rangle \le 2|\langle u, v \rangle| \le 2|u||v|.$$ – Theo Bendit Oct 27 '17 at 13:08
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If $x=0$ or $y=0$ then we have nothing to prove.We assume that x and y are not zero.Then: $||x+y||=||x||+||y||$<=>$(||x+y||)^2=(||x||+||y||)^2$(norm is non-negative)<=>$<x,y>=||x||||y||$.Now we have <x/||x||,y/||y||>=1<==>x/||x||=y/||y|| (**)

(**)Theorem:If you have (H,<.>) inner product space and $||x||=||y||=1$(norm is given from inner product) then $<x,y>=1<=>x=y$

'<==' this direction is obvious

'==>' $||x-y||^2=<x-y,x-y>=||x||^2-2<x.y>+||y||^2=1-2+1=0->x=y$

George Giatilis
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