Let $\langle\cdot,\cdot\rangle$ be a scalar product on a space $X$, and let $\lVert \cdot\rVert$ denote the norm induced by this scalar product. I need to show that for $x,y\in X$, $\lVert x+y\rVert=\lVert x\rVert+\lVert y\rVert $ holds $\Longleftrightarrow$ $x$ and $y$ are linear transformations of each other, i.e. $\exists \alpha\geq 0$ with $x = \alpha y$. While one direction is clear ($\Longleftarrow$) I am struggling with $\Longrightarrow$. How can I construct such an $\alpha$?

  • $\begingroup$ I think you actually meant "..$\,x\,,\,y\,$ are linearly dependent, or a scalar multiple of each other. $\endgroup$ – DonAntonio Jun 23 '13 at 12:33

Do you know that the Cauchy-Schwarz inequality $$\lvert \langle x, y \rangle \rvert \leq \lVert x \rVert \lVert y \rVert$$ is an equality if and only if $x = \alpha y$ for some $\alpha$? (If not, here's a proof.)

With that, you can prove the other direction by noting that $$\lVert x \rVert^2 +\lVert y \rVert^2 + 2\lVert x \rVert \lVert y \rVert = (\lVert x \rVert + \lVert y \rVert)^2 = \lVert x + y \rVert^2 =\langle x+y , x+y \rangle \\ = \langle x , x \rangle + \langle y , y \rangle + 2\langle x, y \rangle = \lVert x \rVert^2 + \lVert y \rVert^2 + 2 \langle x , y \rangle.$$ This implies that $\langle x , y \rangle = \lVert x \rVert \lVert y \rVert$, which by the first claim means that $x = \alpha y$ for some $\alpha$.

  • $\begingroup$ I did not know that. Thanks. $\endgroup$ – smi Jun 23 '13 at 11:54
  • $\begingroup$ @smi: You should accept and upvote an answer which answers your question. Simply click on the check on the left (as well as on the arrow pointing upwards). $\endgroup$ – Dominik Jun 23 '13 at 13:23
  • $\begingroup$ I'm working the same problem and just had to delete a similar thread of my own. Thanks for your help, fuglede. However, I have a problem with the equality: $$\langle x+y , x+y \rangle \\ = \langle x , x \rangle + \langle y , y \rangle + 2\langle x, y \rangle.$$ We just have $$\langle x+y , x+y \rangle \\ = \langle x , x \rangle + \langle y , y \rangle + \langle x, y \rangle + \langle y, x \rangle \\ = \langle x , x \rangle + \langle y , y \rangle + \langle x, y \rangle + \overline{\langle x, y \rangle},$$ don't we? So an additional argument is required. $\endgroup$ – Amarus Jun 25 '13 at 17:40
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    $\begingroup$ @Amarus: I figured that OP was working over a real vector space, so I did not include the extra argument. In your case $$2\lVert x \rVert \lVert y \rVert = \langle x , y \rangle + \overline{\langle x , y \rangle} = 2\mathrm{Re}\langle x , y \rangle.$$ Since also $$\mathrm{Re} \langle x, y \rangle \leq \lvert \langle x , y \rangle \rvert \leq \lVert x \rVert \lVert y \rVert,$$ the first line implies that both of these inequalities are actually equalities. This means that also in this case $\lVert x \rVert \lVert y \rVert = \lvert \langle x , y \rangle \rvert$, which proves the claim. $\endgroup$ – fuglede Jun 25 '13 at 17:53

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