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
  • 1
    $\begingroup$ @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

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.