The following is a proof from Appendix C (Linear Spaces Review) of Introduction to Laplace Transforms and Fourier Series, Second Edition, by Phil Dyke:

enter image description here

enter image description here

With regards to the proof for property 4 (triangle inequality), I'm confused about the following:

  1. In the last section of the proof, the author goes from $\langle \mathbf{a}, \mathbf{b} \rangle + \overline{\langle \mathbf{a}, \mathbf{b} \rangle}$ to $2||\mathbf{a}|| ||\mathbf{b}||$. I'm assuming the author is using the preceding derivation of $|\langle \mathbf{a}, \mathbf{b} \rangle + \overline{\langle \mathbf{a}, \mathbf{b} \rangle}| \le 2||\mathbf{a}|| \cdot ||\mathbf{b}||$. However, I don't understand how this substitution is valid, since, in this derivation, the author uses $|\langle \mathbf{a}, \mathbf{b} \rangle + \overline{\langle \mathbf{a}, \mathbf{b} \rangle}|$, which is the absolute value, instead of just $\langle \mathbf{a}, \mathbf{b} \rangle + \overline{\langle \mathbf{a}, \mathbf{b} \rangle}$, as would be required. In fact, I'm not sure why the author uses the expression with absolute values instead of without, since we originally had the one without -- the absolute values were just thrown in with no justification.

  2. For the triangle inequality, we need to show that $||\mathbf{a} + \mathbf{b}|| \le ||\mathbf{a}|| + ||\mathbf{b}||$. But in the last section of the proof, I cannot see anything that resembles this. All we have is $||\mathbf{a} + \mathbf{b}||^2 \le ||\mathbf{a}||^2 + 2||\mathbf{a}||||\mathbf{b}|| + ||\mathbf{b}||^2$. I just don't see how this is proving the triangle inequality?

I would greatly appreciate it if people could please take the time to clarify these two points.

  • $\begingroup$ Your question is about $|\langle a,b\rangle |\leq \|a\| \|b\|$? $\endgroup$ – HK Lee Jul 2 '18 at 14:45

$(1)$: As the author noted, $\langle a, b\rangle + \overline{\langle a, b\rangle}$ is real. For every real number $r$ we have $r\leqslant |r|$.

$(2)$: You missed the power of two in the LHS. It's ${\lVert a + b\rVert}^2$.

  • $\begingroup$ Thanks for the response. I fixed it. $\endgroup$ – The Pointer Jul 2 '18 at 14:52
  • $\begingroup$ For (2), is there a typo in the proof? Should it say $(||\mathbf{a}|| + ||\mathbf{b}||)^2$ instead of $(||\mathbf{a} + \mathbf{b}||)^2$ $\endgroup$ – The Pointer Jul 2 '18 at 14:58
  • 2
    $\begingroup$ Yes, yes, that's right, it's a typo. $\endgroup$ – Berci Jul 2 '18 at 15:00
  • $\begingroup$ @Berci Ok, that makes sense then. I was confused because it was saying that $|| \mathbf{a} + \mathbf{b} ||^2 = (|| \mathbf{a} + \mathbf{b} ||)^2$, which is true but isn't the triangle inequality, haha. $\endgroup$ – The Pointer Jul 2 '18 at 15:02
  • 1
    $\begingroup$ Ooops, that's ddefinitely a typo. Missed it since the end result was already expected. $\endgroup$ – Fimpellizieri Jul 2 '18 at 15:04
  1. You are partly right, but we can use that $\alpha:=\langle a, b\rangle+\langle b, a\rangle$ as any real number satisfies $\alpha\le|\alpha|$.
    Note also that we want to use the Cauchy-Schwartz inequality $|\langle a, b\rangle|\le\|a\|\cdot\|b\|$, and that justifies introducing the absolute value sign.
  2. It's $\|a+b\|^{\bf 2}$ which is less or equal than the right hand side, which amounts to $(\|a\|+\|b\|)^2$.
  • $\begingroup$ For (2), is there a typo in the proof? Should it say $(||\mathbf{a}|| + ||\mathbf{b}||)^2$ instead of $(||\mathbf{a} + \mathbf{b}||)^2$ $\endgroup$ – The Pointer Jul 2 '18 at 14:57
  • $\begingroup$ I'm not sure I can follow.. We have $(\|a\|+\|b\|)^2=\|a\|^2+2\|a\|\cdot\|b\|+\|b\|^2\ \ge\ \|a+b\|^2$. $\endgroup$ – Berci Jul 2 '18 at 14:59
  • $\begingroup$ So just to confirm, (2) is a typo? It should be $(||\mathbf{a}|| + ||\mathbf{b}||)^2$ instead of $(||\mathbf{a} + \mathbf{b}||)^2$ in the last section of the proof? $\endgroup$ – The Pointer Jul 2 '18 at 15:03
  • $\begingroup$ Ok, typo confirmed. Thank you for the assistance. $\endgroup$ – The Pointer Jul 2 '18 at 15:05

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.