# Showing that $\sqrt{\langle \mathbf{a}, \mathbf{a} \rangle}$ satisfies triangle inequality.

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

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.

• Your question is about $|\langle a,b\rangle |\leq \|a\| \|b\|$? – 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 ${\lVert a + b\rVert}^2$.

• Thanks for the response. I fixed it. – The Pointer Jul 2 '18 at 14:52
• For (2), is there a typo in the proof? Should it say $(||\mathbf{a}|| + ||\mathbf{b}||)^2$ instead of $(||\mathbf{a} + \mathbf{b}||)^2$ – The Pointer Jul 2 '18 at 14:58
• Yes, yes, that's right, it's a typo. – Berci Jul 2 '18 at 15:00
• @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. – The Pointer Jul 2 '18 at 15:02
• Ooops, that's ddefinitely a typo. Missed it since the end result was already expected. – 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 $(\|a\|+\|b\|)^2$.
• For (2), is there a typo in the proof? Should it say $(||\mathbf{a}|| + ||\mathbf{b}||)^2$ instead of $(||\mathbf{a} + \mathbf{b}||)^2$ – The Pointer Jul 2 '18 at 14:57
• I'm not sure I can follow.. We have $(\|a\|+\|b\|)^2=\|a\|^2+2\|a\|\cdot\|b\|+\|b\|^2\ \ge\ \|a+b\|^2$. – Berci Jul 2 '18 at 14:59
• 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? – The Pointer Jul 2 '18 at 15:03
• Ok, typo confirmed. Thank you for the assistance. – The Pointer Jul 2 '18 at 15:05