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

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

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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.

Best Answer

$(1)$: 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 ${\lVert a + b\rVert}^2$.