Proof involving norm of an integral

I know this is a very old question, but I thought it would be nice to have an answer using the approach suggested by the textbook mentioned in the original post.

Let $\mathbf{v} = (v_1, \cdots, v_n) = \int_a^b \mathbf{g}(x)\;d x$. Then by definition $v_j = \int_a^b g_j(x)\;d x$. If $\mathbf{v} = \mathbf{0}$, then we are done. Otherwise, we have \begin{align*} \|\mathbf{v}\|_2^2 &= \sum_{j = 1}^n v_j^2 = \sum_{j = 1}^n v_j \int_a^b g_j(x)\;d x = \int_a^b \sum_{j = 1}^n (v_j g_j(x))\;d x = \int_a^b \mathbf{v}\cdot \mathbf{g}(x)\;d x\\ &\leq \int_a^b \|\mathbf{v}\|_2\|\mathbf{g}(x)\|_2\;d x = \|\mathbf{v}\|_2\int_a^b \|\mathbf{g}(x)\|_2\;d x. \end{align*} where the inequality is by Cauchy-Schwarz. Divide by $\|\mathbf{v}\|_2$ and we are done.

$\rm\bf GUIDE:\quad$ Riemann integrals are defined with Riemann sums. The triangle inequality applies to, you guessed it, finite sums. Non-strict inequalities are preserved through taking limits.

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Alright, it seems you need more help to see how to apply all of this. The triangle inequality tells us

$$\left\|\sum_{i=1}^n g(x_i)\Delta x_i \right\| \le \sum_{i=1}^n \|g(x_i)\|\Delta x_i.$$

Now nostrict inequalities are preserved by limits, i.e. $a_n\le b_n\implies \lim\limits_{n\to\infty}a_n\le\lim\limits_{n\to\infty}b_n.$ If we take limits of both sides of the above, though, we end up with integrals and thus original formula!

$$\left\|\int_a^b g(x)dx\right\|\le \int_a^b \|g(x)\|dx.$$

QED.