Proof involving norm of an integral I am totally stuck and have no idea whatsoever on how to prove the following inequality (by the way this is a problem from an undergraduate book in multivariable advanced calculus at Junior/Senior level ):
Let $g=\left ( g_{1},g_{2},...,g_{n} \right ): \left [ a,b \right ]\rightarrow \mathbb{R}^{n}$ is a continuous function, then we define: $\int_{a}^{b}g\left ( x \right )dx=\left \langle \int_{a}^{b}g_{1}\left ( x \right )dx,...,\int_{a}^{b}g_{n}\left ( x \right ) \right \rangle$
Prove that: $\left \| \int_{a}^{b}g\left ( x \right )dx \right \|\leq \int_{a}^{b}\left \| g\left ( x \right ) \right \|dx$
In the book, there is a hint saying that I should use the Cauchy Schwarz inequality, but I have no clue how to use it. The only I was able to prove is:
Left hand side= $\sqrt{\left (\int_{a}^{b}g_{1}\left ( x \right )dx  \right )^{2}+...+\left ( \int_{a}^{b}g_{2}\left ( x \right )dx \right )^{2}}$
Right hand side is= $\int_{a}^{b}\sqrt{\left (g_{1}\left ( x \right )  \right )^{2}+...+\left ( g_{n}\left ( x \right ) \right )^{2}}dx$
I am looking forward for your suggestions and answers.
 A: 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.
A: $\rm\bf GUIDE:\quad$ 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.

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.
A: I'm not sure if this approach is suitable for a problem in an undergrad level, but because of the triangle inequality, any norm is a convex function; moreover, (a,b) is a bounded domain; hence by Jensen's inequality, one can easily prove the result.
