# 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= $\int_{a}^{b}\sqrt{\left (g_{1}\left ( x \right ) \right )^{2}+...+\left ( g_{n}\left ( x \right ) \right )^{2}}dx$

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$\rm\bf GUIDE:\quad$ 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.

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Can you please give a detailed answer please? What you wrote above is obvious and I don't know how this can used to solve the prblem. – M.Krov Jan 30 '12 at 0:17
@m_p2009: Write the formula you're supposed to prove but with Riemann sums in place of the integrals. The truth of this formula follows from CS. Now show that the limit of each side of this inequality are the sides of the original formula (with the actual integrals)... – anon Jan 30 '12 at 0:22
@m_p2009: Sorry, I meant the triangle inequality, the hint had me saying CS inequality even thought it's not what I was thinking of. – anon Jan 30 '12 at 0:54
@anaon: Can you prove your statement: $$$\left\|\sum_{i=1}^n g(x_i)\Delta x_i \right\| \le \sum_{i=1}^n \|g(x_i)\|\Delta x_i. because You said that the triangular inequality involves this:$$\left\|\sum_{i=1}^n g(x_i)\Delta x_i \right\| \le \sum_{i=1}^n \|g(x_i)\|\Delta x_i.$ – M.Krov Jan 30 '12 at 1:02
@m_p2009: You should be able to see how at your level. The triangle inequality tells us $\|a+b\|\le\|a\|+\|b\|$. By induction this generalizes to $$\|a_1+\cdots+a_n\|\le\|a_1\|+\cdots\|a_n\|.$$ We just apply this to the Riemann sum on the left, and use the fact that $\|\lambda a\|=\lambda\|a\|$ for $\lambda>0$. Also, can you please delete the comment with the bad LaTeX? – anon Jan 30 '12 at 1:12