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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.

3 Answers 3

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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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    @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?2012-01-30
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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.

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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.

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    I agree that the approach may not be appropriate for an undergraduate, but that doesn't mean that it can't be explained in a way that is more appropriate for an undergraduate. Could you perhaps flesh out your answer a little and provide a few details?2018-09-06