5
$\begingroup$

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

7

$\rm\bf GUIDE:\quad$ 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.

  • 0
    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.2012-01-30
  • 0
    @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)...2012-01-30
  • 0
    @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.2012-01-30
  • 0
    @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.$2012-01-30
  • 0
    @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
5

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.

0

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.

  • 0
    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