Proving $\left\| \int_a^b f(x) \,dx\right\| \leq \int_a^b \| f(x) \|\, dx,$ for $f = (f_1,f_2,\dots,f_n):[a,b] \to \mathbb{R}^n$

Question

Let $v = (v_1, \dots, v_n)$ and $\| v \| = (\sum | v_i |^2)^{1/2}.$ I would like to show that $$ \left\| \int_a^b f(x)\, dx\right\| \leq \int_a^b \| f(x) \| \, dx, $$ where $f = (f_1,f_2,\dots,f_n):[a,b] \to \mathbb{R}^n,$ and $$ \int_a^b f(x) \, dx = \left(\int_a^b f_1(x) \,dx,\; \dots, \int_a^b f_n(x) \, dx \right) $$

My method is below, but I hide it in order to not "pollute" any potential ideas/ answers others may have. Thanks in advance!

A sketch is as follows: Let $y = \int f(x)\, dx, \; y_i = \int f_i(x)\, dx,$ for $i = 1,\dots, n.$ All integrals will be from $a$ to $b.$
$\|y\|^2 = \sum y_i^2 = \sum y_i \int f_i\, dx = \int \left(\sum y_if_i\right) dx$ Recognizing that $\sum y_i f_i \leq \|y\|\|f\|,$
$\|y\|^2 =\int \left(\sum y_if_i\right) dx \leq \int \|y\| \|f\| \, dx = \|y\| \int \|f\|\, dx$ So, $\|y\| \leq \int \|f\| \, dx.$ This argument seems pretty sketchy to me... hence why I'm interested in seeing other's thoughts on how to go about showing this inequality.

Answer 1

I don't know whether you have worked out the proof using Riemann sums, but I guess that method would be hopelessly complicated. Instead, I will illustrate your proof using specific examples to make it less "sketchy".

We look at $n=3$. Let $f=(f_1,f_2,f_3)$, and let $v=\left(\int_a^bf_1,\int_a^bf_2,\int_a^bf_3\right)$. We have $$\begin{split}\|v\|^2&=\left(\int_a^bf_1\right)^2+\left(\int_a^bf_2\right)^2+\left(\int_a^bf_3\right)^2\\ &=\left(\int_a^bf_1\right)\left(\int_a^bf_1\right)+\left(\int_a^bf_2\right)\left(\int_a^bf_2\right)+\left(\int_a^bf_3\right)\left(\int_a^bf_3\right).\end{split}$$

Now since each $\left(\int_a^bf_i\right)$ is a number, we can move it inside the integral sign, so the above equation becomes $$\int_a^b\left(\int_a^bf_1\right)f_1+\int_a^b\left(\int_a^bf_2\right)f_2+\int_a^b\left(\int_a^bf_3\right)f_3.$$ Since now we have addition of three integrals all from $a$ to $b$, we can sum over all the integrands and put them under one integral sign, so the above becomes $$\int_a^b\left[\left(\int_a^bf_1\right)f_1+\left(\int_a^bf_2\right)f_2+\left(\int_a^bf_3\right)f_3\right].$$ Inside the bracket is the inner product of $v=\left(\int_a^bf_1,\int_a^bf_2,\int_a^bf_3\right)$ and $f=(f_1,f_2,f_3)$, so that we can write it compactly as $$\int_a^{b}v\cdot f.$$ Now, $v\cdot f$ is a function from $[a,b]$ to $\mathbb{R}$, as you can easily recognize from the bracket, so we have from calculus

$$\int_a^{b}v\cdot f\,\leq\int_a^b|v\cdot f|.$$

And this is the place where you need the Riemann sum definition. Our proof is built on the similar case for real-valued function! We don't have to run through the trouble again. Now, according to Cauchy-Schwartz inequality, $|v\cdot f|\leq\|v\|\|f\|$, so, again by elementary properties of Riemann integral for real-valued functions, $$\begin{split}\|v\|^2\leq\int_a^{b}|v\cdot f|\,&\leq\int_a^b\|v\|\|f\| \\ &=\|v\|\int_a^b\|f\|.\end{split}$$

Dividing $\|v\|$ we now get $$\|v\|\leq\int_a^b\|f\|.$$

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Summarizing the proof, we have used three elementary facts:

1. If two real-valued functions $f,g$ are Riemann integrable on $[a,b]$, and $f\leq g$ on $[a,b]$, then $\displaystyle\int_a^bf\leq\int_a^bg$.
2. If $f$ is Riemann integrable, then $|f|$ is Riemann integrable, and $\displaystyle\int_a^bf\leq\int_a^b|f|$.
3. Cauchy-Schwartz inequality: for $v,w\in\mathbb{R}^n$, we have $\displaystyle|v\cdot w|\leq\|v\|\|w\|$.

Proofs of the first two are directly from the definition, namely the Riemann sum. This is where we use it. Since now we are dealing with vector-valued functions, the vector space structure should come into play. Thus there should be no surprise, on a second thought, that inner product and the Cauchy-Schwartz inequality pop up in our proof. And this also demonstrates the power of linear algebra in analysis. Using Riemann sum of the whole function $f=(f_1,f_2,\ldots,f_n)$ to prove the inequality would then ignore the vector space structure, which is not a wise choice, I fear.

Answer 2

I found one approach with induction and Cauchy-Schwarz :

For $n=1$, it is easy. For $n=2$, by Cauchy-Schwarz inequality, \begin{align} \left(\int_{a}^{b} f_{1}(x)dx\right)^{2}&\leq \left(\int_{a}^{b}\left( \sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}+f_{2}(x)\right)dx\right)\left(\int_{a}^{b}\frac{f_{1}(x)^{2}}{\sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}+f_{2}(x)}\right)dx\\ &=\left(\int_{a}^{b}\left( \sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}+f_{2}(x)\right)dx\right)\left(\int_{a}^{b}\left( \sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}-f_{2}(x)\right)dx\right) \\ &=\left(\int_{a}^{b} \sqrt{f_{1}(x)^{2}+f_{2}(x)^{2}}dx\right)^{2}-\left(\int_{a}^{b}f_{2}(x)\right)^{2} \end{align}

Suppose it hold for some $n$. Then for $n+1$, \begin{align} \sum_{k=1}^{n}\left(\int_{a}^{b}f_{k}(x)dx\right)^{2}&\leq \left(\int_{a}^{b} \sqrt{\sum_{k=1}^{n}f_{k}(x)^{2}}dx\right)^{2} \\ &\leq \left(\int_{a}^{b} \sqrt{\sum_{k=1}^{n+1}f_{k}(x)^{2}}dx\right)^{2}-\left(\int_{a}^{b}f_{n+1}(x)dx\right)^{2} \end{align}