# Proving that $\|x\|_2 \geq \|x\|_1$?

How would you prove that $\|x\|_2 \geq \|x\|_1$, or in other words that $$\sqrt{\int_0^1|f(x)|^2 dx} \geq \int_0^1|f(x)|dx \quad?$$

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Dou you know Cauchy-Schwarz's inequality? –  Julián Aguirre Apr 7 '12 at 9:39
That $\left(\sum_{i=1}^n a_i b_i \right)^2 = \left(\sum_{i=1}^n a_i^2\right) \left(\sum_{i=1}^n b_i^2\right)$? –  user26069 Apr 7 '12 at 9:47
$\left(\sum_{i=1}^n a_i b_i \right)^2 = \left(\sum_{i=1}^n a_i^2\right) \left(\sum_{i=1}^n b_i^2\right)$ is Cauchy's inequality ... Schwarz has a generalization that applies to integrals, and even to abstract inner products. –  GEdgar Apr 7 '12 at 13:00
The integral form (see robjohn's comment to anon's answer) –  Julián Aguirre Apr 7 '12 at 20:29

Answer: I would 1. note that $\displaystyle\iint g(x,y)\mathrm dx\mathrm dy\geqslant0$, where $g(x,y)=(f(x)-f(y))^2$, because $g\geqslant0$ everywhere, 2. expand the square defining $g(x,y)$, 3. use the linearity of the integral, and 4. conclude.

Additionally, this method would give me the equality case in the inequality, namely that $f$ equals almost everywhere a constant.

And of course, of course, I would refer you to this gem of a book called The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele.

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Cauchy-Schwarz inequality applied to Riemann sums, taken in the limit$^\dagger$:

$$\left(\sum_{i=1}^n \frac{1}{n}|f(x_i)|\right)^2\le \left(\sum_{j=1}^n \frac{1}{n^2}\right)\left(\sum_{k=1}^n |f(x_k)|^2\right)$$

$^\dagger$Remark: If $a_n\le b_n$ for all $n\ge 1$ and the limits exist, we have $\lim\limits_{n\to\infty}a_n\le\lim\limits_{n\to\infty}b_n$.

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Or simply use the integral form of Cauchy-Schwarz: $$\int_0^1|f(x)|\cdot1\;\mathrm{d}x\le\left(\int_0^1|f(x)|^2\; \mathrm{d}x\right)^{\frac12}\left(\int_0^11^2\;\mathrm{d}x\right)^{\frac12}$$ –  robjohn Apr 7 '12 at 12:55