# Show that $\left(\int_{0}^{1}\sqrt{f(x)^2+g(x)^2}\ dx\right)^2 \geq \left(\int_{0}^{1} f(x)\ dx\right)^2 + \left(\int_{0}^{1} g(x)\ dx\right)^2$ [closed]

Show that $$\left( \int_{0}^{1} \sqrt{f(x)^2+g(x)^2}\ \text{d}x \right)^2 \geq \left( \int_{0}^{1} f(x)\ \text{d}x\right)^2 + \left( \int_{0}^{1} g(x)\ \text{d}x \right)^2$$ where $f$ and $g$ are integrable functions on $\mathbb{R}$.

That inequality is a particular case. I want to approximate the integral curves using some inequalities who imply this inequality.

• Please provide some context – robjohn Jan 5 '16 at 10:38
• This inequality is a particular case.I want to use this inequality to aproximate the integral curves. – alexb Jan 5 '16 at 10:50
• clarifications and improvements should be applied to the question, not left in comments – robjohn Jan 5 '16 at 11:05
• I think there other useful answers to this question and this inequality is useful by itself. I would like to see it reopened. – robjohn Jan 6 '16 at 16:00
• Do you mean that you want to use inequalities implied by this inequality? – robjohn Jan 6 '16 at 16:48

Let a curve $C\in\mathbb{R}^2$ be defined by the parametrisation $\displaystyle x(t)=x(0)+\int_0^t f(x)dx$ and $\displaystyle y(t)=y(0)+\int_0^t g(x)dx$, $t\in [0, 1]$. Then the LHS is the square of the arc length of $C$ joining $(x(0), y(0))$ and $(x(1), y(1))$, whereas the RHS is the square of the shortest distance between $(x(0), y(0))$ and $(x(1), y(1))$.

• But if I have n functions I can to proof the inequality în the same way? – alexb Jan 5 '16 at 13:00
• Yes. You can generalise to an $n$-dimensional curve. – Alex Fok Jan 5 '16 at 13:02

Definition: $u:\mathbb{R}^n\to\mathbb{R}$ is convex if for all $a\in\mathbb{R}^n$, there is a $v(a)\in\mathbb{R}^n$ so that for all $x\in\mathbb{R}^n$ $$u(x)-u(a)\ge v(a)\cdot(x-a)\tag{1}$$ Theorem (Extension of Jensen): If $u:\mathbb{R}^n\to\mathbb{R}$ is convex, $f:\Omega\to\mathbb{R}^n$, and $\int_\Omega\mathrm{d}\omega=1$, then $$\int_\Omega u(f)\,\mathrm{d}\omega\ge u\left(\int_\Omega f\,\mathrm{d}\omega\right)\tag{2}$$ Proof: Let $a=\int_\Omega f\,\mathrm{d}\omega$. Then $(1)$ becomes $$u(f)-u\left(\int_\Omega f\,\mathrm{d}\omega\right) \ge v\left(\int_\Omega f\,\mathrm{d}\omega\right) \cdot\left(f-\int_\Omega f\,\mathrm{d}\omega\right)\tag{3}\\$$ Since $\int_\Omega\mathrm{d}\omega=1$, integrating $(3)$ over $\Omega$ gives \begin{align} \int_\Omega u(f)\,\mathrm{d}\omega-u\left(\int_\Omega f\,\mathrm{d}\omega\right) &\ge v\left(\int_\Omega f\,\mathrm{d}\omega\right) \cdot\left(\int_\Omega f\,\mathrm{d}\omega-\int_\Omega f\,\mathrm{d}\omega\right)\\[3pt] &=0\tag{4} \end{align} QED

Claim: $u(x)=\left\|x\right\|$ is convex.

Proof: \begin{align} \left\|a\right\|\left\|x\right\|&\ge a\cdot x\tag{5}\\[6pt] \left\|x\right\|&\ge\frac{a}{\left\|a\right\|}\cdot x\tag{6}\\ \left\|x\right\|-\left\|a\right\|&\ge\frac{a}{\left\|a\right\|}\cdot(x-a)\tag{7} \end{align} Explanation:
$(5)$: Cauchy-Schwarz
$(6)$: divide both sides by $\left\|a\right\|$
$(7)$: subtract $\left\|a\right\|$ from both sides

QED

The theorem and the claim prove that $$\int_0^1\left\|f(x)\right\|\mathrm{d}x\ge\left\|\int_0^1f(x)\,\mathrm{d}x\right\|\tag{8}$$ which, in $\mathbb{R}^2$, is the inequality in the question.

Suppose that $f$ and $g$ are continuous functions. Define $$\phi(t) = \left(\int_0^t \sqrt{f(s)^2 + g(s)^2} ds\right)^2 -\left(\int_0^t f(s) ds\right)^2 - \left(\int_0^t g(s) ds\right)^2.$$ It is obvious that $\phi(0) =0$ and $$\phi'(t) = 2\left[\int_0^t \sqrt{f(t)^2 + g(t)^2}\sqrt{f(s)^2 + g(s)^2}ds - \int_0^t (f(s)f(t)+g(s) g(t))ds\right].$$ By Cauchy-Schwartz inequality, we have $$\sqrt{f(t)^2 + g(t)^2}\sqrt{f(s)^2 + g(s)^2} \geq f(t)f(s) + g(t) g(s).$$ Hence $\phi'(t) \geq 0$. This implies that $\phi(1) \geq \phi(0) =0$. This is the inequality in question.

In general case when $f$ and $g$ are integrable, we can approximate them by continous functions, and hence finish the proof.