Proving $\int_a^b\sqrt{f^2+g^2}\ge\sqrt{(\int_a^b f)^2 + (\int_a^b g)^2 } $ without using $|\int_a^b z(t)\,dt|\le\int_a^b |z(t)|\,dt$

Question

I have tried to prove this inequality $$ \int_a^b \sqrt{f^2+g^2} \ge \sqrt{ \left( \int_a^b f \right)^2 + \left( \int_a^b g \right)^2 } $$ without the use the following theorem in complex analysis $$\left| \int_a^b z(t)\,dt \right| \le \int_a^b |z(t)|\,dt$$

I failed. Can You help me with it?

Answer 1

This is quite easy. Let $c,d \in \mathbb R$ with $c^{2}+d^{2}=1$. Then $c\int_a^{b}f +d\int_a^{b} g=\int_a^{b} (cf+dg)\leq \int_a^{b} \sqrt {f^{2}+g^{2}}$ by C-S inequality. Now just take $c=\frac {\int_a^{b}f} {\sqrt {(\int_a^{b}f)^{2}+(\int_a^{b}g)^{2}}}$ and $d=\frac {\int_a^{b}g} {\sqrt {(\int_a^{b}f)^{2}+(\int_a^{b}g)^{2}}}$

Answer 2

Suppose $z(r)=f(r)+ig(r)$, where $r$ is a real variable and $f$ and $g$ are real valued functions. Then:

$(f(s)g(t)-f(t)g(s))^2\geq0 \quad\;\forall\;\;s,t\in\mathbb R \implies f(s)^2g(t)^2+f(t)^2g(s)^2\geq2f(t)g(t)f(s)g(s)\\\implies f(s)^2g(t)^2+f(s)^2f(t)^2+f(t)^2g(s)^2+g(t)^2g(s)^2\geq f(s)^2f(t)^2+2f(t)g(t)f(s)g(s)+g(t)^2g(s)^2\\\implies(f(s)^2+g(s)^2)(f(t)^2+g(t)^2)\geq(f(s)f(t)+g(t)g(s))^2\\\implies\sqrt{(f(s)^2+g(s)^2)}\sqrt{(f(t)^2+g(t)^2)}\geq f(s)f(t)+g(t)g(s)$

Now, we'll need to use a generalized form of inequalities between functions for two variable functions. That'll give us (I'm going to assume $b>a$ in the following):

$$\int_{s=a}^{s=b}\int_{t=a}^{t=b}\sqrt{(f(s)^2+g(s)^2)}\sqrt{(f(t)^2+g(t)^2)}\quad dtds\geq \int_{s=a}^{s=b}\int_{t=a}^{t=b} (f(s)f(t)+g(t)g(s))\quad dtds\\\implies\int_{s=a}^{s=b}\int_{t=a}^{t=b}\sqrt{(f(s)^2+g(s)^2)}\sqrt{(f(t)^2+g(t)^2)}\quad dtds\geq \int_{s=a}^{s=b}\int_{t=a}^{t=b} (f(s)f(t))\quad dsdt\quad + \int_{s=a}^{s=b}\int_{t=a}^{t=b} (g(t)g(s))\quad dsdt$$

These integrals may be simplified into:

$$\Bigl(\int_{r=a}^{r=b}\sqrt{(f(r)^2+g(r)^2)}\quad dr\Bigr)^2\geq \Bigl(\int_{r=a}^{r=b} f(r)\quad dr\Bigr)^2 + \Bigl(\int_{r=a}^{r=b} g(r)\quad dr\Bigr)^2$$

see this
So, finally, we have:

$$\int_{r=a}^{r=b}\sqrt{(f(r)^2+g(r)^2)}\quad dr\geq \sqrt{\Bigl(\int_{r=a}^{r=b} f(r)\quad dr\Bigr)^2 + \Bigl(\int_{r=a}^{r=b} g(r)\quad dr\Bigr)^2}$$

Or:

$$\int_a^b|z(r)|\quad dr\geq \Biggl|\int_a^bz(r)\quad dr \Biggr|$$