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Suppose that $f,g:[0,1] \to \mathbb{R}^{+}$ are smooth functions, such that $$ 0< L_1 \le \int_0^1 f \, dx,0

I am looking for a "direct proof" that

$$ \sqrt{L_1^2+L_2^2} \le \int_0^1 \sqrt{f^2 + g^2 }\, dx.$$

Note: For the interested, This inequality has a geometric content, namely that the product of minimizing geodesics (in a product of Riemannian manifolds) is also minimizing.

Here is a non-direct proof:

Define $F(t)=\int_0^t f(t)dt,G(t)=\int_0^t g(t)dt$, and look at the path $\gamma:[0,1] \to \mathbb{R}^2$ defined by $$ \gamma(t)=(F(t),G(t))$$

Then $\dot \gamma(t)=(f(t),g(t))$, and

$$\| \gamma(1)-\gamma(0)\|=\sqrt{(F(1)-F(0))^2+(G(1)-G(0))^2}$$

$$ = \sqrt{\big(\int_0^1 f\big)^2+\big(\int_0^1 g\big)^2} \le L(\gamma)= \int_0^1 \sqrt{f^2+g^2} \tag{1}$$

Where the inequality follows from the geometric fact that $\| \gamma(1)-\gamma(0)\| \le L(\gamma)$. (There is also a direct argument for that here).

By item $(1)$, we deduce that

$$ \sqrt{L_1^2+L_2^2} \le \sqrt{\big(\int_0^1 f\big)^2+\big(\int_0^1 g\big)^2} \le \int_0^1 \sqrt{f^2 + g^2 }\, dx. $$

1 Answers 1

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This is a consequence of the Minkowski inequality for $p<1$, which says that $$\left(\int_0^1|u|^p\right)^{1/p}+\left(\int_0^1|v|^p\right)^{1/p}\leq \left(\int_0^1|u+v|^p\right)^{1/p},$$ when $u$ and $v$ are nice enough and $p\in(0,1)$.

In your case, apply the inequality above for $u=f^2$, $v=g^2$, and $p=1/2$.