$0\leq f,g\in C^{1},\int_{a}^{b}\sqrt{f}\geq\int_{a}^{b}\sqrt{g}\implies\int_a^b f\geq\int_{a}^{b}g$?

Question

$f,g$ are differentiable non-negative functions on $[a,b]$ with $ \int_{a}^{b}\sqrt{f(t)}dt\geq\int_{a}^{b}\sqrt{g(t)}dt $.

So do we have that $\int_{a}^{b}f(t)dt\geq\int_{a}^{b}g(t)dt$ ?

Does anyone have any clue about this? I tried to make a counterexample but I couldn't find. If it is true, any hint on how to prove would be appreciated.

P.S.: I hope it is true because it would make easier to find the shortest path between two points on a surface using Euler-Lagrange equation since we would have simpler differential equations to solve. If it is not true, does anyone has any tip on evaluating this paths?

Answer 1

Concerning the geodesics problem, you may set, for any curve $f\in C^1([a,b])$, $$L(f)=\int_a^b\|f'(t)\|dt,\qquad E(f)=\int_a^b\|f'(t)\|^2dt $$ The goal is to minimize $L$, but as you say, it is easier to minimize $E$. By Cauchy-Schwartz, we immediately see that $L(f)^2\leq (b-a)E(f)$. The main idea is that any regular curve $f\in C^1([a,b])$ may be reparametrized (through a multiple of the arc length parameter) into a curve $\tilde{f}\in C^1([a,b])$ such that $$\|\tilde{f}'(t)\|=\frac{L(f)}{b-a},\qquad \forall t\in [a,b] $$ which also implies that $$L(f)^2=L(\tilde{f})^2=E(\tilde{f})(b-a) $$ So if $E(f)\leq E(h)$ for all $h$, you have $$L(f)^2\leq (b-a)E(f)\leq (b-a)E(\tilde{g})=L(g)^2,\qquad \forall g $$ Hence $L(f)\leq L(g)$ for all $g$ and $f$ minimizes curve length.

Answer 2

No, that implication does not hold in general. A counter-example is $f(x) = 1$, $g(x) = 4x^2$ on the interval $[0, 1]$: $$ \int_{0}^{1}\sqrt{f(t)} \, dt = 1 = \int_{0}^{1}\sqrt{g(t)} \, dt \, , $$ but $$ \int_{0}^{1}f(t) \, dt = 1 < \frac 43 = \int_{0}^{1}g(t) \, dt \, . $$