Show that $\int_0^1 h(t)dt\geq\left(\int_0^1 f(t)dt \right)^a\left(\int_0^1 g(t)dt \right)^{1-a}$

Solution 1:

This is the Prékopa–Leindler inequality (see e.g. here). Also here

Edit: The proof that can be found in the paper of Gardner, in the setting of the OP, is at the level of a Calculus class. Let us consider the functions $$ F(u) := \int_0^u f(x)\, dx, \qquad G(u) := \int_0^u g(x)\, dx, \qquad u \in [0,1]. $$ Since $f,g$ are continuous and strictly positive functions, we have that $F$ and $G$ are continuously differentiable and strictly increasing, with $F' = f$ and $G' = g$. If we set $F_1 := F(1)$, $G_1 := G(1)$, then $F$ is a bijection from $[0,1]$ to $[0, F_1]$ and $G$ is a bijection from $[0,1]$ to $[0, G_1]$. Let $u,v\colon [0,1] \to [0,1]$ be the functions defined by $$ u(t) := F^{-1}(F_1\, t), \qquad v(t) := G^{-1}(G_1\, t), \qquad t\in [0,1]. $$ Since $u'(t) = F_1 / f(u(t))$ and $v'(t) = G_1 / g(v(t))$ for every $t\in [0,1]$, using the AM-GM inequality we deduce that $$ w'(t) := a\, u'(t) + (1-a)\, v'(t) \geq [u'(t)]^a [v'(t)]^{1-a} = \frac{F_1^a}{f(u(t))^a}\cdot \frac{G_1^{1-a}}{g(v(t))^{1-a}} $$ and finally $$ \int_0^1 h(x)\, dx = \int_0^1 h(w(t))\, w'(t)\, dt \geq \int_0^1 f(u(t))^a g(v(t))^{1-a} \frac{F_1^a}{f(u(t))^a}\cdot \frac{G_1^{1-a}}{g(v(t))^{1-a}}\, dt = F_1^a G_1^{1-a}. $$

Solution 2:

As we will see, (a) there is nothing particular about the choice of the interval $[0,1]$. The result holds if the condition holds in an interval $I$, bounded or unbounded; (b) continuity is not crucial here. However we assume that $f$ and $g$ are measurable functions such that $f^{-1}(U)+g^{-1}(U)$ is measurable whenever $U$ is a Borel subset of $\mathbb{R}$. Certainly this is satisfied when $f$ and $g$ are continuous, and more generally when $f$ and $g$ are Borel measurable (in this case $f^{-1}(U)+g^{-1}(U)$ is universally measurable).

Some notation first. For any function $\Phi:[0,1]\rightarrow[0,\infty)$ and $t\geq0$, we denote $\{\Phi>t\}=\{x\in[0,1]: \Phi(x)>t\}$; for any set $A\subset\mathbb{R}$, and $b\in\mathbb{R}$, $bA=\{ba:a\in A\}$; For any sets $A,B\subset\mathbb{R}$, $A+B=\{a+b:a\in A, b\in B\}$.

• The first claim is that $$a \{f>t\} +(1-a)\{g>t\}\subset\{h>t\}\tag{1}\label{one}$$ To check this, notice that if $u$ belongs to the set in the left-hand side of $\eqref{one}$ then $u=ax+(1-a)y$ for some $x\in \{f>t\}$ and $y\in\{g>t\}$ and so, $h(u)\geq f(x)^ag^{1-a}(y)>t^a t^{1-a}=t$.

• By Brunn–Minkowski's inequality and the homogeneity of Lebesgue's measure $\lambda$ (on $\mathbb{R}$) \begin{aligned} \lambda(\{h>t\})&\geq \lambda\big(a\{f>t\} +(1-a)\{g>t\}\big)\\ &\geq\lambda(a\{f>t\})+\lambda((1-a)\{g>t\})\\ &=a\lambda(\{f>t\})+(1-a)\lambda(\{g>t\}) \end{aligned}

• Fubuni's theorem leads to \begin{aligned} \int^1_0h(s)\,ds&=\int^\infty_0\lambda(\{h>t\})\,dt\\ &\geq a\int^\infty_0\lambda(\{f>t\})\,dt + (1-a) \int^\infty_0\lambda(\{g>t\})\,dt \\ &= a\int^1_0f(s)\,ds + (1-a)\int^1_0 g(s)\,ds \end{aligned}

• Finally, by the arithmetic--geometric inequality ($a\,\alpha+ (1-a)\,\beta\geq \alpha^a\beta^{1-a}$)

\begin{aligned} \int^1_0h(s)\,ds\geq a\int^1_0f(s)\,ds + (1-a)\int^1_0 g(s)\,ds \geq\Big(\int^1_0 f(s)\,ds\Big)^a\big(\int^1_0 g(s)\,ds\Big)^{1-a} \end{aligned}

Edit: We have used a big result, namely the Brunn-Minksowki inequality. In the real line however, this result may be proven without much effort. Here is a short proof.

Brunn-Minkowski's inequality on $\mathbb{R}$. Suppose $A,B\subset\mathbb{R}$ are measurable subsets such that $A+B$ is measurable. Then $$\lambda(A+B)\geq \lambda(A) +\lambda(B)\tag{2}\label{two}$$

Proof: It suffices to assume that both $\lambda(A)$ and $\lambda(B)$ are finite. Due to the inner regularity of Lebesgue's measure, it is enough to show that $\eqref{two}$ holds for $A$ and $B$ compact. If $a^*:=\sup A$ and $b_*:=\inf B$ then $$ A+B\subset (a^*+B)\cup(A+ b_*)\supset\{a^*+b_*\}$$ If $x\in (a^*+B)\cap(A+ b_*)$ then for some $(a,b)\in A\times B$, $x=a^*+b=b_*+a$. Since $0\leq a^*-a=b_*-b\leq0$, it follows that $a=a^*$ and $b=b_*$ and so, $x=a^*+b_*$. Thus \begin{aligned} \lambda(A+B)&\geq \lambda\big((a^*+B)\cup(A+b_*)\big)\\ &=\lambda((a^*+B)+\lambda(A+b_*)-\lambda(\{a^*+b^*\})=\lambda(A)+\lambda(B) \end{aligned}

Notes:

• The result can be generalized to higher dimensions starting from the one dimensional case and proceeding by induction with the help of Fubini's theorem.
• The higher dimensional versions of Brunn-Minkowski's inequality can be obtained form the high dimensional version of Prékopa–Leindler's theorem.

Thanks to Rigel for the reminder of the aforementioned result which in turn reminded me of the Brunn–Minkowski inequality.

I still would like to know how rulergraham's teacher proved the statement in his Calculus class. A much simpler (but wickedly tricky) argument perhaps?