Prove Young's inequality

Question

Suppose that f is continuous increasing function with $f(0)=0$ then prove that for $a,b>0$ we have Young's inequality

$$ab\leq \int_{0}^{a} f(x)dx+\int_{0}^{b}f^{-1}(x)dx$$

My attempt is incorrect: I didn't argue using lower and upper sums which i assume you need to use, any help would be appreciated thanks.

Answer 1

We can make use of the Riemann-Stieltjes integral. Proceeding we enforce the substitution $x=f(t)$ to obtain

$$\begin{align} \int_0^b f^{-1}(x)\,dx&=\int_{0}^{f^{-1}(b)} t\,df(t)\\\\ &=bf^{-1}(b)-\int_0^{f^{-1}(b)}f(t)\,dt\tag 1 \end{align}$$

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ASIDE:

Continuous monotonic functions are differentiable almost everywhere (i.e., everywhere excluding a set of measure $0$). But it is important to stress that in using the Riemann-Stieltjes Integral in $(1)$, we did not assume that $f$ was differentiable.

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Using $(1)$ we see that

$$\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx=\int_{f^{-1}(b)}^af(x)\,dx+bf^{-1}(b)\tag 2$$

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If $a>f^{-1}(b)$, then $\int_{f^{-1}(b)}^af(x)\,dx\ge b(a-f^{-1}(b))$ since $f$ is increasing. Therefore, we assert that

$$\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx\ge ab \tag 3$$

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If $a<f^{-1}(b)$, then $\int_{f^{-1}(b)}^af(x)\,dx=-\int_a^{f^{-1}(b)}f(x)\,dx\ge -b(f^{-1}(b)-a)$ since $f$ is increasing. Therefore, we assert that

$$\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx\ge ab \tag 4$$

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Finally, if $a=f^{-1}(b)$, then we see directly from $(2)$ that

$$\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx\ge ab \tag 5$$

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Putting together $(3)$, $(4)$, and $(5)$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_0^a f(x)\,dx+\int_0^b f^{-1}(x)\,dx\ge ab}$$

as was to be shown!

Answer 2

Not exactly an answer, but too long to comment.

There exists a formula for the antiderivative of an inverse function. If we let $f(x)$ have inverse $f^{-1}(x)$ and antiderivative $F(x)$, then $$xf^{-1}(x)-(F\circ f^{-1})(x)+C$$ and if we plug this in, we get $$F(a)-F(0)+bf^{-1}(b)-(F\circ f^{-1})(b)-0f^{-1}(0)+(F\circ f^{-1})(0)$$ $$F(a)-F(0)+bf^{-1}(b)-(F\circ f^{-1})(b)+(F\circ f^{-1})(0)$$ If $f(x)$ is continuously increasing then it must be one-to-one and it has an inverse, so $f^{-1}(0)=0$ and we get $$F(a)-F(0)+bf^{-1}(b)-(F\circ f^{-1})(b)+F(0)$$ $$F(a)+bf^{-1}(b)-(F\circ f^{-1})(b)$$

Does this simplify it enough for you to take it from here?