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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.

mlc
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    Draw the graph of $f$ and the rectangle $[0,a] \times [0,b]$; then interpret the integrals as areas. – Catalin Zara Apr 26 '17 at 19:07
  • is there any strictly analyitic proofs? –  Apr 26 '17 at 19:16
  • I guess you can complicate the proof to make it look "analytic," but the picture at https://en.wikipedia.org/wiki/Young%27s_inequality should really suffice. – Catalin Zara Apr 26 '17 at 19:21
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    If $f$ is differentiable you can use the change of variable theorem and integration by parts. If $f$ is not differentiable you could regularize and proceed from there. – Umberto P. Apr 26 '17 at 19:26

2 Answers2

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

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

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.

Using $(1)$ we see that

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

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

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

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

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

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

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

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!

Mark Viola
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  • I understand your answer perfectly but did you miss a factor of b off on the last inequality on the rhs? Should it not be $ab$ not $a$? –  Apr 27 '17 at 09:52
  • Pleased to hear. And yes, the $b$ was missing. I've corrected the typographical ommission. – Mark Viola Apr 27 '17 at 14:19
  • If you did not assume $f$ was differentiable, then how did you get (1) since you used integration by parts? –  Mar 01 '19 at 01:49
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    @Diddysmash As stated upfront, we are using the Riemann-Stieltjes Integral for which I provided this reference. – Mark Viola Mar 01 '19 at 03:20
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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?

Franklin Pezzuti Dyer
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