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I am looking at a proof of the converse Young's inequality which is stated as follows:

Let $f$ and $g$ be two continuous, one-to-one and increasing functions such that $f(0) = g(0) = 0, g^{-1}(x) \geq f(x)$ for all non-negative $x$. If for every positive real numbers a and b we have $$ab \leq \int_0^a f(x) dx + \int_0^b g(x) dx$$ then $f$ and $g$ are inverse.

I look at a proof given by Tatsuo Takahashi here: https://www.jstage.jst.go.jp/article/tmj1911/36/0/36_0_99/_pdf/-char/ja

We prove this in the case that $f(x) \neq g^{-1}(x)$ for all non-negative real x, by showing that there exists a pair $a,b$ such that $$ab > \int_0^a f(x) dx + \int_0^b g(x) dx.$$ $f(x) \neq g^{-1}(x)$ for all non-negative real x implies that there exists a positive real c, for which $f(c) \neq g^{-1}(c).$ Therefore, $g^{-1}(x) \geq f(x),$ can be rewritten as $g^{-1}(c) > f(c).$ As f,g are continuous, there exists a $\delta$ such that $g^{-1}(x) > f(x),$ when x is between $(c-\delta,c].$ Consider the function $$\int_0^c f(x) dx + \int_0^{g^{-1}(c)} g(x) dx + \int_0^c g^{-1}(x) - f(x) dx = \int_0^{g^{-1}(c)} g(x) dx + \int_0^c g^{-1}(x) dx.$$ I follow everything up until this point, then we have the line "Using the substitution $g^{-1}(x) = x,$ this is equal to $$\int_0^c x\frac{dg^{-1}(x)}{dx} dx + \int_0^c g^{-1}(x) dx = \int_0^c d(xg^{-1}(x)) = cg^{-1}(c)$$"

Why exactly does this follow? Especially, the second result of the three results.

Mittens
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KeynesianSpaceman
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  • Is $g^{-1}$ differentiable? it is not stated in your assumptions nor in the assumptions in the paper you quoted. – Mittens Mar 08 '23 at 00:22
  • @OliverDíaz I presume so – KeynesianSpaceman Mar 08 '23 at 00:22
  • Then state that as an assumption. My impression is that it is valid without differentiability (the functions are Lebesgue a.s. differentiable but not absolutely continuous so an appropriate version of integration by parts is needed.) – Mittens Mar 08 '23 at 00:30

2 Answers2

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The argument is clearer if we write the substitution in the form $t:=g(x)$, rather than reusing $x$. Then $x = g^{-1}(t)$ and $dx = \frac {dg^{-1}(t)}{dt}dt$ so we can write $$ \int_{x=0}^{g^{-1}(c)}g(x)\,dx=\int_{t=0}^c t \,\frac {dg^{-1}(t)}{dt}\,dt.\tag1 $$ At this point you can change the dummy variable of integration on the RHS of (1) from $t$ back to $x$.

As for the equality $$ \int_0^c x\frac{dg^{-1}(x)}{dx} dx + \int_0^c g^{-1}(x) dx = \int_0^c d(xg^{-1}(x)), $$ this is simply the product rule applied to the function $x\mapsto xg^{-1}(x)$: $$\frac{d}{dx}[xg^{-1}(x)]=x\cdot \frac {dg^{-1}(x)}{dx}+1\cdot g^{-1}(x) $$ and therefore $$\int_0^c \left(x\frac {dg^{-1}(x)}{dx}+g^{-1}(x)\right)dx=\int_0^c \left(\frac{d}{dx}[xg^{-1}(x)]\right)\,dx=\left.xg^{-1}(x)\right|_0^c=cg^{-1}(c). $$

grand_chat
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  • Differentiability is not assumed. – Mittens Mar 08 '23 at 00:22
  • @OliverDíaz Right. The function $g^{-1}$ is continuous and strictly increasing, so in the non-differentiable case a Riemann-Stieltjes integral can handle the change of variables. – grand_chat Mar 08 '23 at 18:04
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    @gran_chat: I understand that. It is just that in the OP there is no such assumption and it is not too difficult to deal with the general case (ie. integration by parts via Lebesgue-Stieltjes or Riemann-Stieltjes). – Mittens Mar 08 '23 at 18:42
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In general, if $g$ is continuous and strictly monotone increasing ($g(x)g(y)$ whenever $x<y$) on an interval $[0,T]$ (which all means that $g^{-1}$ exists, is continuous and and is strictly monotone increasing) and $g(0)=0$, then for all $0\leq a\leq T$ and $0\leq b\leq g(T)$ \begin{align}\int^a_0g +\int^b_0g^{-1}\geq ab\tag{0}\label{zero}\end{align} with equality iff $b=f(a)$.

There are several proofs of this fact, and this has been considered many times in MSE in the past (postings 1 and 2, and 3 for example). Some proofs are based on splitting a bounded region in the plane in two easy to estimate pieces and others are based on Riemann-Stieltjes-Lebsgue integration by parts.

Taking \eqref{zero} as a known, suppose $f,g$ satisfy the assumptions in the OP and that there is $0<c<g(T)$ such that $g^{-1}(c)>f(c)$. Then $\int^c_0 g^{-1}>\int^c_0 f$ and so $$cg^{-1}(c)=\int^c_0g^{-1}+\int^{g^{-1}(c)}_0 g>\int^c_0f+\int^{g^{-1}(c)}_0g$$

contradicting the assumption that $ab\leq\int^a_0f+\int^b_0g$ for all $0\leq a\leq T$ and $0\leq b\leq g(T)$. Therefore, $g^{-1}\equiv f$.

Mittens
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