Are the two definitions of the complementary Young function equivalent?

Question

The description of the problem:

For a Young function I would refer the reader to the book "Function spaces" by Pick Luboš, Kufner Alois, John Oldrich and Fucík Svatopluk, and published by de Gruyter. The following Definition 1 is about the complementary Young function which would be found in this book (see p.115).

Definition 1. Let $\Phi $ be a Young function generated by a function $\varphi$, that is, $$ \Phi(t)=\int_{0}^{t}\varphi(s)ds, \ t \in \left [ 0, \infty\right ). $$ We set $$\psi(t)=\underset{\varphi(s)\leq t}{ \sup}s$$

and $$ \Psi(t)=\int_{0}^{t}\psi(s)ds, \ t\in \left [0,\infty \right ). $$ The function $\Psi$ is called the complementary function to $\Phi$.

However, I have found another kind of definition of $\Psi$ in many occasions. See below:

Definition 2. Geven a Young function $\Phi$, the complementary Young function $\Psi$ is defined by $$ \Psi(t)=\underset{s>0}{\sup}\left \{ st-\Phi(s) \right \}, \ t>0. $$

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My question is:

Are these two definitions equivalent? And if they are, how can we proof this result?

Thank you for your help.

Answer 1

Yes, the two definitions are equivalent. If the functions are smooth (and I am not aware of any reason to use nonsmooth Young functions), verification is easy: in the second definition, the supremum defining $\Psi$ is attained when $t=\Phi'(s)$. Write $\phi=\Phi'$ and $\psi=\phi^{-1}$, so that $s=\psi(t)$. Then
$$\Psi(t) = \sup_{s>0} (st-\Phi(s))= t\psi(t)-\Phi(\psi(t))$$ which implies $$\Psi'(t) = t\psi'(t)+\psi(t)-\phi(\psi(t))\psi'(t) = \psi(t) $$ which conforms to the first definition of $\Psi$.

For general functions, the fact that the definitions agree is the content of Young's inequality $$ab \le \left(\int_0^a \phi\right)+\left( \int_0^b \psi\right)$$ (where $\phi$ and $\psi$ are inverses of each other). The inequality has a nice "proof by a picture", which I took from Wikipedia:

Answer 2

The is YES if $\varphi$ is invertible. In this case, we clearly have $$\psi= \varphi^{-1}.$$

The second definition $$\Psi(t)= \sup_{s>0} [st-\Phi(s)]= at-\varphi(a) \quad\text{for some }a>0$$

Note that $a$ exists by convexity and we have $\Psi(0)=0$. Since $\Phi'=\varphi$ even have

$$\frac{d}{ds}[st-\Phi(s)](a)=0\Longleftrightarrow t=\varphi(a)\Longleftrightarrow a=\varphi^{-1}(t)$$

which means that $$\Psi(t) = at-\varphi(a)= t\varphi^{-1}(t)-\Phi(\varphi^{-1}(t))$$

That is we have $$\Psi'(t) = t(\varphi^{-1})'(t)+\varphi^{-1}(t)-(\varphi^{-1})'(t)\Phi'(\varphi^{-1}(t))= t(\varphi^{-1})'(t)+\varphi^{-1}(t)-(\varphi^{-1})'(t)\varphi(\varphi^{-1}(t))= \varphi^{-1}(t). $$

This shows that $$\Psi(t)=\sup_{s>0} [st-\Phi(s)]= \Psi(0)+\int_0^t \Psi'(s) ds=\int_0^t \varphi^{-1}(s) ds=\int_0^t \psi(s) ds.$$