A mapping $\Phi:[0,\infty)\to[0,\infty)$ is termed an N-function (nice Young function) if
(i) $\Phi$ is continuous on $[0,\infty)$;
(ii) $\Phi$ is convex on $[0,\infty)$;
(iii) $\lim\limits_{t \rightarrow 0}\frac{\Phi(t)}{t}=0$;
(iv)$\lim\limits_{t\rightarrow\infty}\frac{\Phi(t)}{t}=\infty$.
In the book "Robert A. Adams, John J. F. Fournier, Sobolev Spaces" the authors claim that this definition is equivalent to $\Phi$ has an integral representation such that
$$
\Phi(t)=\int_0^t\varphi(s)ds,\quad t\in [0,\infty),
$$
where $\varphi$ be a real valued function defined on $[0,\infty)$ and having the following properties
(i*) $\varphi(0)=0$, $\varphi(t)>0$ if $t>0$, $\lim\limits_{t\rightarrow\infty}\varphi(t)=\infty$;
(ii*) $\varphi$ is nondecreasing;
(iii*) $\varphi$ is right continuous.
My first question is : We know that a convex function on an open inerval is continuous. Using this fact and conditions (ii) and (iii) can we write condition (i) as $\lim\limits_{t \rightarrow 0^+}\Phi(t)=\Phi(0)=0$?
My second question is : The function $\Phi(t)=\left\{
\begin{array}{ccc}
0& , & t\leq 1 \\
t\log t&,& t>1
\end{array}
\right.$ is an N-function or not? It seems like it satisfies the all conditions (i)-(iv). But according to second definition $\varphi(t)=\left\{
\begin{array}{ccc}
0& , & t< 1 \\
\log t+1&,& t\geq1
\end{array}
\right.$ and it doesn't satisfy the condition (i*). Where is the mistake?
Best Answer
The second condition in (i*) implies that $\Phi$ must be strictetly nondecreasing that is if $x <y$ then $\Phi(x)<\Phi(y)$. You must add this condition on $\Phi$ in the first definition. You will note that your example does not satisfy this condition. Therefore it is not a young function (in this sens).