2
$\begingroup$

Assume that $\Phi:[0,\infty)\to [0,\infty)$ is an N-function, i.e., $\Phi(0)=0$, $\Phi$ is convex, strictly increasing, $\Phi(t)/t\to 0$ if $t\to 0$ and $\Phi(t)/t\to \infty$ if $t\to \infty$, or equivalently, $\Phi$ can be written as $$\Phi(t)=\int_0^t \phi(s)ds,$$

where $\phi(t)=0$ if and only if $t=0$, $\phi$ is right continuous, $\phi$ is nondecreasing and $\phi(t)\to \infty$ if $t\to \infty$.

Let $\Omega\subset\mathbb{R}^N$ be a bounded domain ans suppose that $u:\Omega\to \mathbb{R}$ is a measurable function satisfying $$\tag{1}\int_\Omega \Phi(|u|)<\infty.$$

My question: is it true that for small $\alpha>0$, we have that $$\int_\Omega \Phi(|u|+\alpha)<\infty?$$

The class of functions for which $(1)$ is true, is called Orlicz class and it is denoted by $\mathcal{L}_\Phi(\Omega)$. It is a convex space and it is know that (for some N-functions) there are functions $u\in\mathcal{L}_\Phi(\Omega)$ with the property that $\lambda u\notin \mathcal{L}_\Phi(\Omega)$ for some $\lambda>0$, so it is not a vector space.

If $\Phi$ satisfies the $\Delta_2$ condition, i.e., if there are constants $t_0\ge 0$ and $k>0$ such that $\Phi(2t)\le k\Phi(t)$ for $t\ge t_0$ then, the Orlicz class $\mathcal{L}_\Phi(\Omega)$ is a vector space, so we can assume that $\Phi$ does not satisfies the $\Delta_2$ condition.

If $\Phi$ can be compared with a polynomial near the infinity then the above is true, so we can even assume that $\Phi$ grows really fast at the infinity.

My motivation to understand this problem is to calculate the derivative of the functional $\int_\Omega \Phi(|u|)$ in the direction $v$, where $v$ is a bounded function.

Any idea or references is appreciated.

$\endgroup$
3
$\begingroup$

I think the following will work, although I was a bit sloppy with the details:

Let us consider $$ \Phi\left(t\right)=e^{e^{t}}-e-et. $$ We note that (for $t$ large), we have $$ \frac{1}{2}e^{e^{t}}\geq e+et $$ and thus $\Phi\left(t\right)\geq\frac{1}{2}e^{e^{t}}$ for $t$ large.

Now, let $\Omega=\left(0,c\right)$ for some $0<c\ll1$ and define $$ f\left(x\right):=\ln\ln\left[\sum_{n\in\mathbb{N}}\frac{1}{x^{1-\frac{1}{n}}}\cdot\frac{1}{2^{n}\cdot c_{n}}\right] $$ with $c_{n}:=\left\Vert x\mapsto1/x^{1-\frac{1}{n}}\right\Vert _{L^{1}\left(\left(0,1\right)\right)}$. This idea is borrowed from one possible way of constructing a function $f$ with $f \in L^p$ for all $p>1$, but $f \notin L^1$.

We have \begin{eqnarray*} \int_{0}^{c}\Phi\left(\left|f\left(t\right)\right|\right)\, dt & \leq & \int_{0}^{c}e^{e^{f\left(t\right)}}\, dt\\ & = & \int_{0}^{c}\sum_{n\in\mathbb{N}}\frac{1}{x^{1-\frac{1}{n}}}\cdot\frac{1}{2^{n}\cdot c_{n}}\, dx\\ & = & \sum_{n\in\mathbb{N}}\frac{1}{2^{n}\cdot c_{n}}\cdot\int_{0}^{c}\frac{1}{x^{1-\frac{1}{n}}}\, dx\\ & \leq & \sum_{n\in\mathbb{N}}\frac{1}{2^{n}}<\infty. \end{eqnarray*} But for arbitrary $\alpha>0$, we have \begin{eqnarray*} \int_{0}^{c}\Phi\left(\left|f\left(t+\alpha\right)\right|\right)\, dt & \geq & \int_{0}^{c_{\alpha}}\Phi\left(\alpha+\ln\ln\left(\frac{1}{x^{1-\frac{1}{n}}}\cdot\frac{1}{2^{n}c_{n}}\right)\right)\, dx\\ & = & \int_{0}^{c_{\alpha}}e^{e^{\alpha+\ln\ln\left(\frac{1}{x^{1-1/n}}\cdot\frac{1}{2^{n}c_{n}}\right)}}\, dx\\ & = & \int_{0}^{c_{\alpha}}e^{e^{\alpha}\cdot\ln\left(\frac{1}{x^{1-1/n}}\cdot\frac{1}{2^{n}c_{n}}\right)}\, dx\\ & = & \int_{0}^{c_{\alpha}}\left(\frac{1}{x^{1-1/n}}\cdot\frac{1}{2^{n}c_{n}}\right)^{e^{\alpha}}\, dx\\ & = & c_{n,\alpha}\cdot\int_{0}^{c_{\alpha}}\frac{1}{x^{\left[1-\frac{1}{n}\right]\cdot e^{\alpha}}}\, dx \end{eqnarray*} for each $n\in\mathbb{N}$, for some suitable $c_{\alpha}\in\left(0,c\right)$. But since $\alpha>0$, we have $e^{\alpha}>1$, so that there is some $n=n\left(\alpha\right)\in\mathbb{N}$ with $\left(1-\frac{1}{n}\right)\cdot e^{\alpha}>1$. Thus, the above integral diverges.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.