# Perturbation of a function in the Orlicz class by a constant ($\int_\Omega\Phi(|u|)<\infty$ implies $\int_\Omega \Phi(|u|+\alpha)<\infty$?).

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.

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.