Example of a Young function which does not satisfy $\Delta_{2}$ condition

Question

Let me define first
(1) A convex function $\Phi \colon \mathbb{R}\to \mathbb{R}^+$ which satisfies the conditions,
(a) $\Phi(0)=0$
(b)$\Phi(-x)=\Phi(x)$
(c) $\lim_{x \to \infty}\Phi(x)=+\infty$, is called the Young function.

(2) A Young function $\Phi \colon \mathbb{R}\to \mathbb{R}^+$ is said to satisfy the $\Delta_{2}$ condition if, $\Phi(2x)\leq K\Phi(x)$ for $x\geq x_{0}\geq 0$ for some absolute constant $K>0$

Can someone give some example that does not satisfy $\Delta_{2}$ condition?Thanks

Answer 1

Consider the function $\Phi \colon \mathbb R \to {\mathbb R}^+, \, \Phi(x) = \mathrm e^{x^2} - 1.$ Then we have:

• $\Phi(0) = \mathrm e^0 - 1 = 0$.
• $\Phi(-x) = \mathrm e^{(-x)^2} - 1 = \mathrm e^{x^2} - 1 = \Phi(x)$ for all $x \in \mathbb R$.
• $\lim_{x \to \infty} \Phi(x) = \lim_{x \to \infty} \mathrm e^{x^2} - 1 = \infty$ since $\mathrm e^x \to \infty$ and $x^2 \to \infty$ if $x \to \infty$.
• Finally, $\Phi$ is convex as $\Phi''(x) = (x) = 2 \mathrm e^{x^2} (1 + 2 x^2) > 0$ for all $x \in \mathbb R$.

Now suppose there exist $K > 0$ and $x_0 \geq 0$ such that $\Phi(2x) \leq K \Phi(x)$ for all $x \geq x_0$. Then $\mathrm e^{(2x)^2} - 1 = \Phi(2 x) \leq K \Phi(x) = K \mathrm e^{x^2} - K$. However, if $x \to \infty$ the above estimate does not hold since $\mathrm e^{4x^2}$ grows way faster than $\mathrm e^{x^2}$ as $x \to \infty$.