constructing a specific (real-) analytic function

Question

Im searching for an example of a special-behaving analytic function. Maybe you can beat me to constructing such one. The criterias are

• $g :\mathbb{R}\rightarrow \mathbb{R^+}$ is analytic
• $g$ is $\mathcal{L}^1$ (implicitly $\liminf_{x\rightarrow \infty}xg(x)=0$)
• $\limsup_{x\rightarrow \infty} xg(x)>0$

If you can give an example with a easy expression for the antiderivative, double the love is awarded

Answer 1

a solution is $$g(x) = \sum_{n=-\infty, n \ne 0}^\infty e^{-n^4 (x-n)^2}$$

• it is real analytic since it is complex analytic

• it is $L^1$ since $$\|g\|_{L^1} = \sum_{n=-\infty}^\infty \int_{-\infty}^\infty e^{-n^4 (x-n)^2} dx = \sum_{n=-\infty, n \ne 0}^\infty \frac{1}{n^2}\int_{-\infty}^\infty e^{-x^2} dx = 2\frac{\pi^2}{6}\sqrt{2\pi}$$

• and clearly $g(x) \ge 0$ and $\displaystyle\underset{|x| \to \infty}{\lim \sup} g(x) = 1$

Answer 2

The function

$$f(x) = \left ( \frac{2+\cos x}{3} \right) ^{x^4}$$

has this property. This follows from the fact that $\int_0^{2\pi} [(2+\cos x)/3]^n\, dx \sim 1/\sqrt n$ as $n\to \infty$ (by Laplace's method).