Finiteness of an integral given an [itex]L^2[/itex] function

[lmedin02]

Homework Statement

Let $\Omega$ be a torus and $g\in L^2(\Omega)$ be a scalar value function. Is $\int_{\Omega}{e^g}dx<\infty$?

Homework Equations

The Attempt at a Solution

Not sure where to start. However, if $g\in W^{1,2}(\Omega)$ then I can show that the answer is yes by applying Moser-Trudinger inequality and decomposing $W^{1,2}$ as the direct sum of $\mathbb{R}$ and functions of average value zero on $W^{1,2}$. Any hints are greatly appreciated.

 

[mighty2000]

Hello, thank you for your forum post. To answer your question, we first need to understand what the notation and terminology mean.

The symbol \Omega typically represents a domain or region in a mathematical space, in this case, a torus. This means that the function g is defined on this torus. The symbol L^2(\Omega) represents the space of square integrable functions on the torus. This means that the function g is "well-behaved" in the sense that it is not too oscillatory or too singular.

Now, the question is asking whether the integral of the exponential function with g as an exponent, over the torus, is finite. In other words, is the function e^g also square integrable on the torus?

To answer this, we need to consider the regularity of g. As you mentioned, if g is in the Sobolev space W^{1,2}(\Omega), then we can apply the Moser-Trudinger inequality to show that the integral is finite. This is because the Sobolev space captures the regularity of functions, and in this case, g is regular enough for the integral to converge.

However, if g is not in W^{1,2}(\Omega), then we cannot guarantee the finiteness of the integral. This is because g may have too many oscillations or singularities, making the exponential function too oscillatory or too singular, and thus not square integrable.

In summary, the answer to the question depends on the regularity of g. If g is in W^{1,2}(\Omega), then the integral is finite, but if g is not in W^{1,2}(\Omega), then we cannot guarantee the finiteness of the integral. I hope this helps you approach the problem. Good luck with your solution!