Property of a limit of functions of average value zero in L^2 space

lmedin02
Messages
52
Reaction score
0

Homework Statement



Let f_k\rightarrow f in L^2(\Omega) where |\Omega| is finite. If \int_{\Omega}{f_k(x)}dx=0 for all k=1,2,3,\ldots, then \int_{\Omega}{f(x)}dx=0.

Homework Equations





The Attempt at a Solution


I started by playing around with Holder's inequality and constructing examples where this is the case. Since every other example I create usually does not converge to a function. I used functions defined on a bounded symmetric interval like \dfrac{1}{k}x, \sin(\dfrac{1}{k}x). However, I do not see how to actually set up to make this conclusion.
 
Physics news on Phys.org


What if you write
$$\begin{align} \left| \int_{\Omega} f(x) dx \right| &=
\left| \int_{\Omega} f(x) dx - \int_{\Omega} f_k(x) dx \right| \\
&= \left|\int_{\Omega} (f(x) - f_k(x)) dx \right| \\
\end{align}$$
and apply an appropriate inequality to the right hand side?
 


Got it. Thanks. Using Holder's inequality and noting that the size of \Omega is finite the results follows since f_k\rightarrow f in L^2(\Omega).
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top