Hi Abdelhak,

It might be a stupid question but what measure do you consider when defining your L2 space ? (Lebesgue) , and what set do you consider (R) ? I assume hereunder you take L2mu(R,R).

Anyway, if you take a function like this : t^2/sqrt(1+x^2) , This function is, for all t, in L2(R), see integration in arctan. But the mean cannot be defined (using Lebesgue measure on R), whatever x chosen.

Please correct me if I did not understand what you are trying to explain.

Best regards !

Sounds good, but how to prove it?  First, we need to know what is the relation between g(x,t) and G(x).  Not in words, but expressed as a precise formula as I have no idea what is meant by "averaqe" in this context.  This formula should cover the "obvious" case g(x,t)=const(x) as well.

Read the attached file please.

Ok, so if I understand this correctly, you work with bounded open sets.

And what is L^2 is the function x -> g(x,t). If you consider an example ]t_1,t_2[ = ]0,1[, and the function g(x,t) = x/t . It is clearly in L^2(\Omega), with Omega an open bounded set included in R,  However, the mean (which is not really a mean by the way) cannot be defined as the function (x,t) -> x/t  integral in t when t in the neighborhood of 0 have no sense.

So, it seems that you have to add some assumptions to make it work ! (on the sets your consider and/or the function t -> g(x,t) for all x)

Best regards, 

I don't know what is the regularity of g(t,x) with respect to the time, but, if you can get the sup of g in t, then 

$\int_{\Omega} G(x)^{2} dx = \int_{\Omega} (\int_{t_{1}^{t_{2}} g(t,x) dt)^{2} dx 

\le (t_{2}-t_{1})^{2} \int_{\Omega} sup_{t \in [t_{1}, t_{2}]} |g(t.x)|^{2} dx < \infty$, since $g \in L^{2}(\Omega)$. Then yes!

If you give your question and problem  (you face) in a simplified language ,it  will not be very  difficult to give my suggestions. Perhaps something is in your mind ,specify it 

clearly

Prof Rath

Assuming the measure of \Omega is positive and finite, then a simple sufficient condition for the average $G(x)=[1/mes(\Omega}]\int_{\Omega} g(x,t) dx$ to be in L^2(t_1,t_2) is that g should be in L^2(\Omega\times(t_1,t_2))$.  [This condition can be weakened, but this seems simplest.]

Using your definition of G (G(x) = \int_t0^t1 g(x,s) ds) you have by the Cauchy-Schwartz inequality

G^2(x) \le (t1-t0) \int_t0^t1 g^2(x,s) ds

hence

\int_Omega G^2(x) d\Omega \le (t1-t0) int_\Omega \int_t0^t1 g^2(x,s) ds d\Omega

You need this to be bounded for G to be in L_2(\Omega). Changing the order of integration (justification left as an exercise :) ) you see that you need the L_2(\Omega)  norm of g to be in L_2(t0,t1) as a function of  t.

Much like Thomas Seidman said. But I think you cannot weaken this condition.

To say the condition g\in L^2(\Omega\times(t_0,t_1)) "cannot be weakened" would just be to say that it is necessary as well as sufficient.  An easy counterexample to that is to look at g(x,t) of the form a(x)b(t) with nontrivial a having average 0 and b not in L^2(t_0,t_1).

Heh, nice catch, Sir. But since Cauchy-Schwartz is sharp for constants, I think you'd be safe to restrict your attention to functions  g(x,t)  with the property that \int_\Omega g d\Omega  \neq 0 on a set S \subset (t0,t1) of measure >0.