Holder inequality for double integrals

[sarrah1]

Hi

I have a double integral

$\iint(f(x) g(x,y)dxdy$ over $x\in[a,y]$ , $y\in[a,b]$

and I wish to bound the integral in terms of integral f times integral g.
I suppose there must exist a form of holder inequality to do that ?
many thanks
Sarrah

 

[jvicens]

Hi Sarrah,

Yes, there is indeed a form of Holder's inequality that can help you bound your double integral in terms of the individual integrals of $f$ and $g$. Holder's inequality states that for two functions $f$ and $g$ and their respective integrals $I_f$ and $I_g$, the following holds true:

$\iint(f(x)g(x,y)dxdy \leq (\int f(x)dxdy)^{\frac{1}{p}}(\int g(x,y)dxdy)^{\frac{1}{q}}$

where $p$ and $q$ are positive real numbers such that $\frac{1}{p} + \frac{1}{q} = 1$. This is known as the Holder's inequality with $p$ and $q$ conjugates.

In your case, you can use this inequality with $p=q=2$ to get the following bound:

$\iint(f(x)g(x,y)dxdy \leq (\int f(x)dxdy)^{\frac{1}{2}}(\int g(x,y)dxdy)^{\frac{1}{2}}$

Since you have specified that $x\in[a,y]$ and $y\in[a,b]$, you can rewrite the individual integrals as follows:

$\int f(x)dxdy = \int_a^b \int_a^y f(x)dxdy$

$\int g(x,y)dxdy = \int_a^b \int_a^y g(x,y)dxdy$

Substituting these into the inequality, you get the desired bound:

$\iint(f(x)g(x,y)dxdy \leq (\int_a^b \int_a^y f(x)dxdy)^{\frac{1}{2}}(\int_a^b \int_a^y g(x,y)dxdy)^{\frac{1}{2}}$

Hope this helps! Let me know if you have any further questions.