# Prove that $\lim \limits_{k\rightarrow \infty} \int \limits_E f_k g = \int \limits_E fg$

I want to prove that $$\lim \limits_{k\rightarrow \infty} \int \limits_E f_k g = \int \limits_E fg$$ using Holders inequality. Assume that all of the conditions for Holders inequality are met ($f,g$ are Lebesgue measurable, $E$ is Lebesgue measurable, etc). I know that the norm is defined as $$\parallel f \parallel_{p,E} = \left( \int \limits_E |f|^p \right)^{1/p}$$ So what I have is $$\parallel f_k g \parallel_{1,E} \leq \parallel f_k \parallel_{p,E} \parallel g \parallel_{q,E}$$ Then I can take the limit of both sides $$\lim \limits_{k\rightarrow \infty} \parallel f_k g \parallel_{1,E} \leq \lim \limits_{k\rightarrow \infty} \parallel f_k \parallel_{p,E} \parallel g \parallel_{q,E} = \parallel g \parallel_{q,E} \lim \limits_{k\rightarrow \infty} \parallel f_k \parallel_{p,E}$$ I shown before that $\lim \limits_{k\rightarrow \infty} \parallel f_k \parallel_{p,E} = \parallel f \parallel_{p,E}$ So then I have $$\lim \limits_{k\rightarrow \infty} \parallel f_k g \parallel_{1,E} \leq \parallel f \parallel_{p,E} \parallel g \parallel_{q,E}$$ I can also say that $$\parallel f g \parallel_{1,E} \leq \parallel f \parallel_{p,E} \parallel g \parallel_{q,E}$$ but I don't know if I am on the right track or even how to continue. Any ideas?

• It makes much more sense to write the difference of integrals as $\int_E(f_k-f)g$ first. – fedja Nov 12 '12 at 3:59

Assuming that your hypothesis is that $f_k\to f$ in $L^p$ (you don't say), $$\left|\int_Ef_kg-\int_Efg\right|=\left|\int_E(f_k-f)g\right|\leq\int_E|f_k-f|\,|g|\leq \|f_k-f\|_p\,\|g\|_q \to0,$$ so $$\lim_k\int_Ef_kg=\int_Efg.$$
• So I am trying to understand what you are saying but am having some difficulty. I get that on the left hand side, you are starting with the definition of a limit. Are you then trying to prove that is less than some $\epsilon > 0$. Or by saying that the right handside goes to $0$ then the limit converges? Also, I am not 100% sure how justified that the right hand side goes to $0$. – rioneye Nov 12 '12 at 16:34
• It's an example of what they call the "squeeze property" in Calculus. Proving that your limit exists means showing that the left-most absolute value is arbitrarily small as $k$ grows. So, if you fix $\varepsilon>0$, choose $k_0$ such that for all $k>k_0$ you have $\|f_h-f\|_p<\varepsilon/\|g\|_q$. Then the inequalities in the answer show that $$\left|\int_Ef_kg-\int_Efg\right|<\varepsilon$$ for any $k>k_0$. – Martin Argerami Nov 12 '12 at 16:37
• Regarding your assertion about the right-hand-side, you never included any hypothesis about the $\{f_k\}$ in your question. My assumption was that $f_k\to f$ in $L^p$; this means exactly that $\|f_k-f\|_p\to0$. – Martin Argerami Nov 12 '12 at 16:39
• I understand. All I know about $f, f_k$ is that both $f, f_k \in L ^p(E)$ – rioneye Nov 12 '12 at 16:51
• But then your question makes no sense, or at least there is not the slightest possibility of your limit happening in general. Take $E=[0,1]$, $f=g=1$, $f_k=0$ for all $k$. Then $\int_Ef_kg=0$ for all $k$, $\int_Efg=1$. – Martin Argerami Nov 12 '12 at 18:01