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?
Prove that $\lim \limits_{k\rightarrow \infty} \int \limits_E f_k g = \int \limits_E fg$
0
$\begingroup$
real-analysis
analysis
measure-theory
lebesgue-integral
-
0It makes much more sense to write the difference of integrals as $\int_E(f_k-f)g$ first. – 2012-11-12
1 Answers
2
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 $$ \lim_k\int_Ef_kg=\int_Efg. $$
-
0So 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 $0$. – 2012-11-12
-
1It'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 $k>k_0$. – 2012-11-12
-
1Regarding 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 $\|f_k-f\|_p\to0$. – 2012-11-12
-
0I understand. All I know about $f, f_k$ is that both $f, f_k \in L ^p(E)$ – 2012-11-12
-
0But 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 $k$, $\int_Efg=1$. – 2012-11-12
-
0Would it change anything if I said $$\lim \limits_{k \rightarrow \infty} \parallel f_k - f \parallel_{p,E} = 0$$? – 2012-11-13
-
0Of course! That's exactly what I assumed for my answer. – 2012-11-13
-
0Gosh, I apologize for the trouble. Thanks for the help! – 2012-11-13
-
0You are welcome! – 2012-11-13