The Limit of Integrals Is the Integral of the Limit

**ragnar** · August 1st 2010, 06:14 PM

I am asked to prove that, if $\displaystyle \lim_{n \rightarrow \infty} ||f_{n} - f|| = 0$ and $\displaystyle \lim_{n \rightarrow \infty} ||g_{n} - g|| = 0$ , prove that $$\displaystyle \lim_{n \rightarrow \infty} \int_{I} f_{n} \cdot g_{n} = \int_{I}f \cdot g$.$

I doubt that I want to do this by brute force by providing some increasing sequence of upper functions that approach $f \cdot g$ , though maybe I do since I know that $\displaystyle \lim_{n \rightarrow \infty} ||f_{n}|| = ||f||$ and likewise for $g$ .

In any case, the alternative seems to be to use to the Dominated Convergence Theorem, in which case I want to show that $\{ f_{n} \cdot g_{n} \}$ converges almost everywhere (presumably to $f \cdot g$ ), and that there is a non-negative function which is integrable and dominates $|f_{n} \cdot g_{n}|$ . But the first part seems like it might be false, since I don't know that the interval over which these functions are integrated is bounded.

Maybe I want to try an epsilon-delta argument?

**Vlasev** · August 1st 2010, 09:31 PM

What about uniform convergence? It seems that you are not given enough information though.

**ragnar** · August 2nd 2010, 04:34 AM

Nope, I'm not given that the functions are uniformly convergent, just that they are in L(I) and measurable.

**Jose27** · August 2nd 2010, 03:51 PM

Originally Posted by ragnar

I am asked to prove that, if $\displaystyle \lim_{n \rightarrow \infty} ||f_{n} - f|| = 0$ and $\displaystyle \lim_{n \rightarrow \infty} ||g_{n} - g|| = 0$ , prove that $$\displaystyle \lim_{n \rightarrow \infty} \int_{I} f_{n} \cdot g_{n} = \int_{I}f \cdot g$.$

I doubt that I want to do this by brute force by providing some increasing sequence of upper functions that approach $f \cdot g$ , though maybe I do since I know that $\displaystyle \lim_{n \rightarrow \infty} ||f_{n}|| = ||f||$ and likewise for $g$ .

In any case, the alternative seems to be to use to the Dominated Convergence Theorem, in which case I want to show that $\{ f_{n} \cdot g_{n} \}$ converges almost everywhere (presumably to $f \cdot g$ ), and that there is a non-negative function which is integrable and dominates $|f_{n} \cdot g_{n}|$ . But the first part seems like it might be false, since I don't know that the interval over which these functions are integrated is bounded.

Maybe I want to try an epsilon-delta argument?

Actually you only need that one of the sequences converges in norm, the other can converge weakly and the result still holds (assuming you're working in $L^2$ , otherwise the products $f_ng_n$ and $fg$ may not be integrable):

$\displaystyle| \int_I f_ng_n - \int_I fg | \leq |\int_I f_ng_n - \int_I f_ng | +| \int_I f_ng - \int fg| \leq \| f_n\| \| g_n-g\| + | \int_I f_ng - \int_I fg | \rightarrow 0$ .

Notice then that we only require $f_n$ to coverge weakly to $f$ and $g_n$ to converge in norm. The proof also shows that th result holds if $f_n,f \in L^p$ and $g_n,g\in L^q$ with $\frac{1}{p} + \frac{1}{q} =1$ and $1<p,q<\infty$ .

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