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Let $f_n,f \in L^p$ and $f_n \to f$ uniformly. Further let $g \in L^q$. Now, clearly $\int (f_n - f) g$ is defined. How do I show that

$$\int |f_n - f||g|$$

can be made arbitrary small if I choose $n$ large enough? If we just naively apply Hölder we will not be able to make it small but just finite. Further, we cannot just replace $|f_n - f|$ by the bound "$\epsilon$" because that is larger than this. Does someone have a hint?

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Try choosing a set $A$ of finite measure for which $\int_{A^c} |g|^q$ is small. –  Nate Eldredge Nov 21 '10 at 14:44

1 Answer 1

up vote 6 down vote accepted

Outside of a sufficiently large compact set, $\|g\|_q <\epsilon$. Use Hölder naively, and on that portion the integral is bounded above by $(\|f_n\|_p + \|f\|_p)\epsilon$. On the inside of the compact set $\|g\|_1 \leq C \|g\|_q$ where $C$ is given by the volume of the compact set. Then you can replace $|f_n - f|$ by $\epsilon$.

Note that this requires $\|f_n\|_p$ be uniformly bounded, which would be true if you additionally assume $f_n \to f$ in $L^p$. If you don't assume that:

Consider over $\mathbb{R}$, take $f_n = \frac1n \chi_{[2^n,2^{n+1}]}$, where $\chi$ denotes the characteristic function. $f_n \to 0$ uniformly as a sequence of functions (i.e. $f_n\to f$ in $L^\infty$). For this sequence $\|f_n\|_p = \frac1n \cdot 2^{n/p} \nearrow \infty$ (unless $p = \infty$). (Note that since it is not a bounded sequence, you cannot assert that it has a weakly converging subsequence.)

Now take $g$ be the function such that $g(x) = 2^{-n/(q-\delta)}$ if $2^n < |x| \leq 2^{n+1}$. By construction it is a function in $L^q$.

But $|f_n - f| |g| = \frac1n \cdot 2^{-n / (q - \delta)} \chi_{[2^n,2^{n+1}]}$. If you integrate it, you get

$$ \frac1n \cdot 2^{n( 1 - \frac{1}{q - \delta}) }$$

So unless $q = 1$ and $p = \infty$, you can always choose $\delta$ sufficiently small such that the exponent in the integrated expression is positive, and provide a counterexample to what you want proved. (And of course, in the case where $p = \infty$, you can just replace $|f-f_n|$ by $\epsilon$ and be done with it.

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Thanks. I got the case where the $f_n$ are uniformly bounded in $L^p$ but not where this is not true. Is it possible to give an intuitive meaning why it should be false for the $f_n$ not uniformly bounded in $L^p$? –  Jonas Teuwen Nov 21 '10 at 15:54
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Hum, that's a good question. It has to do with the interplay between compactness of the domain and the compactness inside the function space. See, if instead of $\mathbb{R}^d$, you worked on a space with finite total measure, then uniform convergence in $L^\infty$ will imply boundedness in $L^p$ also. This means that any possible modes of failure must take advantage of the fact that the support of $f_n$ can run off to infinity. –  Willie Wong Nov 21 '10 at 16:01
    
Now, if $f_n$ are uniformly bounded, and their supports run off to infinity, then they converge weakly to zero, which means that what you want is also true. So necessarily a counterexample will require $f_n$ to be unbounded. As to an intuitive meaning: I guess one way to think about it is that while on sets of finite measure, the topology given by $L^p$ is finer than that of $L^q$ if $p > q$, on sets of infinite measure, it is possible that the topology given by $L^p$ and $L^q$ be not comparable for $p\neq q$. –  Willie Wong Nov 21 '10 at 16:06
    
(Sorry for the poor explanation. The extent to which this is intuitive to me is just that there are certain lack of inclusions between Lebesgue spaces on non-compact domains, and a bit of concentration compactness for sequences in Lebesgue spaces. Which is to say, not very intuitive at all.) –  Willie Wong Nov 21 '10 at 16:10

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