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Stuck with this problem from Zgymund's book.

Suppose that $f_{n} \rightarrow f$ almost everywhere and that $f_{n}, f \in L^{p}$ where $1<p<\infty$. Assume that $\|f_{n}\|_{p} \leq M < \infty$. Prove that:

$\int f_{n}g \rightarrow \int fg$ as $n \rightarrow \infty$ for all $g \in L^{q}$ such that $\dfrac{1}{p} + \dfrac{1}{q} = 1$.

Right, so I estimate the difference of the integrals and using Hölder end up with:

$$\left|\int f_{n} g - \int fg\right| \leq \|g\|_{q} \|f_{n} - f\|_{p}$$

From here I'm stuck because we are not assuming convergence in the seminorm but just pointwise convergence almost everywhere. How to proceed?

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1  
@Sivaram: but by pointwise convergence almost everywhere don't they mean convergence with respect the Euclidean norm (and not the seminorm?) perhaps that's my mistake. Here $f,f_{n}$ are real-valued so it makes sense to take the Euclidean norm. –  student Nov 20 '10 at 0:15
    
Take the witch hat (triangle) for $f_n$, $f = 0$. Take $g = 1$ on $[0,1]$ and $0$ otherwise, then $\int f_n g = \int f_n = 1 \not \to \int f = 0$, what am I missing? I think you need to get a dominating function from somewhere. –  Jonas Teuwen Nov 20 '10 at 0:19
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I am sorry. I am wrong. Almost sure convergence doesn't mean convergence in $p^{th}$ mean. Sorry for my wrong comment. –  user17762 Nov 20 '10 at 0:21
    
Isn't $\|f_n\|_p<\infty$ redundant? It's already given that $f_n\in L^p$. Do you mean $\sum_n \|f_n\|_p<\infty$ –  AgCl Nov 20 '10 at 0:24
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The OP asked 25 questions with all being answered, but accepted none of these answers! –  Paul Dec 9 '11 at 4:53

4 Answers 4

HINT

By Egorov's theorem, convergence a.e. implies for every $\epsilon$ there exists $B$ with $|B| < \epsilon$ such that $f_n\to f$ uniformly on $X\setminus B$ (where $X$ is "almost" the whole space).

Split $\int (f_n - f)g$ in two pieces, one over $B$ and one over $X\setminus B$. On $X\setminus B$ uniform convergence implies the integral can be made as small as you want. Holder's inequality implies the integral on $B$ is controlled by $(M + \|f\|_p)\|g\|_{L^q(B)}$. Taking $B\searrow$ a measure zero set, then the integral on $B$ of $g$ goes to zero.

Finish by taking a diagonalizing sequence as usual.

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@ Willie: Why doesn't Jonas T example work? –  user17762 Nov 20 '10 at 0:58
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Food for thought for the interested reader Why does the proof not work for $p = 1$ and $p = \infty$? –  Willie Wong Nov 20 '10 at 1:00
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So while it is true that $|f_n|$ can diverge, the restriction on $\|f_n\|_p \leq M$ guarantees that it cannot diverge too fast on smaller and smaller sets. In fact, this points to a different way of proving the result: you can use Chebyshev's inequality which bounds the size of some set $A$ over which $|f_n|$ is allowed to be large, instead of Egorov's theorem. –  Willie Wong Nov 20 '10 at 1:14
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Pardon the ignorance, but what is a witch hat? (googling witch hat doesn't really get anywhere, and it's a few weeks past halloween..) –  user1736 Nov 20 '10 at 3:02
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@user1736: The Witch's hat is the following triangle with vertices (0,0), (1/(2n), 2n), (1/n, 0). –  Jonas Teuwen Nov 20 '10 at 11:24

Here is a proof which is not based on Egoroff's theorem. As Jonas T points out in an other answer, Fatou's Lemma imply $$\int|f|^p=\int\lim |f_n|^p =\int\liminf |f_n|^p \le \liminf\int |f_n|^p \le M^p$$ and hence $f\in L^p$.

At this point we may assume $f=0$ (consider the problem with $f_n$ replaced by $f_n-f$), and we must show $\int f_ng\to 0$.

Fix a number $A\gt0$, and let $E_n=\{x: |f_n(x) g(x)|\le A|g(x)|^q\}$ where $q=p/(p-1)$ (the conjugate to $p$). By dominated convergence we have $$\int_{E_n}f_ng\,dx\to0$$ On the complement, $E_n^c =X-E_n$, we have $|g|^{q}\le A^{-1}|f_n g|$. Hence, by Hölder's inequality $$\int_{E_n^c}|f_ng|\,dx\le\left(\int_{E_n^c} |f_n|^p\right)^{1/p}\left(\int_{E_n^c} |g|^q\right)^{1/q}\le M A^{-1/q}\left(\int_{E_n^c}|f_ng|\,dx\right)^{1/q}$$ In other words $$\left(\int_{E_n^c}|f_ng|\,dx\right)^{1-1/q}\le MA^{-1/q}$$ that is $$\limsup_n\int_{E_n^c}|f_ng|\,dx\le M^pA^{-p/q}.$$ In total we have shown $$\limsup_n\int|f_ng|\,dx\le M^pA^{-p/q}$$ for all $A\gt0$, from which be conclude $$\lim_n\int|f_ng|=0.$$

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Nice, +1 for a proof without Egoroff. –  Jonas Teuwen Nov 21 '10 at 15:20

I'm not really convinced by Willie Wong's argument, although I don't doubt that it is correct, but I cannot completely fill in the details (especially the Egoroff part) so I made my own (inspired by his).

First note that the hypothesis that $f \in L^p$ is redundant by Fatou's lemma. Now note that we could as well consider $f_n \to 0$ weakly (the conclusion actually means weak-convergence) by linearity.

Now note that: \begin{align*} F_n &: L^q \to \mathbf{K}\\ g & \mapsto \int f_n g. \end{align*} defines a bounded operator linear operator on $L^q$ by Hölder with operatornorm smaller or equal to $M$.

Remember that the stepfunctions are dense in $L^q$. So by linearity and continuity it is sufficient to consider indicatorfunctions $1_A$ with $A$ having finite measure. Now we have a finite measure space so we can use Egoroff.

We want to show that

$$\int_A f_n \to 0.$$

So, now find a $B$ with $|B| \leq \epsilon$ such that $f_n$ converges uniformly to $0$ on $A \setminus B$.

So, for sufficiently large $n$ we have that

$$\int_{A \setminus B} |f_n| \leq \epsilon |A \setminus B| \leq \epsilon |A|.$$

And,

$$\int_{B} |f_n| \leq |B|^{1/q} \|f_n\|_p \leq \epsilon^{1/q} \|f_n\|_p.$$

KTHXBYE.

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This is just another answer trying to produce a set of finite measure to apply Egorov's Theorem to the sequence ${f_k}$.

You are using the book Measure and Integral by Richard L. Wheeden and Antoni Zygmund, me too, so I will refer some parts of that book.

Lemma 1. Suppose that $f\in L(\mathbb{R}^d)$, then $$\lim_{n\to\infty}\int \vert f\chi_{B(0,n)^c} \vert=0,$$ where $B(0,n)=\{\mathbf{x}\in \mathbb{R}^d:\Vert \mathbf{x}\Vert\lt n\}$ and the superscript $c$ is for complement.

Proof. Consider the sequence $\{f_k\}$ given by $$f_k=f\chi_{B(0,k)^c}.$$ Then $\vert f_k\vert\searrow 0$ and $\vert f_k\vert\leq \vert f\vert$ since $\vert f_k\vert\to 0$ as $k\to\infty$, and $\vert f\vert\in L$ by the Monotone Convergence Theorem (5.32(ii)) we get the desired result.

Lemma 2. Let $f\in L(E)$. Given $\epsilon\gt 0$, there exist $\delta \gt 0$ s.t. if $A\subseteq E$ and $m(A)\lt \delta$ then $$\int_A f\lt \epsilon.$$

The above Lemma is just the Theorem (7.1). The background for this theorem is at the beginning of the Chapter 7.

Solution to the exercise. Let $g\in L^q$. First note that if $\Vert g\Vert_q=0$ then $g=0$ a.e. and there is nothing to prove. Suppose that $\Vert g\Vert_q\gt 0$. Let $\epsilon\gt 0$. As a consequence of Fatou's Lemma, and Minkowski's Inequality, we get $\Vert f-f_k\Vert_p\leq 2M$.

By the Lemma 1, there exist $N\in \mathbb{N}$ s.t. $$\int_{B(0,N)^c} \vert g\vert^q\lt \left( \frac{\epsilon}{6M} \right)^q.$$ Let $B=B(0,N)$.

By Lemma 2, there exist $\delta\gt 0$ s.t. if $m(A)\lt\delta$ then $$\int_A \vert g\vert^q\lt \left( \frac{\epsilon}{6M} \right)^q.$$

Since $f_k\to f$ pointwise a.e. and $m(B)\lt \infty$, by Egorov's Theorem there exist $E\subseteq B$ such that $f_k\to f$ uniformly in $E$ and $m(B\setminus E)\lt \delta.$

Since $f_k\to f$ uniformly in $E$, there exist a $N'\in\mathbb{N}$ s.t. if $k\geq N'$ then $$\Vert f_k-f\Vert_{\infty,E}\lt \frac{\epsilon}{3\Vert g\Vert_q m(E)}.$$

Then if $k\geq N'$, we get $$\begin{align*} \int \vert f_k-f\vert\vert g\vert &= \int_{E} \vert f_k-f\vert\vert g\vert + \int_{B\setminus E} \vert f_k-f\vert\vert g\vert + \int_{B^c} \vert f_k-f\vert\vert g\vert\\ &= \int (\vert f_k-f\vert\chi_E)\vert g\vert + \int \vert f_k-f\vert\vert g\chi_{B\setminus E}\vert + \int \vert f_k-f\vert\vert g\chi_{B^c}\vert \end{align*}$$ apply Hölder's inequality to the three terms $$\begin{align*} \int \vert f_k-f\vert\vert g\vert &= \Vert (f_k-f)\chi_E\Vert_p\Vert g\Vert_q + \Vert f_k-f\Vert_p\Vert g\chi_{B\setminus E}\Vert_q+\Vert f_k-f\Vert_p\Vert g\chi_{B^c}\Vert_q\\ &\leq \Vert f_k-f\Vert_{\infty,E}m(E)\Vert g\Vert_q + 2M\Vert g\chi_{B\setminus E}\Vert_q+2M\Vert g\chi_{B^c}\Vert_q\\ &\lt \epsilon. \end{align*}$$ All this says that $\Vert f_kg-fg\Vert_1\to 0$.

Note that for $p=1$ the result fails. Take $f_k=k\chi_{[0,1/k]}$ and $g=\chi_{[0,1]}\in L^\infty$. Then $f_k\to 0$, $\Vert f_k\Vert_1=1$ for each $k$, and $\int fg=0$ but $\int f_kg=\int f_k=1$ for each $k$.

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