Can Uniformly Bounded Functions Converge Weakly to Zero in $\mathscr{L}^p$?

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  • Thread starter Euge
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    2016
In summary, a function is uniformly bounded if its absolute value is always less than or equal to a constant for all points in its domain. Weak convergence in $\mathscr{L}^p$ means that a sequence of functions converges to a limit function in the $\mathscr{L}^p$ space, but not necessarily pointwise. It is possible for uniformly bounded functions to converge weakly to zero in $\mathscr{L}^p$. The conditions for weak convergence in $\mathscr{L}^p$ are that the limit function must be in the $\mathscr{L}^p$ space and the sequence of functions must be uniformly bounded and have a finite $\mathscr{L}^p
  • #1
Euge
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Here is this week's POTW:

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Let $1 < p < \infty$, and let $(f_n)$ be a sequence of real-valued functions in $\mathscr{L}^p(-\infty, \infty)$ which converges pointwise a.e. to zero. Show that if $\|f_n\|_p$ is uniformly bounded, then $(f_n)$ converges weakly to zero in $\mathscr{L}^p(-\infty,\infty)$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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  • #2
No one answered this week's problem. You can read my solution below.
Let $q$ be the Hölder conjugate of $p$. Take $g\in \mathscr{L}^q(-\infty,\infty)$. By density of compactly supported functions in $\mathscr{L}^q$, it suffices to assume $g$ has compact support, and $\int f_n g \to 0$ as $n\to \infty$.

Suppose $g$ is supported on a compact set $K$. By Egorov's theorem, given $\epsilon > 0$ there exists a measurable subset $F$ of $K$ with $m(K\setminus F) < \epsilon$ such that $f_n \to f$ uniformly on $F$. Thus
$$\int_F \lvert f_ng\rvert \le \left(\sup_{x\in F} \lvert f_n(x)\rvert\right)\int_F\lvert g\rvert \quad \text{and}\quad \int_{K\setminus F} \lvert f_n g\rvert \le M\|g\|_\infty\epsilon^{1/q}$$
where $M = \sup_n\|f_n\|_p$. Hence
$$\varlimsup_{n\to \infty} \int \lvert f_n g\rvert \le 2M\|g\|_\infty \epsilon^{1/q}$$
Since $\epsilon$ was arbitary, the $\mathscr{L}^p$-weak limit of $f_n$ is zero.
 

1. What does it mean for a function to be uniformly bounded?

A function is uniformly bounded if there exists a constant M such that the absolute value of the function is always less than or equal to M for all points in its domain. In other words, the function is bounded by the same value at all points.

2. What is a weak convergence in $\mathscr{L}^p$?

A weak convergence in $\mathscr{L}^p$ means that a sequence of functions converges to a limit function in the $\mathscr{L}^p$ space, but not necessarily pointwise. In other words, the integral of the absolute value of the difference between the limit function and the sequence of functions approaches zero as the sequence approaches infinity.

3. Can uniformly bounded functions converge weakly to zero in $\mathscr{L}^p$?

Yes, it is possible for uniformly bounded functions to converge weakly to zero in $\mathscr{L}^p$. This means that the sequence of functions approaches the zero function in the $\mathscr{L}^p$ space, but not necessarily pointwise.

4. What are the conditions for weak convergence in $\mathscr{L}^p$?

In order for a sequence of functions to converge weakly to a limit function in $\mathscr{L}^p$, the limit function must be in the $\mathscr{L}^p$ space and the sequence of functions must be uniformly bounded and have a finite $\mathscr{L}^p$ norm.

5. Is weak convergence in $\mathscr{L}^p$ the same as pointwise convergence?

No, weak convergence in $\mathscr{L}^p$ is not the same as pointwise convergence. While pointwise convergence means that a sequence of functions converges to a limit function at every point in the domain, weak convergence only requires that the integral of the absolute value of the difference between the limit function and the sequence of functions approaches zero.

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