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Let ${\{f_n\}}_{n = 1}^{\infty}$ be a function sequence, where $f_n = {\chi}_{[n , n + 1]}$ for all $n = 1 , 2 , \ldots$. Does ${\{f_n\}}_{n = 1}^{\infty}$ converge in $L^p$, $p \in [1 , \infty)$? And in $L^{\infty}$? I have just thought that $f_n \in L^p \cap L^{\infty}$ for all $n = 1 , 2 , \ldots$, but I haven't found any $f \in L^p$ and $g \in L^{\infty}$ such that $$ \lim_{n \to \infty} {\|f_n - f\|}_p = 0 \qquad \mbox{ and } \qquad \lim_{n \to \infty} {\|f_n - g\|}_{\infty} = 0\mbox{.} $$ Does exist such functions $f$ and $g$? Thank you.

4 Answers 4

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Since $\|f_{n+1}-f_n\|_p = 2^{1/p}$ for all $n$ (even when $p=\infty),$ $(f_n)$ cannot be Cauchy in $L^p,$ hence cannot converge in $L^p.$

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    Very neatly done!2017-01-20
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If you want to find a candidate function for convergence, you need to ask yourself this: given $x \in \mathbb{R}$ what happens to $f_n(x)$ as $n \to \infty$. If $x=1$ for instance $f_1(1) = 1$, $f_2(1)=0$, etc.

Another question to ask is if $\{f_n\}$ is a Cauchy sequence in $L^\infty$ or $L^p$. Remember that every convergent sequence is a Cauchy sequence. That is, given $\epsilon >0$, is there an $N \in \mathbb{N}$ such that for all integers $n,m > N$ we have $\|f_n - f_m\|_p < \epsilon$ (or $\|f_n - f_m\|_\infty < \epsilon$)?

In particular, what is $\|f_1 - f_2\|_\infty$ or more generally $\|f_n - f_m\|_\infty$? This should give you a good idea of what's going on.

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Hint: If $f_n$ converges in $L^p(0, \infty)$ to $f$, then the restriction $f_n|_{K}$ also converge in $L^p$ to $f|_{K}$. Set $K= [0,L]$, it is clear then what $f$ would be.

Then check if $f_n$ really converges in $L^p$ to that $f$.

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We have $f_n \to 0$ pointwise. Hence, if $(f_n)$ is to converge in $L^{\infty}$, it must converge to $0$, but this is clearly impossible. Also, if $(f_n)$ converges to some $f$ in $L^p$, there's a subsequence of it converging to $f$ pointwise a.e., so $f = 0$. But this is clearly impossible.