Sequence, Lebesgue Space

• Dec 13th 2009, 02:07 PM
eskimo343
Sequence, Lebesgue Space
Show that if $\displaystyle || f_n ||_1 \leq 2^{-n}$ for every $\displaystyle n \geq 1$ then $\displaystyle (f_n)_n$ converges to zero a.e.

If $\displaystyle (f_n)_n$ is a sequence in $\displaystyle L^1[0, 1]$ is such that $\displaystyle \sum_{n=1}^{\infty} || f_n ||_1 < \infty$ then $\displaystyle \sum_{n=1}^{\infty} | f_n(s) | < \infty$ for almost every $\displaystyle s \in [0, 1]$.

However, I don't see how to show this. I would appreciate some hints on how to proceed.
• Dec 15th 2009, 09:32 AM
putnam120
Quote:

Originally Posted by eskimo343
Show that if $\displaystyle || f_n ||_1 \leq 2^{-n}$ for every $\displaystyle n \geq 1$ then $\displaystyle (f_n)_n$ converges to zero a.e.

If $\displaystyle (f_n)_n$ is a sequence in $\displaystyle L^1[0, 1]$ is such that $\displaystyle \sum_{n=1}^{\infty} || f_n ||_1 < \infty$ then $\displaystyle \sum_{n=1}^{\infty} | f_n(s) | < \infty$ for almost every $\displaystyle s \in [0, 1]$.

However, I don't see how to show this. I would appreciate some hints on how to proceed.

Not true, take $\displaystyle f_n(x)=1$ if $\displaystyle x=0$ and $\displaystyle f_n(x)=0$ otherwise. Then $\displaystyle \sum\parallel f_n\parallel_1<\infty$ but $\displaystyle \sum|f_n(0)|=\infty$
• Dec 15th 2009, 10:46 AM
Opalg
Quote:

Originally Posted by putnam120
Not true, take $\displaystyle f_n(x)=1$ if $\displaystyle x=0$ and $\displaystyle f_n(x)=0$ otherwise. Then $\displaystyle \sum\parallel f_n\parallel_1<\infty$ but $\displaystyle \sum|f_n(0)|=\infty$

That does not contradict the assertion that $\displaystyle \sum_{n=1}^{\infty} | f_n(s) | < \infty$ for almost every $\displaystyle s \in [0, 1]$.

Quote:

Originally Posted by eskimo343
Show that if $\displaystyle || f_n ||_1 \leq 2^{-n}$ for every $\displaystyle n \geq 1$ then $\displaystyle (f_n)_n$ converges to zero a.e.

Let $\displaystyle g_N = \sum_{n=N+1}^\infty|f_n|$. Then $\displaystyle \|g_N\|_1\leqslant\sum_{n=N+1}^\infty\|f_n\|_1\leq slant2^{-N}$ (sum of geometric series). Define $\displaystyle S_N = \{x:g_N(x)\geqslant2^{-N/2}\}$. Then $\displaystyle \mu(S_N)\leqslant2^{-N/2}$, where $\displaystyle \mu$ denotes the measure.

If $\displaystyle x\notin S_N$ then $\displaystyle |f_n(x)|<2^{-N/2}$ for all n>N. Therefore if $\displaystyle x\notin\limsup_{N\to\infty}S_N$ then $\displaystyle f_n(x)\to0$ as $\displaystyle n\to\infty$. But $\displaystyle \limsup_{N\to\infty}S_N$ is a null set.