# The Unapologetic Mathematician

## Some Banach Spaces

To complete what we were saying about the $L^p$ spaces, we need to show that they’re complete. As it turns out, we can adapt the proof that mean convergence is complete, but we will take a somewhat different approach. It suffices to show that for any sequence of functions $\{u_n\}$ in $L^p$ so that the series of $L^p$-norms converges

$\displaystyle\sum\limits_{n=1}^\infty\lVert u_n\rVert_p<\infty$

the series of functions converges to some function $f\in L^p$.

For finite $p$, Minkowski’s inequality allows us to conclude that

$\displaystyle\int\left(\sum\limits_{n=1}^\infty\lvert u_n\rvert\right)^p\,d\mu=\left(\sum\limits_{n=1}^\infty\lVert u_n\rVert_p\right)^p<\infty$

The monotone convergence theorem now tells us that the limiting function

$\displaystyle f=\sum\limits_{n=1}^\infty u_n$

is defined a.e., and that $f\in L^p$. The dominated convergence theorem can now verify that the partial sums of the series are $L^p$-convergent to $f$:

$\displaystyle\int\left\lvert f-\sum\limits_{k=1}^n u_k\right\rvert^p\,d\mu\leq\int\left(\sum\limits_{l=n+1}^\infty\lvert u_l\rvert\right)^p\to0$

In the case $p=\infty$, we can write $c_n=\lVert u_n\rVert_\infty$. Then $\lvert u_n(x)\rvert except on some set $E_n$ of measure zero. The union of all the $E_n$ must also be negligible, and so we can throw it all out and just have $\lvert u_n(x)\rvert. Now the series of the $c_n$ converges by assumption, and thus the series of the $u_n$ must converge to some function bounded by the sum of the $c_n$ (except on the union of the $E_n$).

August 31, 2010 -