The Unapologetic Mathematician

Mathematics for the interested outsider

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<c_n 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<c_n. 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 - Posted by | Analysis, Functional Analysis, Measure Theory

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