0
$\begingroup$

The following theorem is taken from the book 'An Introductory Course in Functional Analysis' by Bowers and Kalton, page $25.$

Theorem $2.26:$ Let $(\Omega,\Sigma,\mu)$ be a positive measure space. The space $L_{\infty}(\Sigma,\mu)$ is a Banach space when given the essential supremum norm $\| \cdot \|_{\infty}.$

Proof: Again we use Lemme $2.24$ (Cauchy Summability Criteria). Let $(f_k)_{k}$ be a sequence of functions in $L_{\infty}(\Omega,\mu)$ and let $M = \sum_k\| f_k\|_{\infty}.$ Suppose $M<\infty.$ By assummption, for each $k \in \mathbb{N},$ we have $|f_k| \leq \| f_k \|_{\infty}$ a.e. $(\mu).$ Therefore, for each $k \in \mathbb{N},$ there exists a measurable set $N_k$ such that $\mu(N) = 0$ and $|f_k(\omega)| \leq \| f_k\|_{\infty}$ for all $\omega \in \Omega \setminus N_k.$ Let $N = \cup_{k}N_k$. Then $\mu(N) = 0$ and $f_k(\omega) < \infty$ for all $\omega \in \Omega \setminus N$ and all $k \in \mathbb{N}.$ Consequently, $f = \sum_{k}f_k$ exists a.e. and ....

Question: why does $f = \sum_{k}f_k$ exist? Do we need to show that $f$ is the limit of $\sum_{k=1}^nf_k$ in the essential supremum norm?

For first question, I think $f$ exists due to the Weierstrass M-test.

I ask the second question because the authors actually prove it. However, I thought we have proven that $f$ is the limit of $\sum_{k=1}^nf_k$ from the proof above?

$\endgroup$
  • $\begingroup$ The function $f$ is the point wise limit $\sum_k f_k$ however using the Cauchy Summability Criteria you want $\sum_k f_k$ to converge to something using the norm of the space which is the essnetial supremum norm $\endgroup$ – clark Jun 8 '17 at 1:10
1
$\begingroup$

I agree that the authors worded it kind of confusingly. The authors prove that for every $\omega \in \Omega \setminus N$, the sum converges $\sum_{k=1}^\infty f(\omega)$ converges. Thus if you define $f(\omega) = \sum_{k=1}^\infty f_k(\omega)$, the function $f$ is a.e. everywhere defined. Up to this point the authors have only proven that $\sum_{k=1}^n f_k$ converges "pointwise" to $f$. So they still need to an argument to prove that it indeed converges in norm too.

Of course you could avoid this altogether by using the M-test. Since the M-test says this series will not only converge but also converge uniformly. And uniform convergence is equivalent to convergence in the supremum norm.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.