0
$\begingroup$

Let $f_k:X \to B$ be a sequence of functions defined on an arbitrary space $X$ and taking values in a Banach space $B$. Suppose there exist a set $A_n \subset X$ such that $|f_{k+1}(x)-f_k(x)|<\frac{1}{2^k}$ for all $k\geq n$ and $x\in X\setminus A_n$. The exercise asks to show that $(f_k)$ converges absolutely and uniformly on $ X\setminus A_n$.

I did the following:

$\sum_{k=n}^{\infty} |f_{k+1}(x)-f_k(x)| \leq \sum_{k=n}^{\infty} \frac{1}{2^k}=\frac{1}{2^{n-1}} $ for all $x\in X\setminus A_n$, so the series $\sum_{k=n}^{\infty} (f_{k+1}-f_k)$ is absolutely convergent on $ X\setminus A_n$. Since $B$ is complete, the series is also convergent on $ X\setminus A_n$. It is also uniformly convergent because

$$\bigg|\sum_{k=n}^{m} (f_{k+1}-f_k)- \sum_{k=n}^{\infty} (f_{k+1}-f_k)\bigg|= \bigg|\sum_{k=m+1}^{\infty} (f_{k+1}-f_k)\bigg|\leq \sum_{k=m+1}^{\infty} |f_{k+1}-f_k|\leq \frac{1}{2^{m}}\to 0 \text{ as }m\to\infty$$

uniformly on $X\setminus A_n$. Then we note that $m> n$ implies $$f_m=f_m-f_n+f_n=\sum_{k=n}^{m-1} (f_{k+1}-f_k) +f_n\to \sum_{k=n}^{\infty} (f_{k+1}-f_k) + f_n$$ as $m\to \infty$ on $X\setminus A_n$, and so the uniform convergence of $\sum_{k=n}^{m} (f_{k+1}-f_k)$ on $X\setminus A_n$ implies the uniform convergence of $(f_k)$ on $X\setminus A_n$. Finally, since the norm $|\cdot|$ is continuous, we also have that $|f_k|$ converges on $X\setminus A_n$.

Is this proof ok? Am I missing something? Thanks a lot for your feedback.

$\endgroup$
0

Your Answer

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

Browse other questions tagged or ask your own question.