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(Background definitions)

$A(\mathbb{T})$ is the space of all $2\pi$ periodic functions $f$ such that $\sum\limits_{k=-\infty}^{\infty}|\widehat{f}(k)| < \infty$. It is a normed space when normed by $\|\cdot\|_{A(\mathbb{T})} = \sum\limits_{k=-\infty}^{\infty}|\widehat{f}(k)|$.

Also, $\widehat{f}(k) = \int\limits_{0}^{2\pi}f(t)e^{-ikt}dt$.

I have a homework problem which says the following:

Let $f_{n}\in A(\mathbb{T})$ for each $n\geq 1$, and let $\|f_{n}\|_{A(\mathbb{T})}\leq 1$ for each $n\geq 1$. Assume $f_{n}\to f$ in $\|\cdot\|_{\infty}$.

a) Show that $f\in A(\mathbb{T})$ and $||f||_{A(\mathbb{T})}\leq 1$.

b) Show that the given hypothesis are not enough to guarantee that $\|f_{n} - f\| _{A(\mathbb{T})}\rightarrow 0$.

My progress:

I have solved the first part by showing that each partial sum $\sum\limits_{k=-n}^{n}|\widehat{f}(k)|\leq 1 + \epsilon$ for every $\epsilon > 0$.

But this has given me little insight for the second problem. Every example I come up with satisfies $\|f_{n} - f\| _{A(\mathbb{T})}\rightarrow 0$. Even as I am writing this it has just occured to me that I've only considered continuous functions. So I will try some discontinuous ones next but so far this question has completely eluded me.

Any suggestions of what type of functions to focus on?

If I take $f_{n} = \frac{1}{n}$, then $f_{n}\rightarrow 0$ in $\|\cdot\|_{\infty}$ and $\|\cdot\|_{A(\mathbb{T})}$.

If I take $f_{n} = \frac{(-1)^{n}}{n}$, then then $f_{n}\rightarrow 0$ in $\|\cdot\|_{\infty}$ and $\|\cdot\|_{A(\mathbb{T})}$.

If I take $f_{n} = \sum\limits_{j=1}^{n}\frac{(-1)^{j}}{j}$, then $f_{n}\rightarrow 0$ in $\|\cdot\|_{\infty}$ and $\|\cdot\|_{A(\mathbb{T})}$.

If I take $f_{n} = \sum\limits_{j=1}^{n}\frac{1}{j}$, then $f_{n}$ diverges in any norms.

In fact I think that a sequence of constant functions converges to $f$ in $\|\cdot\|_{\infty}$ if and only if it converges to $f$ in $\|\cdot\|_{A(\mathbb{T})}$.

So in conclusion, I am back where I started.

NOTE: I had asked this question previously but didn't get any real answer yet. But since the question was technically "answered", there were no more views. So I deleted/reposted. I hope this is OK.

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    @t.b., I'm inclined to see whether Kyle wants the question undeleted before I vote to do so. I don't feel strongly enough in favor of undeletion to just do it. But thanks for pointing out the option.2011-10-24

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The series $\sum_{k=1}^\infty\frac{2}{k}\sin(k\,t)$ converges pointwise to the function $ f(x)=\pi-x \text{ if }0 Althoug the convergence is not uniform on $[\,0,2\,\pi\,]$, the partial sums are uniformly bounded. Let $ f_m=\frac{1}{\log m}\sum_{k=1}^m\frac{2}{k}\sin(k\,t)=\frac{1}{i\,\log m}\sum_{k=-m,k\ne0}^m\frac{e^{ikt}}{k},\quad m>1. $ Then $f_m\in A(\mathbb{T})$ and $ \lim_{m\to\infty}\|f_m\|_{A(\mathbb{T})}=\lim_{m\to\infty}\frac{1}{\log m}\sum_{k=1}^m\frac{2}{k}=2, $ but $ \lim_{m\to\infty}\|f_m\|_{L^\infty(\mathbb{T})}=\lim_{m\to\infty}\frac{1}{\log m}\Bigr\|\sum_{k=1}^m\frac{2}{k}\sin(k\,t)\Bigr\|_{L^\infty(\mathbb{T})}=0. $

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    Sorry for taking so long to accept. This is a very nice example. And ignore my comment a$b$out the essenti$a$l supremum; it doesn't make sense.2011-10-28