# Example of a sequence of uniformly but not absolutely convergent functions

(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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I don't think it's OK to delete a question if someone has taken the time to post an answer. If you didn't like the answer, you can leave a comment saying so, and you can make some trivial edit to the question to shoot it back up to the front page and higher visibility. People do read questions, even if they have answers, and they post new answers if OP says what's wrong with the old answers. Well, they do that if they have new answers to post. –  Gerry Myerson Oct 24 '11 at 0:26
I agree with you. It was however only a very minor comment which was originally not posted as an answer. Upon reading the comment I prematurely assumed I understood what the answer will be. Because of this I was so happy I invited the commentator to re post the comment as an answer so I could give him credit for it. When I later realized it actually didn't work, it was too late. I wasn't sure if the question gets removed from the list when it has an answer or not but noticed the view count did not change in about 18 hours. Also the option to add a bounty was removed. –  Kyle Schlitt Oct 24 '11 at 0:37
I didn't do this out of impatience. Nonetheless I see your point and I will not do it again. –  Kyle Schlitt Oct 24 '11 at 0:38
There's no such thing as too late. Having accepted an answer, you can unaccept it. If you give a compelling reason why you felt obliged to unaccept it, there should be no hard feelings (indeed, the answerer should be happy that he has learned something about his answer that he didn't know, although this may be asking a bit much). Having unaccepted and left a soothing and/or self-deprecating comment, you have reduced your dilemma to a problem already solved. It may be possible for a moderator to restore your deleted question, should you wish to go this way. –  Gerry Myerson Oct 24 '11 at 2:48
@Gerry: you can see the deleted question here and vote for undeletion (I just did; I'll vote to close this question after undeletion). –  t.b. Oct 24 '11 at 4:11

The series $\sum_{k=1}^\infty\frac{2}{k}\sin(k\,t)$ converges pointwise to the function $$f(x)=\pi-x \text{ if }0<x<2\pi,\quad 0 \text{ if }x=0,2\,\pi.$$ 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 $$\lim_{m\to\infty}\|f_m\|_{L^\infty(\mathbb{T})}=\lim_{m\to\infty}\frac{1}{\log m}\Bigr\|\sum_{k=1}^m\frac{2}{k}\sin(k\,t)\Bigr\|_{L^\infty(\mathbb{T})}=0.$$