# For a given sequence $(a_k)$, there is no Riemann integrable function f such that $\hat{f}(k) = a_k \forall k$

I'm working out of Stein's Fourier Analysis: An Introduction, and am on chapter 3. There is an exercise that gives us a specific sequence $(a_k)$ and asks us to show that $\sum\nolimits_{k=-\infty}^{\infty} |a_k|^2$ converges but that no Riemann integrable function has $k$th Fourier coefficient equal to $a_k \forall k$

My question is, how do I show the second part? It isn't clear from the text what necessary conditions are on Fourier coefficients, but maybe I'm missing something.

Edit - The sequence in question is

$a_k = \begin{cases} 1/k, & k \geq 1 \\ 0, & k \leq 0 \end{cases}$

$(a_k) \in l^2(\mathbb{Z})$ since the sum $\sum_{k=1}^{\infty} \frac{1}{k^2}$ converges, but I'm not sure about the second part. If $(a_k)$ was the Fourier coefficients for $f$, then $f(x) - S_n(f)(x)$ converges to $0$ in the $L^2$ norm where $S_n = \sum_{k=-n}^{n}a_ke^{ikx} = \sum_{k=1}^{n}f(x)e^{ikx}/k$

Should I show that $f-S_n$ does not converge to 0 in the $L^2$ norm? Or is there some other condition I should be checking?

• If you're dealing with a specific sequence, then actually stating the sequence you're dealing with would be helpfull...
– fgp
Feb 18, 2014 at 1:46
• @fgp I'll do so, but I wanted to talk about the general case rather than just a specific one.
– Lost
Feb 18, 2014 at 1:47
• @fgp Added the sequence in question.
– Lost
Feb 18, 2014 at 1:56
• Clearly the sequence of partial sums is Cauchy. The issue is whether or not the function you get out is integrable. Feb 18, 2014 at 2:27

Besides being square-summable, there are no easy necessary conditions for $a_k$ to be the Fourier coefficients of a Riemann integrable function.

Should I show that $f−S_n$ does not converge to $0$ in the $L^2$ norm?

No, because it does converge. Every square-summable sequence of coefficients is the Fourier series of some $L^2$ function $f$, to which the partial sums converge in the $L^2$ norm. The thing is, our $f$ is not Riemann integrable because it's not bounded.

Think of the domain of $f$ as the unit circle on the complex plane. The series in question is $\sum_{k=1}^\infty k^{-1}e^{ik\theta}$, which we can write as $\sum_{k=1}^\infty k^{-1}z^k$, with $z=e^{i\theta}$. The latter series converges in the open unit disk and defines a holomorphic function $F$ there. We can find $F$ explicitly as follows: $$F'(z) = \sum_{k=1}^\infty z^{k-1} = \frac{1}{1-z}$$ $$F(z) = \int_0^z F'(\zeta)\,d\zeta = -\log (1-z)$$ The boundary values of $F$ are not bounded: $$\operatorname{Re}\log (1-e^{i\theta}) =\frac12 \log|1-e^{i\theta}|^2 = \frac12 \log(2-2\cos \theta)$$

Of course, some justification is needed for the claim that the boundary values of $F$ (taken as radial limits) recover the function $f$ that we started with. I don't know what the author of the book wanted the readers to do here, because I don't have the book.

• Thanks for the answer. The book uses strictly real analysis, but the same idea applies: I used Abel summability to show that the Abel means of f would be bounded, yet the alternate definition given by the convolution of f and the Poisson kernel was unbounded, a contradiction.
– Lost
Feb 20, 2014 at 6:02
• Could you please post your solution? Sep 7, 2023 at 12:39