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Suppose we have a sequence $\{a_k\}_k$ of complex numbers, where the index $k$ belongs to $\mathbb Z$.

Is there any general criterion to establish if there exists an $L^1[-\pi,\pi]$ function $f$ such that $a_k$ is the $k$-th complex Fourier coefficient of $f$ for every $k$?

Notice that here I don't care about the convergence of the Fourier series to $f$, but only about its existence. In fact, I know the bounds on $a_k$ to obtain a $C^k$ function $f$. I know also that $L^2$ functions are identified with $\ell^2$ sequences of coefficients, but I need something more general.

In case it is too difficult to answer the above question, I'll write also the problem I am interested in particular.

Given $\{a_k\}_k$ the Fourier coefficients of the $L^1[-\pi,\pi]$ function $f$, let $b_k$ be complex numbers that are either equal to $a_k$ or zero. Is it true that there exists an $L^1[-\pi,\pi]$ function $g$ whose Fourier coefficients are $b_k$ for every $k$?

Bernard
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Exodd
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  • I think the idea is to look at $k_r(x)=Im(\frac{1}{1-r e^{2i \pi x}}) =\frac12 \sum_{n\ne 0} r^{|n|} sign(n) e^{2i \pi nx}$ then $\lim_{r \to 1^-}|k_r|{L^1} = \infty$ implies $f \mapsto \frac12\sum{n\ne 0} r^{|n|}sign(n) c_n(f) e^{2i \pi nx}= f \ast k_r$ isn't continuous $L^1[-1/2,1/2]\to L^1[-1/2,1/2]$. As $r \to 1^-, k_r \to pv( Im(\frac{1}{1- e^{2i \pi x}})) \approx pv(\frac{1_{|x| < 1/2}}{-2 \pi x})$. In contrary for $\Re(\frac{1}{1-r e^{2i \pi x}})$ everything works well as for the Fejer kernel. – reuns Jun 06 '19 at 21:28

1 Answers1

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No, there is no simple necessary and sufficient condition for a sequence to be the Fourier coefficients of an $L^1$ function.

The answer to the specific question at the end is also no, even in the apparently simple case $$b_k=\begin{cases}a_k,&(k\ge0), \\0,&(k<0).\end{cases}$$

Edit: Of course I knew the OP would ask for an example. I don't know that the following quite counts as an "explicit" example, but it's a simple proof that an example exists (simpler than what I might have said yesterday.)

The one-line proof, in terms of things "everybody knows", is just this: The Hilbert transform is not bounded on $L^1$.

A several-line proof, assuming no background: Define the operator $T_N$ in terms of formal trig series: $$T_N\sum_{k=-\infty}^\infty a_ke^{ikt}=\sum_{k=N}^\infty a_ke^{ikt}.$$I'm asserting that there exists $f\in L^1$ with $T_0f\notin L^1$. Assume the contrary.

The Closed Graph Theorem shows that $T_0:L^1\to L^1$ is bounded. Now $$(T_Nf)(t)=e^{iNt}T_0(e^{-iNt}f(t)),$$ so every $T_N$ is bounded on $L^1$, with $||T_N||=||T_0||$.

Hence if $s_Nf$ is the $N$-th partial sum then $$||s_Nf||_1\le 2||T_0||\,||f||_1.$$It's easy to give an explicit counterexample to that: Let $K_N$ be the Fejer kernel and let $$f_N=2K_{2N}-K_N.$$Then $||f_N||_1\le3$ but $s_Nf_N=D_N$, the Dirichlet kernel, so $||s_Nf_N|||_1\to\infty$.

Details: Recall that $$K_N=\frac1N(D_1+\dots+D_N);$$hence $$f_N=\frac1N(D_{N+1}+\dots+D_{2N}).$$And of course $s_nD_k=D_n$ for $k\ge n$.

David C. Ullrich
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