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Let $C(T)$ be the space of continuous functions on the circle, and let $\left \{ \gamma _n \right \}_{n\in \mathbb Z}$ be a sequence of complex numbers. Then, there is a $\Lambda:C(T)\to C(T)$ such that $\Lambda f$ is the function,whose Fourier coefficients are $\hat {\Lambda f(k)}=\gamma_k\cdot \hat f(k)$, if and only f there is a Borel measure $\mu $ on $T$ such that$\gamma_k=\int_Te^{ikt}d\mu$.

Note that in the wording of the exercise, $\Lambda f$ is not required to be continuous. In fact, without some conditions on $\gamma_n,\ \Lambda $ will $not$ be continuous. For example, if $\gamma_n=n$. But, the assignment is $always$ possible. So my questions are: is my proof of the forward direction correct? How can one make sense of the reverse direction?

Here is my attempt:

$(\Rightarrow ): C(T)$ is a Banach space with the sup norm. . Suppose there is such a function, $\Lambda$ and that $f_n\to f$. Then,

$\vert \hat {\Lambda f(k)}-\hat {\Lambda f_n(k)}\vert =\vert \gamma_k\cdot (\hat f_n(k)-\hat f(k))\vert=\left | \gamma_k\int_{-\pi}^{\pi}(f_n(t)-f(t))\cdot e^{ikt}dt \right |\le 2\pi \vert \gamma_k\vert \left \| f_n-f \right \|\to 0\quad \text {as}\ n\to \infty\ \text{so}\ \Lambda f_n\to \Lambda f.$

Now define $\phi: \mathcal T\to \mathbb C$ by $\phi (f)=\sum \gamma _k\hat f(k)$, where $\mathcal T$ is the space of all trigonometric polynomials on $T$. $\phi $ is clearly linear and it is also continuous. Indeed, by the preceding paragraph, we have

$\vert \phi (f)\vert =\vert (\Lambda f)(0)\vert\le \left \| \Lambda f \right \|\to 0\ \text {as}\ f\to 0.$

Since $\mathcal T$ is dense in $C(T), \phi$ extends to all of $C(T)$ and now the Riesz Theorem gives the result: a Borel measure $\mu$ such that $\phi (f)=\int_Tfd\mu$. Thus,

$\phi (e^{-ikt})=\sum_j \gamma_j \cdot \widehat{ (e^{ijt})} =\gamma_k=\int_T e^{-ikt}d\mu. $

$(\Leftarrow):$ Suppose there is a sequence $\left \{ \gamma_n \right \}_{n\in \mathbb Z}$ and a measure $\mu$ such that $\gamma_k=\int_T e^{ikt}d\mu.$ Define $\phi_n :C(T)\to \mathbb C$ by $\phi_n (f)=\int_Tf(t)e^{-int}d\mu$, take $f\in \mathcal T$ so that $f(t)=\sum_{-N}^{N} \hat f(k)\cdot e^{-ikt}$ and

define $\Lambda:\mathcal T\to \mathcal T$ by $\Lambda f=\sum \phi_n(1)\cdot f=\sum \gamma_n\cdot f$. Then $\widehat {\Lambda f(k)}=\gamma_k\cdot \hat f(k)$,so the result is true if $f\in \mathcal T$. The problem is that $\Lambda$ is not bounded and it is here that I am stuck.

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  • $\begingroup$ $f_n\in C(T)$. Why should they be smooth? $\phi $ is initially defined only on the trig polynomials and it is (uniformly) continuous. Since $\mathcal T$ is dense in $C(T)$, $\phi$ extends to all of $C(T)$. $\endgroup$ – Matematleta Nov 27 '16 at 2:38
  • $\begingroup$ if $f \in C(T)$ then $\sum_k \hat{f}(k)e^{ikx}$ doesn't have to converge but $(T_n f)(x) = \sum_k \hat{f}(k)e^{ikx}e^{-k^2/n^2}$ converges and is $C^\infty$ and $\|T_n f - f\|_\infty \to 0$. That's why I think you should consider this sequence $T_n f \to f$ $\endgroup$ – reuns Nov 27 '16 at 3:00
  • $\begingroup$ $\Lambda$ sends an $f\in C(T)$ to the $C(T)$ functions whose fourier coefficients are as given. True, the fourier series for $f$ may not converge, but that does not affect the continuity of $\phi$, which comes from the calculation in the preceding paragraph,using the def of $\Lambda$ $\endgroup$ – Matematleta Nov 27 '16 at 3:07
  • $\begingroup$ isn't this clear since $\vert\gamma_n\vert\le \mu (T)$ and the fact that $\vert f(k)\vert\to 0$? $\endgroup$ – Matematleta Nov 27 '16 at 3:09
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    $\begingroup$ I'd say the condition on $\mu$ is $\lim_{n \to \infty} \|g_n\|_\infty = 0 \implies \lim_{n \to \infty}\int_0^{2\pi} g_n(x)d\mu(x)= 0$ or equivalently thanks to the uniform boundedness principle $|\int_0^{2\pi} f(x)d\mu(x)|< \alpha \|f\|_\infty$. In that case $(\Lambda f)(y) = \int_0^{2\pi} f(y-x)d\mu(x)$ and $\Lambda$ is clearly continuous $C(T) \to C(T)$. And if there is no such $\alpha$ then $\Lambda$ is clearly not continuous. $\endgroup$ – reuns Nov 27 '16 at 3:48
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In your attempt, you only show that the Fourier coefficients of $\Lambda f_n$ converge to the corresponding Fourier coefficient of $\Lambda f$. That does not guarantee that $\Lambda f_n \to \Lambda f$, consider for example $f_n = \frac{1}{2n+1}\cdot D_n$, where $D_n$ is the Dirichlet kernel, $D_n(x) = \sum_{k = -n}^n e^{ikx}$. Then $f_n(0) = 1$ for all $n$, so $f_n \not\to 0$ in $C(T)$, but the Fourier coefficients of $f_n$ converge uniformly to $0$.

We get the continuity of $\Lambda$ from the closed graph theorem. Suppose we have a sequence $(f_n)$ in $C(T)$ such that $f_n \to f$ and $\Lambda f_n \to g$. For every $k\in \mathbb{Z}$ we have $\hat{f}_n(k) \to \hat{f}(k)$, and thus

$$\widehat{\Lambda f}(k) = \gamma_k \hat{f}(k) = \lim_{n\to\infty} \gamma_k\hat{f}_n(k) =\lim_{n\to\infty} \widehat{\Lambda f_n}(k) = \widehat{\lim_{n\to\infty} f_n}(k) = \hat{g}(k).$$

It follows that $g = \Lambda f$, so the graph of $\Lambda$ is closed, and $\Lambda$ is continuous.

Now we need not bother with defining $\phi$ on the space of trigonometric polynomials, we can simply define $\phi(f) = (\Lambda f)(0)$, and as the composition of two continuous maps (evaluation at $0$ and $\Lambda$), $\phi$ is continuous. By the Riesz representation theorem, there is a regular Borel measure $\mu$ with $\phi(f) = \int_T f\,d\mu$, and with $f_k(x) = e^{ikx}$, we have

$$\int_T f_k\,d\mu = \phi(f_k) = (\Lambda f_k)(0) = (\gamma_k f_k)(0) = \gamma_k.$$

For the converse direction, consider the map $\Lambda$ given by

$$\Lambda f \colon x \mapsto \int_T f(x+t)\,d\mu(t).$$

For any fixed $f \in C(T)$, by the uniform continuity of $f$, the map $x \mapsto g_x$, where $g_x(t) = f(x+t)$ is continuous, thus $\Lambda \colon C(T) \to C(T)$. Clearly $\Lambda$ is linear, and we have

$$\lvert (\Lambda f)(x)\rvert \leqslant \int_T \lvert f(x+t)\rvert\, d\lvert\mu\rvert \leqslant \lVert \mu\rVert\cdot \lVert f\rVert,$$

so $\Lambda$ is continuous. And

\begin{align} \widehat{\Lambda f}(k) &= \frac{1}{2\pi} \int_{T} (\Lambda f)(x) e^{-ikx}\,dx \\ &= \frac{1}{2\pi} \int_T \int_T f(x+t)\,d\mu(t) e^{-ikx}\,dx \\ &= \frac{1}{2\pi} \int_T \int_T f(x+t) e^{-ikx}\,dx\, d\mu(t) \tag{Fubini} \\ &= \frac{1}{2\pi} \int_T \int_T f(u) e^{-ik(u-t)}\,du\,d\mu(t) \\ &= \frac{1}{2\pi} \int_T \int_T f(u) e^{-iku}\,du\; e^{ikt}\,d\mu(t) \\ &= \gamma_k \hat{f}(k). \end{align}

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  • $\begingroup$ Clear and thorough. Thank you. Is there an inuitive way to see that one should define $\Lambda f$ as you did? $\endgroup$ – Matematleta Nov 27 '16 at 22:20
  • $\begingroup$ Kind of. Depends on what one knows about Fourier series how intuitive it is. If you know that the sequence of Fourier coefficients of the convolution of two functions is the pointwise product of the coefficient sequences of the two functions, that is $\widehat{f \ast g}(k) = \hat{f}(k)\cdot \hat{g}(k)$, then you know that you should look at the convolution of $f$ with the measure whose Fourier coefficients are the $\gamma_k$. If you don't know that, you still see that you need $\Lambda(e^{ikx}) = \gamma_k \cdot e^{ikx}$, and if you have the good idea to write $\gamma_k$ as the integral, you $\endgroup$ – Daniel Fischer Nov 27 '16 at 22:28
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    $\begingroup$ get $$\Lambda(e^{ikx}) = e^{ikx} \int_T e^{ikt}\,d\mu = \int_T e^{ik(x+t)}\,d\mu.$$ Then one may conjecture that this works for all $f$. $\endgroup$ – Daniel Fischer Nov 27 '16 at 22:29
  • $\begingroup$ Yes,this makes sense. I wonder if you could look at a proof here if you have a chance, I am trying to prove a Riesz Thm in a different way. Thanks-- $\endgroup$ – Matematleta Nov 28 '16 at 14:22

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