Application of the Riesz Theorem to Fourier Series 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.
 A: 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}
