= Solution
For each fixed $n$, the coefficient functional is bounded since
$$
|\widehat f(n)|\leq\frac1{\sqrt{2\pi}}\int_{-\pi}^{\pi}|f(t)|\,dt
\leq\sqrt{2\pi}\|f\|_\infty.
$$
Suppose $f_j\to f$ and $Tf_j\to g$ uniformly. Taking coefficients and passing to the limit gives
$$
\widehat g(n)=\lim_j\widehat{Tf_j}(n)
=\alpha_n\lim_j\widehat f_j(n)=\alpha_n\widehat f(n)
$$
for every fixed $n$. No uniform bound on $\alpha_n$ is needed for this passage. Uniqueness gives $g=Tf$, so $T$ has a closed graph. The space $X$ is Banach, being the closed subspace defined by $f(-\pi)=f(\pi)$ inside the <continuous functions> with the uniform <norm>. The <closed graph theorem> therefore proves \b[$T$ is bounded].
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