Why does $f: \mathbb{T} \rightarrow \mathbb{C} \in C^k$ imply that $f$ is bounded?

Question

From a book on Harmonic Analysis:

Question: Why does $f^{(j)}(- \pi) = f^{(j)}(\pi)$ in this context imply that $f$ is bounded?

EDIT: Does here the author just mean that the function's domain is bounded? If so, I've mistaken a trivial statement with something deep :)

Answer 1

Some hand-wavy reasoning

The statement $f^{(j)}(\pi)=f^{(j)}(-\pi)$ implies that $\Bbb T$ should be used with the unit circle topology. It might me more common to define $f:S^1\to \Bbb C$. However, $\Bbb T$ with the unit circle topology is compact and $f$ is continuous. So the image of $f$ is compact too, which means closed and bounded.

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Still hand-wavy, but different

The author made $f$ to be defined on $[-\pi,\pi)$ which excludes $\pi$, so the expression $f^{(j)}(\pi)$ seems technically meaningless. But he means to extend $f$ to $[-\pi,\pi]$ in a $C^k$-compatible way. Now $[-\pi,\pi]$ is still compact and $f$ continuous. So the above reasoning still applies.