Corollary 2.4 in Stein and Shakarachi Fourier Analysis

Question

The book states that: "Suppose that $f$ is a twice continuously differentiable function on the circle. Then $$\hat{f}(n) = O\bigg(\frac{1}{|n|^{2}}\bigg) \text{ as } |n|\to \infty,$$ So that the Fourier Series of $f$ converges uniformly and absolutely to f."

The book proved the case for $n \ne 0$ and left the case for $n = 0$ as an exercise, I know that $$2\pi\hat{f}(0) =\int_{0}^{2\pi}f(\theta)\,d\theta, $$ I know that the value of a definite integration is a constant and I know that $$\hat{f}(n) = O\bigg(\frac{1}{|n|^{2}}\bigg) \text{ as } |n|\to \infty,$$ Means, $$ |\hat{f}(n)| \leq \frac{C}{|n|^{2}},$$ for all large |n|, where $C$ is a constant but then How can I complete if $n = 0$. What am I supposed to prove I couldn`t see could anyone help me?

Answer 1

There is really nothing to prove. It suffices to show there is a constant $C$ such that $$n^2 |\hat f(n)| \le C$$ for all $n \in \mathbb Z$. When $n = 0$ this hold trivially for any nonnegative $C$.

Answer 2

It doesn't matter. Intuitively $\hat{f}(n)=O(1/|n|^2)$ is saying something about what happens $n$ gets bigger, what happens for a finite number of $n$ is irrelevant. You say yourself it means the inequality is satisfied for all large $|n|$. 0 is not large.