Use Dominated convergence theorem to show that $f(x):=\sum_{k=1}^{\infty}\frac{\cos(kx)}{k^3}$ is differentiable

Question

Let

$$f(x):=\sum_{k=1}^{\infty}\frac{\cos(kx)}{k^3},$$

how can we show that f is differentiable everywhere by using the Lebesgue dominated convergence theorem? I know this theorem as saying that the integral and the limit can change each other under the dominating condition. But how can we use it in this question?Thanks!

Answer 1

$f(x) = \int_\mathbb{N} \frac{\cos (kx) }{ k^3 } dk$

$f'(x) = \lim_{h\rightarrow 0} \int_\mathbb{N} \frac{\cos (k(x+h)) - \cos(kx)}{ hk^3 } dk$
We bound the numerator by $kh$ using mean value theorem. Hence we have a dominating function $1/k^2$ and apply the DCT to pull the limit through.

Answer 2

For any sequence $x_n \to a$, write $$g_n(k) = \frac{1}{k^3} \frac{\cos kx_n - \cos ka}{x_n-a}.$$

The sequence $g_n(k)$ converges pointwise to $g(k) = -\frac{1}{k^2} \sin ka$.

Uniformly in $n$, we have by the mean-value theorem that $|g_n(k)| \leq \frac{1}{k^2}$, and the function $\frac{1}{k^2}$ is integrable with respect to counting measure $dk$. Therefore by the dominated convergence theorem, we have $$\lim_{n \to +\infty} \frac{f(x_n) - f(a)}{x_n-a} = \lim_{n \to +\infty} \sum_k g_n(k) = \lim_{n \to +\infty} \int g_n \, dk = \int g \, dk = -\sum_k \frac{1}{k^2} \sin ka.$$

Since the sequence $x_n$ was arbitrary, we have $$f'(a) = -\sum_k \frac{1}{k^2} \sin ka.$$