Lecture 2: Differentiable Test

Uniform convergence of derivatives

Let $(f_n)$ be a sequence of differentiable functions on $[-\pi,\pi]$. Suppose that:

There exists $x_0\in[-\pi,\pi]$ such that $ \bigl(f_n(x_0)\bigr)_{n=1}^{\infty} $ converges.
The sequence of derivatives $(f_n')$ converges uniformly on $[-\pi,\pi]$ to a function $g$.

Then $(f_n)$ converges uniformly on $[-\pi,\pi]$ to a differentiable function $f$, and \[ f'(x)=g(x) \] for every $x\in[-\pi,\pi]$.

Term-by-term differentiation

Let $(u_n)$ be a sequence of differentiable functions on $[-\pi,\pi]$. Suppose that:

There exists $x_0\in[-\pi,\pi]$ such that $\sum_{n=1}^{\infty}u_n(x_0)$ converges.
The derivative series $\sum_{n=1}^{\infty}u_n'(x)$ converges uniformly on $[-\pi,\pi]$.

Then $\sum_{n=1}^{\infty}u_n(x)$ converges uniformly on $[-\pi,\pi]$ to a differentiable function $f$, and \[ f'(x)=\sum_{n=1}^{\infty}u_n'(x). \]

In the specific case of our Fourier series, if we have already proved that

$$ \begin{align*} S_N\to f \end{align*} $$

uniformly. Then we need to show that each $S_N$ is differentiable. If it is, then what remains is to prove that

$$ \begin{align*} S_N'\to g \end{align*} $$

uniformly for some function $g$. If we prove that, the theorem gives that $f$ is differentiable and $f'=g$.