Uniform convergence and differentiation

Uniform convergence and differentiation: Let $f_n:[a,b]\to\mathbb{R}$ be differentiable on $[a,b]$ (or on $(a,b)$ with suitable endpoint control). Assume:

• there exists $x_0\in[a,b]$ such that the sequence $(f_n(x_0))$ converges in $\mathbb{R}$, and
• the derivatives $f_n'$ converge uniformly on $[a,b]$ to a function $g$.

Then $f_n$ converges uniformly on $[a,b]$ to a differentiable function $f$, and $f'(x)=g(x)\quad \text{for all } x\in[a,b].$

This theorem provides the rigorous justification for differentiating a sequence (or series) term-by-term, under uniform convergence of derivatives.

Proof sketch: For each $n$ and $x\in[a,b]$, $f_n(x)=f_n(x_0)+\int_{x_0}^x f_n'(t)\,dt$ by the fundamental theorem of calculus . Since $f_n(x_0)$ converges and $f_n'\to g$ uniformly, the right-hand side converges uniformly in $x$ to $f(x)=\lim_{n\to\infty} f_n(x_0)+\int_{x_0}^x g(t)\,dt.$ Differentiate $f$ using the fundamental theorem of calculus to obtain $f'=g$.