Let's begin by looking at a bad example. Let g = 0 on [0,2], and let fn squash down to 0 uniformly. However, each fn has a small bump at one. The height of the bump is bounded by ε, as the functions approach 0, and the width of the bump decreases accordingly. Still, the bump is always rising at 1, so that the derivative at 1 is always 1. The bump reaches its maximum shortly thereafter, then drops back down towards the x axis, looking a bit like the bell curve. As n approaches infinity, the derivatives approach 0 everywhere, except for x = 1, which always has a derivative of 1. Obviously this does not agree with g. In summary, the original functions converge uniformly, and the derivatives converge pointwise; yet the limit of the derivatives is not g′. Something more is needed. As we shall see, the derivatives must converge uniformly. In this case, the derivatives are continuous on [0,2], and the limit function is not, thus convergence is not uniform.
The easiest way to prove this theorem is to combine the previous theorem with the fundamental theorem of calculus.
Assume each fn is differentiable on a fixed interval [a,b], and assume these derivatives converge to a function h, at least on the interval [a,b].
Obviously each function fn′ is integrable, its integral being fn. Actually its integral, evaluated at x, is fn(x)-fn(a), for any x in [a,b].
Now assume we can prove, using the previous theorem, that the integral of h, from a to x, is the limit of the integrals of fn′, from a to x, for every x in [a,b]. This is usually done by proving the sequence of derivative functions fn′ is uniformly convergent over [a,b], though there are other methods.
The integral of h, from a to x, is now the limit of f(x)-f(a), which is equal to g(x)-g(a). Take derivatives, and h = g′. In other words, the limit of the derivatives equals the derivative of the limit function.
The above describes a sequence of convergent, differentiable functions, but this can be extended to a series of functions in the usual way. If we are given a series of functions, we can interchange differentiation and summation, provided the sequence of partial sums meets the criteria outlined above.
Since fn converges uniformly on the region, it converges uniformly on a small circle about p. As z moves around the circle, let d = z-p. Divide each fn by d2. When fn is within ε of its limit g, fn/d2 is within ε/r2 of g/d2. In other words, convergence remains uniform, even after dividing by d2.
With uniform convergence, the limit of the integral is the integral of the limit, hence the contour integral of g/d2, around the circle, is the limit of the contour integrals of fn/d2. Therefore g′ is the limit of fn′.
But how do we know g is differentiable? It is the uniform limit of continuous functions, hence continuous. Also, contour integrals of f, which are 0, converge to contour integrals of g, which must also be 0. This satisfies the conditions of morera's theorem, hence g is analytic.
What about the second derivative at p? Choose r small, so that the disk of radius 2r, centered at p, lies entirely inside our simply connected region. By the above, the sequence fn′ converges pointwise to g′. If it converges uniformly, on the disk of radius r, apply the theorem again, and g′′(p) is the limit of fn′′(p).
To show uniform convergence, return to the definition of a derivative as a contour integral. The circles that define these contour integrals will always have radius r. Take a contour integral around a point x, close to p, and the circle lies within the disk of radius 2r centered at p. We know that fn converges uniformly on this larger disk of radius 2r. The same holds for fn/d2. The contour integrals that define fn′(x) converge uniformly to g′(x), as long as x is within r of p. Therefore the first derivatives converge uniformly on a small disk about p, and g′′(p) is the limit of fn′′(p).
By induction, the kth derivative of any point p in the interior of our simply connected region is the limit of the kth derivatives of fn, where each fn is analytic, and the sequence fn uniformly converges to g.
Note that the functions only need converge uniformly on a small disk about p. Find such a disk for each p, and the theorem remains valid. In other words, we only need local uniform convergence, i.e. uniform convergence on a small open set about each point p.