## 8.2. Uniform Convergence

### Theorem 8.2.11: Uniform Convergence and Differentiation

Let

*f*be continuously differentiable functions defined on the interval_{n}(x)*[a, b]*. If:Then

- the sequence
fconverges pointwise to a function_{n}f, and- the sequence of derivatives
fconverges uniformly_{n}'

*f*is differentiable andfor allf'(x) = f_{n}'(x)

*x [a, b]*. In other words, the limit process and the differentiation process can be switched in this case.We already saw that neither pointwise nor uniform convergence by themselves were sufficient to switch limits and differentiation. But together things will work out, as we will see.

The sequence of derivatives *f _{n}'* converges uniformly and
each

*f*is continuous. Thus the function defined as:

_{n}'g(x) = f_{n}'

is well defined and continuous. We also have by the fundamental theorem of calculus that:

f_{n}(x) - f_{n}(a) = f_{n}' (x) dx

By a previous theorem we know that:

f_{n}' (t) dt = f_{n}' (t) dt = g(t) dt

But then, together with the previous line, we get:

because thef_{n}(x) - f_{n}(a) = f(x) - f(a) = g(t) dt

*f*'s converge pointwise to the function

_{n}*f*. Thus:

f(x) = g(t) dt + f(a)

But the function on the right, being defined as an integral of a
continuous function, is differentiable and therefore *f*
must be differentiable. Indeed, we get that:

which is what we wanted to prove.f'(x) = g(x) = f_{n}'