# Uniform convergence and differentiability

Let $$f_n:[a, b] \rightarrow \Bbb R$$ be continuous differentiable functions that converge pointwise to a function $$f:[a, b] \rightarrow \Bbb R$$. The sequence of the derivatives $$f'_n:[a, b] \rightarrow \Bbb R$$ converges uniformly. Then $$f$$ is differentiable and

$$f'(x) = \lim_{n\to \infty} f'_n(x)$$ for all $$x \in [a, b]$$.

Proof:

The proof starts the following way:

Let $$f* = \lim f'_n$$. $$f*$$ is continuous on $$[a, b]$$. For all $$x \in [a, b]$$ is

$$f_n(x) = f_n(a) + \int_a^x f'_n(t) dt.$$

Question:

Where does the last equation come from? What does $$f_n(x)$$ have to do with that specific integral?

• There's a typo, the upper limit should be $x$, at which point it is just the FTC applied to $f_n$.
– Ian
Aug 23 '16 at 13:09
• Fixed it, thanks! Aug 23 '16 at 13:13

The last equation is a direct consequence of the fundamentaly theorem of calculus.

It says that if $$F(x)=\int_a^x f(t) dt$$

then $F'(x) = f(x)$.

So, if $g$ is any continuously differentiable function, you can define $f(x)=g'(x)$ and you can know that the function $F(x)=\int_a^x f(t)dt$ has the same derivative as $g$ (because the derivative of $F$ is $F'=f=g'$, and the derivative of $g$ is obviously $g'$).

So, $F$ and $g$ only differ by a constant.

You also know that $F(a)=0$, which should be enough to conclude that $$g(x) = F(x) + g(a)$$

or in other words $$g(x)=g(a) + \int_a^x f(t)dt$$

which is the same as $$g(x)=g(a) + \int_a^x g'(t)dt$$