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 $x \in [a, b]$.


The proof starts the following way:

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

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


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

  • 1
    $\begingroup$ There's a typo, the upper limit should be $x$, at which point it is just the FTC applied to $f_n$. $\endgroup$
    – Ian
    Aug 23 '16 at 13:09
  • $\begingroup$ Fixed it, thanks! $\endgroup$
    – Julian
    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 $$g(x)=g(a) + \int_a^x g'(t)dt$$


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