Suppose $(f_n)_{n=1}^\infty$ is a sequence of continuous functions on [a, b] and that ${f'_n}$ exists and is continuous on [a, b]. Suppose further that $\lim_{n\to \infty}{f_n}=f$ uniformly and $\lim_{n\to \infty}{f'_n}=g$ uniformly for some continuous functions f and g.

Let $x\in{[a,b]}$$$\int_{a}^{x}g(t)dt={f(x)}-f(a)$$

use the formula above to prove ${f'(x)}=\lim_{n\to \infty}{f'_n(x)}$ for all $x\in{[a,b]}$ (Hint: FTC.)

I am having a really hard time just understanding this question so I'll write out my train of thought hopefully someone can help me figure it out.

So there is a sequence ${f_n}$ in which the derivative is continuous on [a,b], the limit of ${f_n}=f$ and the limit of the derivative of ${f_n}$, ${f'_n}=g$ the functions f and g are continuous.

Thus g is integrable on [a,b] since the derivative ${f'_n}$ is continuous and converges uniformly to g which implies that g is continuous on [a,b] and I have a theorem that all continuous functions are integrable.

Next I was supposed to prove $$\int_{a}^{x}g(t)dt={f(x)}-f(a)$$ for x $\in{[a,b]}$ I did this using a theorem which statesthat if ${f_n\to}$ f uniformly on [a,b] then $$\lim_{n\to \infty}\int_{a}^{b}{f_n(x)dx}=\int_{a}^{b}f(x)dx$$ and the fundamental theorem of calculus to get $$\int_{a}^{x}g(t)dt={f(x)}-f(a)$$ OK this is where I'm stuck, so now I'm supposed to prove ${f'(x)}=\lim_{n\to \infty}{f'_n(x)}$ for all $x\in{[a,b]}$ (Hint: FTC.) I just don't see how the fundamental theorem of calculus is used in this final step. Any help would be greatly appreciated I've been working on this problem for hours now.

Thank you,


Differentiate your last equality on both sides. On the left, you get $g(x)$, which is $\lim_{n \to \infty} f_n'(x)$ by definition of $g$. On the right, you get $f'(x)$.

  • $\begingroup$ Thank you for your answer I'm still not seeing how the f(a) on the right disappeared wouldn't I get $\lim_{n \to \infty} f_n'(x)$ = $f'(x)$ - $f'(a)$ ? @Aloizio Macedo $\endgroup$ – jack Mar 19 '18 at 18:57
  • 1
    $\begingroup$ No, you are differentiating with respect to $x$. $\endgroup$ – Aloizio Macedo Mar 19 '18 at 19:02

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