# Why a uniform limit of a sequence of bounded derivative is a derivative?

Assume that a sequence $(f_n)$ of functions $f_n:[0,1] \rightarrow \mathbb R$ is uniformly convergent to a function $f:[0,1]\rightarrow \mathbb R$. Moreover let (for each $n\in \mathbb N$) $f_n$ be bounded and $f_n$ be a derivative of some $F_n: [0,1]\rightarrow \mathbb R$.

How to show that $f$ is a derivative of some $F:[0,1]\rightarrow \mathbb R$?

My attempt: by assumptions follows easily that $f_n$ (as derivative) is measurable , consequently (as measurable and bounded) $f_n$ is integrable. Hence $F_n$ is absolutely continuous. We have also that $f$ is integrable and bounded (as a consequence of uniformly convergence). Let $$G_n(x)= \int_0^xf_n(t)dt.$$ Then (using definition of uniform convergence) $$\lim_{n\rightarrow \infty} G_n(x)= \int_0^x f(t)dt=:G(x) \textrm{ for } x\in [0,1].$$

I'm not sure whether $G'(x)=f(x)$ for all $x∈[0,1]$., because I know only that absolutely continuous function has derivatives almost everywhere.

• Because of uniform convergance we have that $f$ is continuous on $[0,1]$. $f_n$ are bounded because $[0,1]$ is compact. So $f$ as continuous on the compact $[0,1]$ is bounded and thus is integrable – Haha Nov 7 '13 at 15:22
• @L.T, I believe some background regarding your level of familiarity with calculus/integration-theory will be very helpful in tailoring the answer to your needs. – Jonathan Y. Nov 7 '13 at 15:23
• @Dimitris Why $f$ is continuous? Derivative of function has only the Darboux property. – A.B Nov 7 '13 at 15:23
• correct.i just assumed that $f_n$ are continuous – Haha Nov 7 '13 at 17:36
• @L.T: Hint: you may imitate the proof of that uniform convergence limit of continuous functions is continuous to show $G'=f$. If you need me to elaborate, could you please add the content in your first comment to the body of your question? It would be helpful for me/others to answer your question in details. – Hu Zhengtang Nov 7 '13 at 19:13

To show that $G'(x) = f(x)$ for an $x \in [0,1]$, we must, for every $\varepsilon > 0$, find (establish the existence of) a $\delta > 0$ such that

$$0 <\lvert h\rvert < \delta \land (x+h)\in[0,1] \Rightarrow \left\lvert \frac{G(x+h)-G(x)}{h} - f(x)\right\rvert < \varepsilon.$$

Let $\varepsilon > 0$ be arbitrary. By the uniform convergence of $f_n \to f$, choose an $N\in \mathbb{N}$ such that $\lvert f_N(y) - f(y)\rvert < \varepsilon/3$ for all $y \in [0,1]$. For that $N$, by the assumed differentiability of $F_N$ in $x$, choose $\delta > 0$ such that

$$0 <\lvert h\rvert < \delta \land (x+h)\in[0,1] \Rightarrow \left\lvert \frac{F_N(x+h)-F_N(x)}{h} - f_N(x)\right\rvert < \varepsilon/3.$$

Then, for admissible $h$, we have

\begin{align} \left\lvert \frac{G(x+h)-G(x)}{h} - f(x)\right\rvert &= \left\lvert \frac1h \int_0^h f(x+t)\,dt - f(x) \right\rvert\\ &= \Biggl\lvert \frac1h \int_0^h \bigl(f(x+t) - f_N(x+t)\bigr)\,dt\\ &\qquad + \left(\frac1h \int_0^h f_N(x+t)\,dt - f_N(x)\right)\\ &\qquad + \bigl(f_N(x) -f(x)\bigr)\Biggr\rvert\\ &\leqslant \frac1{\lvert h\rvert} \int_0^{\lvert h\rvert} \lvert f(x+ \sigma(h)t)-f_N(x+\sigma(h)t)\rvert\,dt\tag{*}\\ &\qquad + \left \lvert\frac{F_N(x+h)-F_N(x)}{h} -f_N(x)\right\rvert\\ &\qquad + \lvert f_N(x) - f(x)\rvert\\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3}, \end{align}

so we see that $G$ is differentiable in $x$ with derivative $G'(x) = f(x)$.

In $(*)$, $\sigma(h)$ is the sign of $h$, $\pm 1$.

• Many thanks for answer!!! – A.B Nov 7 '13 at 20:58