7.12. Sequences of Differentiable Functions

We resume the subject from part 6.10 in which we found uniform convergence to be compatible with continuity.

We won't get a similar result however with differentiability as we know, for example from the proof of the Weierstrass approximation theorem, that any function continuous on $\left[0,1\right]$ could be represented as the uniform limit of a sequence of polynomials, and thus as the uniform limit of a sequence of differentiable functions. (see [6.7.2]). The function $|\mathrm{X}-\frac{1}{2}|$ e.g. is such a uniform limit on $\left[0,1\right]$ but is not differentiable at $\frac{1}{2}$.

But even if the limit function proves to be differentiable there is no certainty that its derivative emanates from the derivatives of the sequence members. Due to the comparison test [6.10.7] we have for example

$\frac{1}{n}\mathrm{sin}\circ \left(n\mathrm{X}\right)\underset{uf}{\to }0$,

as $|\frac{1}{n}\mathrm{sin}\left(nx\right)|\le \frac{1}{n}$ for all n and all x, but the sequence $\left(\left(\frac{1}{n}\mathrm{sin}\circ n\mathrm{X}{\right)}^{\prime }\right)=\left(\mathrm{cos}\circ n\mathrm{X}\right)$ even fails to be pointwise convergent: $\left(\mathrm{cos}\left(n\pi \right)\right)=\left({\left(-1\right)}^{n}\right)$ for instance is divergent.

Surprisingly the limit function will be differentiable with an appropriate derivative if the sequence of derivatives converges uniformly. The original sequence itself needs only to be pointwise convergent.

Again, the mean value theorem is an essential argument in the proof, which is thus only valid for intervals, actually for bounded intervals in this case.

Proposition:  Let I be a bounded interval,  ${f}_{n}\in {\mathcal{D}}^{1}\left(I\right)$ and $g:I\to ℝ$ such that ${{f}_{n}}^{\prime }\underset{uf}{\to }g$. If for one $a\in I$, the sequence of numbers $\left({f}_{n}\left(a\right)\right)$ is convergent then the following holds in turn:

 $\left({f}_{n}\right)$ is uniformly convergent [7.12.1]
 $f≔\mathrm{lim}\phantom{\rule{0.3em}{0ex}}{f}_{n}:I\to ℝ$ is differentiable at a and ${f}^{\prime }\left(a\right)=g\left(a\right)$ [7.12.2]

Proof:

1.  For each n the basic representation theorem [7.5.1] yields a function ${r}_{n}:I\to ℝ$ continuous at a such that ${r}_{n}\left(a\right)={{f}_{n}}^{\prime }\left(a\right)$ and

${f}_{n}={f}_{n}\left(a\right)+\left(\mathrm{X}-a\right){r}_{n}$

We now use the Cauchy test [6.10.8] to show that the sequence $\left({r}_{n}\right)$ is uniformly convergent on I. If $\epsilon >0$ is arbitrary the premise on $\left({{f}_{n}}^{\prime }\right)$ guarantees (using the Cauchy test again) that there will be an ${n}_{0}\in {ℕ}^{\ast }$ such that

$|{{f}_{n}}^{\prime }\left(x\right)-{{f}_{m}}^{\prime }\left(x\right)|<\epsilon$

for all $n,m\ge {n}_{0}$ and all $x\in I$. In particular we have for these n, m:

$|{r}_{n}\left(a\right)-{r}_{m}\left(a\right)|=|{{f}_{n}}^{\prime }\left(a\right)-{{f}_{m}}^{\prime }\left(a\right)|<\epsilon$.

Now let $x\in I$ be different from a. Due to the mean value theorem [7.9.5] we get an ${\stackrel{˜}{x}}_{n\text{\hspace{0.17em}}m}$ in between a and x for each $n,m\ge {n}_{0}$ such that

$\frac{\left({f}_{n}-{f}_{m}\right)\left(x\right)-\left({f}_{n}-{f}_{m}\right)\left(a\right)}{x-a}=\left({f}_{n}-{f}_{m}{\right)}^{\prime }\left({\stackrel{˜}{x}}_{n\text{\hspace{0.17em}}m}\right)$.

Using  we thus may estimate as follows:

$\begin{array}{ll}|{r}_{n}\left(x\right)-{r}_{m}\left(x\right)|\hfill & =|\frac{{f}_{n}\left(x\right)-{f}_{n}\left(a\right)}{x-a}-\frac{{f}_{m}\left(x\right)-{f}_{m}\left(a\right)}{x-a}|\hfill \\ \hfill & =|\frac{\left({f}_{n}-{f}_{m}\right)\left(x\right)-\left({f}_{n}-{f}_{m}\right)\left(a\right)}{x-a}|\hfill \\ \hfill & =|\left({f}_{n}-{f}_{m}{\right)}^{\prime }\left({\stackrel{˜}{x}}_{n\text{\hspace{0.17em}}m}\right)|\hfill \\ \hfill & =|{{f}_{n}}^{\prime }\left({\stackrel{˜}{x}}_{n\text{\hspace{0.17em}}m}\right)-{{f}_{m}}^{\prime }\left({\stackrel{˜}{x}}_{n\text{\hspace{0.17em}}m}\right)|<\epsilon \hfill \end{array}$

Now that the uniform convergence of $\left({r}_{n}\right)$ is verified we will prove the actual assertion. Again we take an $\epsilon >0$. As I is bounded, I has a finite diameter $c>0$. For $\frac{\epsilon }{2c}$ there is now an ${n}_{1}\in {ℕ}^{\ast }$such that

$|$  for all $n,m\ge {n}_{1}$ and all $x\in I$.

Furthermore we get an ${n}_{2}\in {ℕ}^{\ast }$ (due to [5.5.7]  $\left({f}_{n}\left(a\right)\right)$ is a Cauchy sequence) such that

$|{f}_{n}\left(a\right)-{f}_{m}\left(a\right)|<\frac{\epsilon }{2}$  for all $n,m\ge {n}_{2}$.

For $n,m\ge {n}_{0}≔\mathrm{max}\left\{{n}_{1},{n}_{2}\right\}$ and every $x\in I$ we now get from  using  and :

$\begin{array}{ll}|{f}_{n}\left(x\right)-{f}_{m}\left(x\right)|\hfill & =|{f}_{n}\left(a\right)+\left(x-a\right){r}_{n}\left(x\right)-{f}_{m}\left(a\right)-\left(x-a\right){r}_{m}\left(x\right)|\hfill \\ \hfill & \le |{f}_{n}\left(a\right)-{f}_{m}\left(a\right)|+|x-a|\cdot |{r}_{n}\left(x\right)-{r}_{m}\left(x\right)|\hfill \\ \hfill & <\frac{\epsilon }{2}+c\cdot \frac{\epsilon }{2c}=\epsilon \hfill \end{array}$

2.  Firstly we find that the limit $r≔\mathrm{lim}\phantom{\rule{0.3em}{0ex}}{r}_{n}:I\to ℝ$ verified in 1 is continuous at a due to [6.10.5]. Secondly from

• ${f}_{n}\left(a\right)+\left(\mathrm{X}-a\right){r}_{n}\underset{pw}{\to }f\left(a\right)+\left(\mathrm{X}-a\right)r$

• ${f}_{n}\left(a\right)+\left(X-a\right){r}_{n}={f}_{n}\underset{pw}{\to }f$

we see that the function $f=\mathrm{lim}\phantom{\rule{0.3em}{0ex}}{f}_{n}$ could be represented as $f=f\left(a\right)+\left(\mathrm{X}-a\right)r$. According to [7.5.1]  f is thus differentiable at a with

${f}^{\prime }\left(a\right)=r\left(a\right)=\mathrm{lim}\phantom{\rule{0.3em}{0ex}}{r}_{n}\left(a\right)=\mathrm{lim}\phantom{\rule{0.3em}{0ex}}{{f}_{n}}^{\prime }\left(a\right)=g\left(a\right)$

[7.12.2] is a local result which could be extended to a global version for ${\mathcal{D}}^{1}$-functions and, in addition, for ${\mathcal{C}}^{1}$-functions as well. Also, the restriction to bounded intervals is no longer needed.

Proposition:  Let I be an arbitrary interval,  ${f}_{n}\in {\mathcal{D}}^{1}\left(I\right)$, $g:I\to ℝ$ such that ${{f}_{n}}^{\prime }|J\underset{uf}{\to }g|J$ for each closed subinterval $J$. If the sequence of numbers $\left({f}_{n}\left(a\right)\right)$ converges for one $a$ we have for the pointwise limit $f$:

 $f\in {\mathcal{D}}^{1}\left(I\right)$ and ${f}^{\prime }=g$ [7.12.3] ${f}_{n}\in {\mathcal{C}}^{1}\left(I\right)\text{ }⇒\text{ }f\in {\mathcal{C}}^{1}\left(I\right)$ [7.12.4]

Proof:  We first show that the sequence $\left({f}_{n}\right)$ is pointwise convergent. If $x\in I$ is arbitrary, without restriction $x\ne a$, x and a span a closed subinterval $J\subset I$. According to the premise $\left({{f}_{n}}^{\prime }|J\right)$ is uniformly convergent. Due to [7.12.1]  $\left({f}_{n}|J\right)$ is thus uniformely and consequently pointwise convergent as well. As $x\in J$ the sequence of numbers $\left({f}_{n}\left(x\right)\right)$ converges eventually.

1.   According to our preliminary consideration $\left({f}_{n}\left(x\right)\right)$ converges for each $x\in I$. As there is a relative ε-neighbourhood  ${I}_{x,\epsilon }$ i is a non-negative number such that is a closed subinterval of I satisfying ${I}_{x,\epsilon }$. __________ *) With the convention $|x±\infty |=|\infty +\infty |=\infty >r$ for each real r we thus take account of the case of an unbounded interval.
for each x being part of a closed subinterval J of I, the function $f|{I}_{x,\epsilon }$ is differentiable at x with $\left(f|{I}_{x,\epsilon }{\right)}^{\prime }\left(x\right)=g\left(x\right)$ due to [7.12.2]. From the local character of differentiability (cf [7.4.2]) we see that this is the assertion.

2.   If all ${{f}_{n}}^{\prime }$ are continuous the locally uniform limit ${f}^{\prime }=g=\mathrm{lim}\phantom{\rule{0.3em}{0ex}}{{f}_{n}}^{\prime }$ will be continuous as well according to [6.10.6]. 