Let $ f_n: (a, b) \to \mathbb R $ be a sequence of functions that satisfies the following hypotheses:

Question

$ f_n $ is twice differentiable in (a, b);

There is $ c \in (a, b) $ such that $ (f_n’(c)) $ is a bounded sequence;

$(f_n'')$ is a uniformly bounded sequence of functions in (a, b).

Prove that if $f_n$ simply converges to a function $f:(a,b) \to \mathbb R$, then $f$ is a class $C^1$.

The class $C^1$ consists of all differentiable functions whose derivative is continuous, and in $C^1$ the zeroth and first derivatives are continuous.

From the hypothesis Isn't it straightforward that $x_n$ is $C^1$?

Thanks for the attention.

Answer 1

Thm: (Arzela-Ascoli). Suppose $(g_n)$ is a sequence of continuous functions on $[a,b]$ such that $(g_n)$ is uniformly equicontinuous and pointwise bounded on $[a,b].$ Then there is a subsequence of $(g_n)$ that converges uniformly on $[a,b].$

This version of A-A will be used for the given problem. A more general version is Thm. 11.6.9 at https://www.jirka.org/ra/html/sec_arzelaascoli.html

For our problem, we assume $f_1,f_2,\dots \in C^2[a,b],$ $f_n'(c)$ is bounded for some $c\in [a,b],$ and $f_n''$ is uniformy bounded on $[a,b].$ Assume further that $f_n$ converges pointwise to $f$ on $[a,b].$ We want to prove that $f\in C^1[a,b].$

Proof: There exists $C$ such that $|f_n''|\le C$ on $[a,b].$ It follows that for $x,y\in [a,b],$ the MVT implies

$$|f'(x)-f_n'(y)|\le C|y-x|.$$

This implies $(f_n')$ is uniformly equitcontinuous on $[a,b].$ Furthermmore, since there is $C_1$ such that $|f_n'(c)|\le C_1$ for all $n,$ we have

$$|f_n'(x)|\le |f_n'(x)-f_n'(c)| + |f_n'(c)| $$ $$\le C|x-c| + C_1 \le C(b-a) + C_1.$$

Thus $f_n'$ is uniformly bounded on $[a,b].$

Apply A-A to see there is a subsequence $f_{n_k}'$ converging uniformly to some $h$ on $[a,b].$ Because each $f_{n_k}'$ is continuous, so is $h.$ Fixing $x_0\in [a,b],$ we have for any $x$ that

$$f_{n_k}(x)= f_{n_k}(x_0) + \int_x^{x_0}f_{n_k}'.$$

Letting $k\to\infty,$ we get

$$f(x)= f(x_0) + \int_x^{x_0}h.$$

Since $h$ is continuous, $f\in C^1[a,b].$

We're not quite done, because your problem occurs on some $(a,b),$ not $[a,b].$ But our work above shows $f\in C^1[a',b']$ for all $a<a'<b'<b,$ and that implies $f\in C^1(a,b).$

Answer 2

Choose $M>0$ such that $|f''_{n}(x)|\leq M$ and $|f'_{n}(c)|\leq M$ for all $x\in(a,b)$ and $n\in\mathbb{N}$.

---

Claim 1: The sequence $(f'_{n})$ is uniformly bounded on $(a,b)$.

Proof of Claim 1: Let $x\in(a,b)$ and $n\in\mathbb{N}$. By mean-value theorem, we have that \begin{eqnarray*} \left|f'_{n}(x)-f'_{n}(c)\right| & = & \left|f''(\xi_{n,x})(x-c)\right|\\ & \leq & M(b-a). \end{eqnarray*} Hence, $|f'_{n}(x)|\leq|f'_{n}(c)|+M(b-a)\leq M(b-a+1).$

---

Claim 2: There exists a subsequence $(f_{n_{k}})$ of $(f_{n})$ such that $\lim_{k\rightarrow\infty}f_{n_{k}}'(x)$ exists for each $x\in(a,b)$.

Proof of Claim 2: Since for each $x\in(a,b)$, $\{f_{n}'(x)\mid n\in\mathbb{N}\}$ is bounded, there exists a strictly increasing function $\phi:\mathbb{N}\rightarrow\mathbb{N}$ such that $(f'_{\phi(n)}(x))_{n}$ is convergent for each $x\in(a,b)\cap\mathbb{Q}$. This is known as Cantor Diagonal Argument. For the sake of completeness and to facilitate those who do not know such argument, I include a complete proof.

Fix an enumeration $\{r_{n}\mid n\in\mathbb{N}\}$ for $\mathbb{Q}\cap(a,b)$. Since $(f_{n}'(r_{1}))_{n}$ is a bounded sequence, it has a convergent subsequence $(f_{\theta_{1}(n)}'(r_{1}))_{n}$, where $\theta_{1}:\mathbb{N}\rightarrow\mathbb{N}$ is a strictly increasing function. For the bounded sequence $(f_{\theta_{1}(n)}'(r_{2}))_{n}$, we can choose a convergent subsequence $(f_{\theta_{1}\circ\theta_{2}(n)}'(r_{2}))_{n}$, where $\theta_{2}:\mathbb{\mathbb{N}\rightarrow\mathbb{N}}$ is a strictly increasing function. Note that $(f_{\theta_{1}\circ\theta_{2}(n)}'(x))_{n}$ converges for $x\in\{r_{1},r_{2}\}$. Repeating this argument recursively, we obtain strictly increasing functions $\theta_1,\theta_2,\ldots,\theta_{m}:\mathbb{N}\rightarrow\mathbb{N}$ such that $(f_{\theta_{1}\circ\theta_{2}\circ\ldots\circ\theta_{m}(n)}'(x))_{n}$ is convergent for $x\in\{r_{1},r_{2},\ldots,r_{m}\}$. Define $\phi:\mathbb{N}\rightarrow\mathbb{N}$ by $\phi(n)=\theta_{1}\circ\ldots\circ\theta_{n}(n)$, then $\phi$ is strictly increasing. Moreover, $(f_{\phi(n)}'(x))_{n}$ is convergent for $x\in\mathbb{Q}\cap(a,b)$.

We include the details of proof. We verify that $\phi$ is strictly increasing: Let $n\in\mathbb{N}$, then \begin{eqnarray*} \phi(n+1) & = & (\theta_{1}\circ\ldots\circ\theta_{n})\circ\theta_{n+1}(n+1)\\ & \geq & (\theta_{1}\circ\ldots\circ\theta_{n})(n+1)\\ & > & (\theta_{1}\circ\ldots\circ\theta_{n})(n)\\ & = & \phi(n). \end{eqnarray*} In the above, we have used the fact that $\theta_{1}\circ\ldots\circ\theta_{n}$ is strictly increasing and $\xi(n)\geq n$ whenever $\xi:\mathbb{N}\rightarrow\mathbb{N}$ is a strictly increasing function.

We verify that $(f'_{\phi(n)}(x))_{n}$ is convergent for all $x\in(a,b)\cap\mathbb{Q}$. Let $x\in(a,b)\cap\mathbb{Q}$, then $x=r_{k}$ for a unique $k$. Let $\varepsilon>0$ be given. Since $(f_{\theta_{1}\circ\ldots\circ\theta_{k}(n)}'(r_{k}))_{n}$ is convergent, there exists $N$ such that $|f_{\theta_{1}\circ\ldots\circ\theta_{k}(n)}'(r_{k})-l|<\varepsilon$ whenever $n\geq N$. Here, $l=\lim_{n\rightarrow\infty}f_{\theta_{1}\circ\ldots\circ\theta_{k}(n)}'(r_{k}).$ Let $N_{1}=\max(k,N)$. Let $n\geq N_{1}$ be arbitrary. Note that \begin{eqnarray*} \phi(n) & = & (\theta_{1}\circ\ldots\circ\theta_{k})\circ(\theta_{k+1}\circ\ldots\circ\theta_{n})(n) \end{eqnarray*} and $(\theta_{k+1}\circ\ldots\circ\theta_{n})(n)\geq n\geq N$, so $|f_{\phi(n)}'(r_{k})-l|<\varepsilon$. This shows that $(f_{\phi(n)}'(x))_{n}$ is convergent.

We further prove that $(f_{\phi(n)}'(x))_{n}$ is convergent for all $x\in(a,b)\setminus\mathbb{Q}$. Denote $g_{n}=f_{\phi(n)}'$. Let $x\in(a,b)\setminus\mathbb{Q}$. Let $\varepsilon>0$ be given. Choose $y\in(a,b)\cap\mathbb{Q}$ such that $|x-y|<\frac{\varepsilon}{4M}$. Note that $(g_{n}(y))_{n}$ is convergent, so we can choose $N$ such that $|g_{m}(y)-g_{n}(y)|<\frac{\varepsilon}{2}$ whenever $m,n\geq N$. Let $m,n\geq N$, then \begin{eqnarray*} \left|g_{n}(x)-g_{m}(x)\right| & = & \left|[g_{n}(y)-g_{m}(y)]+[g_{n}(x)-g_{n}(y)]+[g_{m}(y)-g_{m}(x)]\right|\\ & \leq & \left|g_{n}(y)-g_{m}(y)\right|+\left|g_{n}(x)-g_{n}(y)\right|+\left|g_{m}(y)-g_{m}(x)\right|\\ & \leq & \frac{\varepsilon}{2}+2M|x-y|\\ & < & \varepsilon. \end{eqnarray*} This shows that $(g_{n}(x))_{n}$ is a Cauchy sequence and hence is convergent.

---

Claim 3: Define $g:(a,b)\rightarrow\mathbb{R}$, $g(x)=\lim_{k\rightarrow\infty}f_{n_{k}}'(x)=\lim_{k\rightarrow\infty}f_{\phi(k)}'(x),$ where $\phi$ is as in Claim 2. Then, $g$ is continuous.

Proof of Claim 3: Let $x_{0}\in(a,b)$. Let $\varepsilon>0$ be given. Define $\delta=\frac{\varepsilon}{M}$. Let $x\in(x_{0}-\delta,x_{0}+\delta)\cap(a,b)$ be arbitrary. Choose $n$ such that $|g(x_{0})-g_{n}(x_{0})|<\varepsilon$ and $|g(x)-g_{n}(x)|<\varepsilon$. We have estimation \begin{eqnarray*} \left|g(x)-g(x_{0})\right| & \leq & \left|g(x)-g_{n}(x)\right|+\left|g_{n}(x)-g_{n}(x_{0})\right|+\left|g_{n}(x_{0})-g(x_{0})\right|\\ & < & \varepsilon+M|x-x_{0}|+\varepsilon\\ & \leq & 3\varepsilon. \end{eqnarray*} This shows that $g$ is continuous at an arbitrary point $x_{0}$ and hence $g$ is a continuous function.

---

Now, define $G:(a,b)\rightarrow\mathbb{R}$ by $G(x)=\int_{c}^{x}g(t)dt$. By fundamental theorem of calculus, $G'(x)=g(x)$ for all $x\in(a,b)$. On the other hand, since $(g_{n})_{n}$ is uniformly bounded, by Lebesgue dominated convergence theorem, we have that \begin{eqnarray*} G(x) & = & \int_{c}^{x}g(t)dt\\ & = & \lim_{k\rightarrow\infty}\int_{c}^{x}f_{n_{k}}'(t)dt\\ & = & \lim_{k\rightarrow\infty}\left\{ f_{n_{k}}(x)-f_{n_{k}}(c)\right\} \\ & = & f(x)-f(c). \end{eqnarray*} Hence, $f(x)=G(x)+f(c)$. It follows that $f$ is differentiable and $f'=G'=g$ is continuous. That is, $f\in C^{1}$.