Show that $\{f_n\}$ converges uniformly to some continuously differentiable function $f:[a,b]\to R.$

Question

Q. For a continuously differentiable function $f:[a,b]\to \mathbb{R}$, define $$\|f\|_{C^1}=\|f\|_u+\|f^{\prime}\|_u.$$ Suppose $\{f_n\}$ is a sequence of continuously differentiable functions such that for every $\varepsilon>0$, there exists an $M$ such that for all $n,k\ge M$ we have $$\|f_n-f_k\|_{C^1}<\varepsilon.$$ Show that $\{f_n\}$ converges uniformly to some continuously differentiable function $f:[a,b]\to \mathbb{R}.$

I have no idea of how to solve this question. I just know $f_n$ is uniformly cauchy, but I am not sure it implies $f_n$ converges uniformly to $f$ since we don't know $f_n$ is bounded.

Could you give some hint?

Answer 1

Since $\|\cdot\|_u \le \|\cdot\|_{C^1}$ we see that the sequences $(f_n)_n$ and $(f_n')_n$ are Cauchy w.r.t $\|\cdot\|_u$. Since $\left(C[a,b], \|\cdot\|_u\right)$ is a Banach space, there exist $f,g \in C[a,b]$ such that $f_n \to f$ and $f_n' \to g$ uniformly. We claim that $f$ is differenentiable and $f' = g$.

First note that $\phi_n(x) = \frac{f_n(x) - f_n(c)}{x-c}$ converges uniformly over $x \in [a,b]$ to $\phi(x) = \frac{f(x) - f(c)}{x-c}$.

Let $\varepsilon > 0$ and pick $n_0 \in\mathbb{N}$ such that $m, n \ge n_0 \implies \|f_n' - f_m'\|_u < \frac\varepsilon2$. For $m, n \ge n_0$ and all $x \in [a,b]$ the MVT on the function $f_n - f_m$ gives

$$|\phi_n(x) - \phi_m(x)| = \left|\frac{f_n(x) - f_n(c)}{x-c} - \frac{f_m(x) - f_m(c)}{x-c}\right| = \left|\frac{(f_n-f_m)(x) - (f_n-f_m)(c)}{x-c}\right| = |f_n'(\theta) - f_m'(\theta)|$$

for some $\theta$ between $x$ and $c$. Hence $|\phi_n(x) - \phi_m(x)| \le \|f_n' - f_m'\|_u < \frac\varepsilon2, \forall x \in [a,b]$.

Clearly $\phi_n \to \phi$ pointwise so letting $m\to\infty$ we get $|\phi_n(x) - \phi(x)| \le \frac\varepsilon2 < \varepsilon, \forall x \in [a,b]$. We conclude $\phi_n \to \phi$ uniformly.

Now pick $c \in [a,b]$. We claim that $f$ is differentiable at $c$ and $f'(c) = g(c)$. Let $\varepsilon > 0$. Since $\phi_n \to \phi$ and $f_n' \to g$ uniformly, there exists $n_0 \in \mathbb{N}$ such that $n \ge n_0 \implies \|\phi - \phi_n\|_u, \|f_n' - g\|_u < \frac\varepsilon3$.

The function $f_{n_0}$ is differentiable at $c$ so there exists $\delta > 0$ such that $0 < |x-c| < \delta \implies \left|\frac{f_{n_0}(x) - f_{n_0}(c)}{x-c} - f_{n_0}'(c)\right| < \frac\varepsilon3$.

Putting this together, for $0 < |x-c| < \delta$ we have

$$\left|\frac{f(x) - f(c)}{x-c} - g(c)\right| \le \overbrace{\left|\frac{f(x) - f(c)}{x-c} - \frac{f_{n_0}(x) - f_{n_0}(c)}{x-c}\right|}^{= |\phi(x) - \phi_{n_0}(x)|} + \left|\frac{f_{n_0}(x) - f_{n_0}(c)}{x-c} - f_{n_0}'(c)\right| + |f_{n_0}'(c) - g(c)| < \frac\varepsilon3 + \frac\varepsilon3 + \frac\varepsilon3 = \varepsilon $$

Hence $\lim_{x\to c} \frac{f(x) - f(c)}{x-c} = g(c)$.

Since $c \in [a,b]$ was arbitrary, we conclude $f' = g$. Since $g \in C[a,b]$, we have $f' \in C[a,b]$, so $f$ is continuously differentiable, i.e. $f \in C^1[a,b]$.

It remains to prove that $f_n \to f$ in $\|\cdot\|_{C^1}$, which is trivial:

$$\|f - f_n\|_{C^1} = \|f - f_n\|_{u} + \|f' - f_n'\|_{u} \xrightarrow{n\to\infty} 0$$

In fact, we only used that $f_n \to f$ pointwise (not uniformly), so the stronger claim holds:

Theorem.

Let $(f_n)_n$ be a sequence of functions $f_n : [a,b] \to \mathbb{R}$ differentiable on $[a,b]$ and suppose that $f_n \to f$ pointwise and $f_n' \to g$ uniformly, where $f,g : [a,b] \to \mathbb{R}$.

Then $f$ is differentiable on $[a,b]$ with $f' = g$.

Answer 2

$\{f_n\}$ and $\{f_n'\}$ are Cauchy sequences for the sup norm so there exist continuous functions $f$ and $g$ such that $f_n \to f$ uniformly and $\{f_n'\} \to g$ uniformly. Just take limits in the equation $f_n(x)=f_n(0)+\int_0 ^{x} f_n'(t) \, dt$ to get $f(x)=f(0)+\int_0 ^{x} g(t) \, dt$. This implies that $f$ is differentiable (with derivative $g$).