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

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?

• You know $f_n$ is bounded! Read the question again and break-up definitions. (In fact, you know there is a constant $\alpha > 0$ such that $\|f_n(x)\| \leq \alpha$ for all $x$ and $n.$ Just read definitions of the symbols. – Will M. May 27 '18 at 6:30
• Do you mean convergence in cauchy uniform implies $f_n$ is bounded? – shk910 May 27 '18 at 6:43
• Any fundamental sequence is bounded, hence $\|f_n\|_{C^1}$ is bounded, hence $\|f_n\|_u$ is bounded. – Will M. May 27 '18 at 6:47

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 $$f' = g$$.

$\{f_n\}$ 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 $g$).