# Show that $C^{1}([a,b];\mathbb{R})$ is a Banach space.

Let $C^{1}([a,b];\mathbb{R})$ the vectorial space of the functions (bounded) $f:[a,b]\to\mathbb{R}$ where all $f$ has a continuous derivate (and bounded) in all point of $[a,b]$, with the norm $\vert\vert f\vert\vert_{1}=\displaystyle\sup_{a\leq x\leq b}{(|f(x)|+|f'(x)|)}$. Show that $C^{1}([a,b];\mathbb{R})$ is a Banach space. For all $r\in\mathbb{N}$, define the Banach space $C^{r}([a,b];\mathbb{R})$.

I know that complete vectorial space is a Banach space, this mean that cauchy sequence with the norm $\vert\vert f\vert\vert_{1}$ converges in the space. But I have issues with this. Regards!

• Hint: Uniform convergence of derivatives plus a point convergence of the original functions guarantees the exchange of limit and derivatives. – user111463 Sep 4 '15 at 4:43

## 2 Answers

Hint.

Denote by $\Vert f \Vert =\displaystyle\sup_{a\leq x\leq b}{|f(x)|}$ and suppose that $(f_n)$ is a Cauchy sequence of $C^{1}([a,b];\mathbb{R})$.

Then $(f^\prime_n)$ is a Cauchy sequence for the norm $\Vert \cdot \Vert$ hence converges to a continuous function $g$ as $C^0([a,b];\mathbb{R})$ is a Banach space. Also $(f_n)$ converges to a continuous function $f$.

Then, you can use a theorem that states that if a sequence of differentiable functions $(f_n)$ is such that the sequence of derivatives converge uniformly and $(f_n(a))$ converges for at least one point, $(f_n)$ converges uniformly to a differentiable function.

Hint: If $\{f_n\}_n$ is a Cauchy sequence then for every $\epsilon >0$ there is $n_0\in \mathbb{N}$ such that $||f_n-f_m||<\epsilon$ when $n,m>n_0$. From this $\sup_{x\in [a,b]}|f_n(x)-f_m(x)|<\epsilon$ and $\sup_{x\in [a,b]}|f'_n(x)-f'_m(x)|<\epsilon$. You can use Ascoli-Arzela theorem for obtain a candidate for $\lim_nf_n$ and $\lim_n f'_n$. If $f=\lim_nf_n$($C^0([a,b])$) try to prove that if $g=\lim_nf'_n$ ($C^0([a,b]))$, so $g=f'$.