# Is $(C^1[0,1],\|\cdot\|)$ with $\|f\|:=|f(0)|+\sup_{0\le{t\le{1}}}{|f'(t)|}$ a Banach space?

Let $$(C^1[0,1],\|\space{}.\|)$$ be a normed space where $$C^1[0,1]$$ is the set of functions with continuous derivatives and let $$\|\space{}.\|$$ be the norm on this set defined by: $$\|f\|:=|f(0)|+\sup_{0\le{t\le{1}}}{|f'(t)|}.$$ Is this space Banach?

My attempt:

I don't think it is. Here is my counter example:

Let $$(f_n)_{n=1}^{\infty}$$ be a sequence of functions defined by $$f_n(t)=\sqrt{(t-\frac{1}{2})^{2}+\frac{1}{n}}$$. This clearly belongs to $$C^1[0,1]$$ but its limit does not, namely $$f(t)=|t-\frac{1}{2}|$$. My issue however is this hasn't shown the sequence converges to $$f$$ with respect to the norm $$\|\space{}.\|$$. I don't know how to use this example, since the norm doesn't make sense with $$f$$, since f is not differentiable on t = 1/2. But does this counter example work?

Hint : it's a Banach Space.

Why ? Because $$(C^1 [0,1], \| \cdot \|_{C_1} )$$ where $$\|f\|_{C_1} = \|f\|_{\infty} + \|f^{'} \|_{\infty}$$ is a Banach Space (more classical). And norms of both spaces are equivalent : It's obvious that :

$$\|f\| \leq \|f\|_{C_1}$$

But since : $$|f(x)| = |\int_0^{x}f'(t) dt + f(0)| \leq \|f^{'}\|_{\infty} + |f(0)|$$ we also have : $$\|f\|_{C_1} \leq 2\|f\|$$

• And this shows that $C^1[0,1]$ is complete w.r.t to $\|.\|_{C^1}$ which is equivalent to $\|.\|$ so $C^1[0,1]$ is also complete w.r.t $\|.\|$?
– kam
Apr 5, 2020 at 12:27
• Yes : two equivalent normed vector spaces have the same Cauchy sequences, and convergent sequences. Apr 5, 2020 at 12:43
• In the second to last line how to do you show the inequality, that the integral is less than or equal to the infinity norm of the derivative?
– kam
Apr 5, 2020 at 13:23
• $\int_{0}^{x} f^{'} (t) dt \leq \int_{0}^{1} \|f^{'} \|_{\infty}dt = \|f^{'} \|_{\infty}$ Apr 5, 2020 at 13:27

The space is complete. Your sequence is not a Cauchy sequence.

Let $$(f_n)$$ be a Cauchy sequence. Then $$\lim f_n(0)$$ exists and $$f_n'$$ converges uniformly to some continuous function $$g$$. Now $$f_n(x)=f_n(0)+\int_0^{x} f_n'(t)dt$$. From this we see that $$(f_n)$$ is uniformly Cauchy and hence $$f_n$$ tends to a continuous function $$f$$ uniformly. Uniform convergence of $$f_n$$ to $$f$$ and $$f_n'$$ to $$g$$ implies that $$f$$ is differentiable and $$f'=g$$. It should now be easy to see that $$f_n \to f$$ in the given norm.

Consider the Cauchy sequence $$\{f_n\}$$, i.e. $$\forall\varepsilon>0$$ $$\exists N\in\mathbb{N}:$$ $$\forall n,m>N$$ $$\|f_n-f_m\|<\varepsilon$$. Since $$|f_n(0)-f_m(0)|+\sup\limits_{t\in[0,1]}|f_n'(t)-f_m'(t)|<\varepsilon$$, then $$|f_n(0)-f_m(0)|<\varepsilon$$ and $$\forall t\in[0,1]$$ $$|f_n'(t)-f_m'(t)|<\varepsilon$$. Thus, $$\{f_n(0)\}$$ is Cauchy and $$\{f_n'(t)\}$$ is uniformly Cauchy. Therefore $$\exists\lim\limits_{n\to\infty}f_n(0)$$ and $$\{f_n'(t)\}$$ is uniformly convergent. By the well-known theorem, this means that $$\{f_n(t)\}$$ converges uniformly to $$f(t)$$ and $$\{f_n'(t)\}$$ converges uniformly to $$\{f'(t)\}$$. By definition of uniformly convergence we have $$\forall\varepsilon>0$$ $$\exists N\in\mathbb{N}:$$ $$\forall n>N$$ $$\forall t\in[0,1]$$ $$|f_n(t)-f(t)|<\varepsilon$$ and $$|f_n'(t)-f'(t)|<\varepsilon$$. Thus, $$\|f_n-f\|\leq2\varepsilon$$. So $$\{f_n\}$$ is converges in $$C^1[0,1]$$ and $$C^1[0,1]$$ is Banach.