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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?

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3 Answers 3

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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 : $$\|f\|_{C_1} \leq 2\|f\| $$

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    $\begingroup$ 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 $\|.\|$? $\endgroup$
    – kam
    Apr 5, 2020 at 12:27
  • $\begingroup$ Yes : two equivalent normed vector spaces have the same Cauchy sequences, and convergent sequences. $\endgroup$
    – purple
    Apr 5, 2020 at 12:43
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    $\begingroup$ 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? $\endgroup$
    – kam
    Apr 5, 2020 at 13:23
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    $\begingroup$ $\int_{0}^{x} f^{'} (t) dt \leq \int_{0}^{1} \|f^{'} \|_{\infty}dt = \|f^{'} \|_{\infty}$ $\endgroup$
    – purple
    Apr 5, 2020 at 13:27
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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.

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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.

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