A norm on $C^1([a,b])$ Let $X=C^1([a,b])$. Let 
$$\|f\|_1=\|f\|_{\infty}$$
$$\|f\|_2=\|f\|_{\infty}+\|f'\|_\infty$$
$$ \|f\|_3=|f(a)|+\|f'\|_\infty $$
$$ \|f\|_4=\|f\|_\infty+\left|f'\left(\frac{a+b}{2}\right)\right|$$
$$ \|f\|_5=\int_a^b|f(x)|dx+\int_a^b|f'(x)|dx  $$
Then which of them will make $X$ to Banach space. 


*

*We know that $(X,\|.\|_1)$ is not Banach space (Using Weierstrass Theorem).

*Second one i.e. $(X,\|.\|_2)$ is a Banach space. 

*I have problem with $\|.\|_3$ and $\|.\|_4$ one.


For the fourth one, let $c=\frac{a+b}{2}$. Consider the sequence $f_n=|x-c|^{\frac{n+1}{n}}$, then 
\begin{align*}
\|f_n\|_4=\|f_n\|_\infty+|f_n'(c)|\to |x-c|\notin X \implies X\ \text{is not complete}.
\end{align*}
For the third one, if $(f_n)$ be any Cauchy sequence in $X$ then for given $\varepsilon>0$,   there exists $n_0\in \mathbb{N}$ such that for $m,n\ge n_0$ $$\|f_m-f_n\|_3<\varepsilon\implies |f_m(a)-f_n(a)|<\varepsilon\ \text{and}\ \|f_m'-f_n'\|_\infty<\varepsilon\implies (f_m(a)), \text{and}\ (f_n')\ \text{are Cauchy sequences}.$$
Since, $(f_n(a))\subset \mathbb{R}$ will converge to some point, say $f(a)$. Also, $f_n'\to g\in C[a,b]$. Now I stuck what to do afterwards. 
For, fifth one I conclude that $(f_n) $ is Cauchy with respect to $\|.\|_5 \implies $ It is Cauchy with respect to $ \|.\|_2$. Hence it is also complete. 
 A: The third norm does make a Banach space out of $C^1([a,b])$.  To see this, observe that for any $c\in[a,b]$,
$$
|f(c)| = \left| f(a) + \int_a^c f'(s)\,ds\right|
\leq |f(a)| + \int_a^b |f'(s)|\,ds \leq |f(a)| + (b-a) \|f'\|_{\infty}.
$$
In particular, we have $\|f\|_{\infty} \leq (1+b-a)\|f\|_3$.  Therefore convergence in the third norm implies uniform convergence as well as uniform convergence of the derivative, so we are done.
For the fifth one, it is not the case that being Cauchy with respect to $\|\cdot\|_5$ implies being Cauchy with respect to $\|\cdot\|_2$.  For example, suppose $f_n$ is a bump of height $1/n$ and has compact support in an interval of width $1/n^2$.  Then $\|f_n\|_5 \to 0$ but $\|f_n\|_2\to\infty$.  
This also means that $C^1$ is not complete with respect to $\|\cdot\|_5$.  To see this suppose $f_n(x) = 0$ for $x<\frac{a+b}{2} - \frac{1}{n}$ and $f_n(x) = x - \frac{a+b}{2}$ for $x>\frac{a+b}{2} + \frac{1}{n}$, with a smooth interpolation between the endpoints.  Then $\|f_n-f\|_5\to 0$ where $f$ is the ramp function $f(x) = (x - \frac{a+b}{2})^+$, but $f\not\in C^1([a,b])$.
