Is $i: (C^1, ||·||_{W^{1,2}}) → (C^0, ||·||_∞)$ a linear, continuous, compact map? 
Consider the map 
  $$i:  (C^1[0,1], ||·||_{W^{1,2}}) → (C^0[0,1], ||·||_∞)$$
  which maps every function to itself, and with Sobolev norm defined as
  $$||u||_{W^{1,2}}=||u||_{L^2}+||u'||_{L^2}.$$
  Is $i$ linear, continuous, compact?

Linearity
Consider $u,v\in C^1[0,1]$ and $a,b\in \mathbb{R}$:
$i(au+bv)=au+bv=ai(u)+bi(v)$
Continuity
By fundamental theorem of calculus: $u(x)=u(0)+\int_0^x u'(t)dt$. Then:
$\begin{align*}
|u(x)| &\le |u(0)|+\int_0^x |u'(t)|dt \\
&\le |u(0)|+\int_0^1 |u'(t)|dt \\
&\le C|u(0)|^2+C\int_0^1 |u'(t)|^2dt \\
&\le C\int_0^1 |u(t)|^2dt+C\int_0^1 |u'(t)|^2dt \\
&\le C(||u||_{L^2}+||u'||_{L^2}) \\
&=C||u||_{W^{1,2}}
\end{align*}$
for $C$ large enough and by mean value theorem.
Since $i$ is linear and bounded, it is also continuous.
Compactness
$i$ is defined on an infinite-dimensional space, so by Riesz theorem the closed unit ball $B$ is not compact. If the dual norm of $i$ would be $1$, then we could say that $i(B)\subseteq B$, and so also $i$ would not be compact. But in this exercise I cannot show this is the case.
Are the computations for linearity and continuity correct?
How to check the compactness?
 A: For compactness it has to be shown that a bounded set is relatively compact.
Now let $G$ be a bounded set, i.e. there is an $M$ such that for all $f \in G$ it holds $ ||f||_{L^2} + ||f'||_{L^2} \le M$.
If we can show now that all elements of $G$ satisfy a uniform Hölder condition (see Arzela-Ascoli theorem in https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem#Lipschitz_and_H%C3%B6lder_continuous_functions) then the set is relatively compact and we are done.
And in fact : Let $a, b \in [0,1]$, and $f \in G $. Then we have $ |f(a)-f(b)| = |\int_a^b f' | = |\int_a^b (f' \times 1) | \le ||f'||_{L^2[a,b]} ||1||_{L^2[a,b]} = ||f'||_{L^2[a,b]}|a-b|^{1/2} \le ||f'||_{L^2}|a-b|^{1/2} \le M|a-b|^{1/2}.$
So Arzela-Ascoli is applicable and we have proven the requested property.
For continuity as $i$ is linear, then $i$ is continuous iff there is a constant $C$ such that for all $u$ it holds $||u||_\infty \le C (||u||_{L^2}+||u'||_{L^2})$.
Now let us pick some $u$. Then there is a $r\in [0,1]$ where $u^2$ attains its minimum. And $|u|$ attains its minimum in $r$ as well. This $r$ is clearly dependant on $u$. But for any $x\in [0,1]$ we have $|u(x)| = |\int^x_r u'(t)dt +u(r)| \le |\int^x_r u'(t)dt| +|u(r)| = |\int^x_r u'(t) \times 1 dt| +|\int_0^1 |u(r)| \times 1 dt| \le ||u'||_{L^2} + |\int_0^1 |u(t)| \times 1 dt|\le ||u'||_{L^2} + ||u||_{L^2}$
Thus $C=1$ and we are done.
