Dgdrxrt

I have che following exercise and some dubts:

Let $M>0$ and $mathcalF= f $. Prove that $mathcalF$ is relatively compact in $(C^0([a,b]), | cdot |_infty)$ and $mathcalF$ is not compact in $(C^1([a,b]), | cdot |_C^1 )$.

Just to be clear with $| cdot |_infty$ I mean the uniform norm, hence $| f |_infty=sup_xin [a,b]|f(x)|$, and with $| cdot |_C^1$ i mean the $C^1([a,b])$ norm hence $| f|_C^1=sum_alpha| D^alphaf |_infty$.

• For the relative compactness in $(C^0([a,b]), | cdot |_infty)$ I argued as follows: I have to prove that $overlinemathcalF$ is compact in $(C^0([a,b]), | cdot |_infty)$ that is equivalent to show that $overlinemathcalF$ is closed and bounded in $(C^0([a,b]), | cdot |_infty)$.
  Given the definition of $mathcalF$ I know that exists a positive constant $Nleq M$ such that
  beginequation
  | f |_infty leq Ntext for any $f in overlinemathcalF$,
  endequation
  hence $overlinemathcalF$ is bounded. Additionaly $overlinemathcalF$ is closed, in fact if I take a sequence $(f_n)_nsubset mathcalF$ I have that $f_nrightarrow f$ where $f in overlinemathcalF$ and the convergence of $f_n$ is induced by the norm $| cdot |_infty$. Thus $overlinemathcalF$ is closed in $(C^0([a,b]), | cdot |_infty)$.




• For the non-compactness of $mathcalF$ in $(C^1([a,b]), | cdot |_C^1)$ I used the following counterexample: I choose $[a,b]=[-1,1]$ and the sequence
  beginequation
  f_n(x)=|x|^1+frac1h
  endequation
  I have that $(f_h)_n in C^1([-1,1])$ and $f_nrightarrow f$, where $f(x)=|x|$ that does not belong to $C^1([-1,1])$. So $mathcalF$ is not compact in $(C^1([a,b]), | cdot |_C^1)$.

Is my argument flawed at some point?

The second part is fine. For the first part, use Arzela-Ascoli: a subset $mathcal F$, of Banach space $X$ is relatively compact if and only if it is bounded and equicontinuous. For convenience, take $M=1$ and $[a,b]=[0,1].$ So we are dealing with the closed unit ball $B=B_C^1(f,0).$

$underlinemathcal F textis equicontinuous:$

By definition of the $C^1$ norm, for any $fin B, |f|_infty+|f'|_inftyle 1.$ Therefore, by the mean value theorem, $f$ is Lipschitz bounded on $[0,1]: f(x)-f(y)le 1cdot |x-y|$ for all $x,yin [0,1],$ which implies immediately that $mathcal F$ is equicontinuous on $B$.

$underlinemathcal F textis bounded:$

It is enough to note that $Bsubseteq B_C^0(f,0)$ so $mathcal F$ is bounded $textitin the norm of the larger space C([0,1])$.