# For which n differential operator $C^{(n)}[0, 1] \rightarrow C[0,1]$ is compact

I'm trying to learn functional analysis mostly on my own, and I'm stuck with the following task on operator compactness.

Consider the operator $Tx(t)=\frac{dx}{dt}$, given $T$ maps $C^{(n)}[0,1]$ to $C[0,1]$. The norm in $C^{(n)}[0,1]$ is given by (standard) $||f||=sup|f^{(n)}| + \dots + sup|f^{'}| + sup|f|$.

The task is to find out for which $n$ the operator is compact and bounded.

I have successfully managed to prove that under the given norm, $T$ is always bounded. Now, I'm trying to analyse its compactness.

It may be possible to make use of Arzela-Ascoli theorem (Let $X$ be a compact metric space and $\{f_{n}\}$ a uniformly bounded equicontinuous sequence of real-valued functions on $X$. Then $\{f_{n}\}$ has a subsequence that converges uniformly on $X$ to a continuous function $f$ on $X$), but I'm unable to move forward with this.

$T$ is bounded for any positive integer $n$. I fact $||Tf|| \leq ||f||$. For $n \geq 2$ consider a norm bounded sequence $\{f_n\}$ in $C^{(n)}[0,1]$. Since teh second derivative of teh $f_n$'s are bounded it follows that $\{f_n'\}$ is equi-continuous (by Mean Value Theorem) and uniformly bounded. Hence $\{f_n\}$ has a convergent sub-sequence in $C[0,1]$ by Arzela - Ascoli Theorem. It follows that $T$ is compact if $n \geq 2$. It is not compact for $n=1$: $\{\frac {t^{n}}{ n}\}$ is abounded sequence in $C^{1}[0,1]$ but $T(\frac {t^{n}}{ n})=t^{n-1}$ has no convergent subsequence in $C[0,1]$.
• Could you please explain why $\{f_n^`\}$ is uniformly bounded? It is bounded, surely, but why uniformly? – mathjunkie May 21 '18 at 9:51
• @V.Shakhray a sequence $\{f_n\}$ of functions is uniformly bounded if and only if the sequnce of numbers $\{||f_n||_{\infty}\}$ is bounded (where $\{||f_n||_{\infty}=sup \{|f_n (x)|: 0\leq x \leq 1\}$. – Kabo Murphy May 22 '18 at 6:33