# Show that a sequence of functions is not compact in $(C^1([-1,1]),\lVert\,\cdot\,\rVert_{\infty})$

Let $F:=(f_n)_{n\in\mathbb{N}}$ where $$\forall x\in[-1,1]\,\,\,\,\,\,\,\,\,f_n(x):=\mid x\mid^{1+\frac{1}{n}}$$ I have to prove that $F$ is not compact in $(C^1([-1,1]),\lVert\,\cdot\,\rVert_{\infty})$.

My idea is to prove that $F$ is not sequentially compact w.r.t. the topology induced by the infinity norm.

In particular, if I prove that $f_n\to f$ uniformly on $[-1,1]$, where $f(x):=\lvert x\rvert$, we are done. Infact, every subsequence will also converge in norm to $f\notin C^1([-1,1])$.

Is this reasoning correct? If yes, how can I prove $f_n\to f$ uniformly on $[-1,1]$?

• Sounds like it; you're basically showing that $F$ is not closed, and compactness implies closedness. Note that in a metric space, sequential compactness is the same as compactness. – Hayden May 6 '14 at 22:06
• $F$ is closed, but not complete. – Daniel Fischer May 6 '14 at 22:06
• So my reasoning is totally wrong. How can I proceed? – avati91 May 6 '14 at 22:08
• No, your reasoning is correct. The sequence converges in the larger space $C^0([-1,1])$, and the limit does not lie in the subspace $C^1([-1,1])$. Therefore, the sequence has no subsequences that converge in $C^1$, whence $F$ is not compact. Well, you need to prove the uniform convergence to have a complete proof. – Daniel Fischer May 6 '14 at 22:10
• $F$ is closed in $C^1([-1,1])$, but not in $C^0([-1,1])$. It has no accumulation points in the smaller space, but it has one in the larger. If $F$ were compact, it would be closed in every larger space as long as that is Hausdorff. – Daniel Fischer May 6 '14 at 22:22

## 1 Answer

It's enough to consider $x\geq 0$. Then $$\left|x\right|-\left|x\right|^{1+\frac{1}{n}}= x-x^{1+\frac{1}{n}}= x\left(1-x^\frac{1}{n}\right)=$$ $$=n^n\left(\frac{x^{\frac{1}{n}}}{n}\right)^n\left(1-x^\frac{1}{n}\right)$$ By AM-GM inequality, $$\ldots\leq n^n\left(\frac{n\,\frac{x^{1/n}}{n} + 1 - x^{1/n}}{n+1}\right)^{n+1}= \frac{n^n}{(n+1)^{n+1}}=\frac{1}{n+1}\left(\frac{n}{n+1}\right)^n\leq \frac{1}{n+1}\to0$$