Help showing derivative is compact I'm struggling with proving the following theorem:
$\textit{Let }A:U\subset X\rightarrow Y\textit{ be a completely continuous operator from an open subset U of a normed}$
$\textit{space X into a Banach space Y and assume A to be Fréchet differentiable at }\psi\in U.\textit{ Then the}$
$\textit{derivative }A'_{\psi}\textit{ is compact.}$
I'm not sure how to go about this. I was thinking of trying to do it by contradiction, i.e., assuming that $A'_{\psi}$ is not compact, but I'm not really sure how to proceed from there... I guess what I am asking for is a hint or a strategy for trying to prove $A'_{\psi}$. Any help is highly appreciated.
Thanks in advance.
 A: Let $T=A'_\psi$. By the assumption, we have for some $\delta>0$,
$$
A(\psi + h) = A(\psi) +T h + \epsilon(h),\quad \forall |h|<\delta
$$ where $\frac{\|\epsilon(h)\|}{\|h\|}\to 0$ as $\|h\|\to 0$. Without loss of generality, we may assume that $\psi =0$ and $A(\psi)=0$. Let $(x_n)\subset X$ be a sequence with $\sup_n \|x_n\|\le 1$. Our goal is to show that there exists a subsequence $n(k)$ such that $
Tx_{n(k)}
$ is convergent. By the assumption on $\epsilon(\cdot)$, for any given $\eta >0$, there exists (sufficiently large) $M\in\mathbb{N}$ such that$$\Big\|\epsilon\left(\frac{1}{M}x_n\right)\Big\|\le \eta\Big\|\frac{1}{M}x_n\Big\|$$ for all $n\ge 1$. This impiles that$$
M\Big\|\epsilon\left(\frac{1}{M}x_n\right)\Big\|\le \eta\|x_n\|\leq \eta.
$$ 
Now, observe that by the assumption that $A$ is completely continuous, (i.e. every bounded sequence $(u_n)$ admits a subsequence that makes $A(u_{n(k)})$ norm-convergent in $Y$) the diagonalization method gives us a subsequence $n(k)$ such that for each $j=1,2,\ldots$, the sequence
$$
A\left(\frac{1}{j}x_{n(k)}\right)
$$ converges to some $y_j\in Y$.
From the fact that$$
A\left(\frac{1}{M}x_n\right) = \frac{1}{M}Tx_n+\epsilon\left(\frac{1}{M}x_n\right),
$$ it follows
$$\begin{eqnarray}
\|Tx_{n(k)}-My_M \|&=& \Big\|MA\left(\frac{1}{M}x_{n(k)}\right)-My_M-M\epsilon\left(\frac{1}{M}x_{n(k)}\right)\Big\|\\
&\le&M\Big\|A\left(\frac{1}{M}x_{n(k)}\right)-y_M\Big\|+\eta.
\end{eqnarray}$$ By taking $k\to\infty$, we get
$$
\limsup_{k\to\infty}\|Tx_{n(k)}-My_M \|\le \eta
$$ and
$$
\limsup_{k,l\to\infty}\|Tx_{n(k)}-Tx_{n(l)} \|\le 2\eta.
$$Since $\eta>0$ was arbitrary, This proves that $Tx_{n(k)}$ is Cauchy in $Y$. Thus $Tx_{n(k)}$ is convergent, and the compactness follows as a result.
