The derivative of a real valued function whose domain in a function space I am struggling to follow an example of the derivative of a function whose domain is a function space. My question concerns in particular Example 1.1.6 taken from Pierre Schapira's online lecture notes for differential calculus on Banach spaces, which can be found here
https://webusers.imj-prg.fr/~pierre.schapira/lectnotes/
The passage I am struggling with is at the top of page 10; in particular, I cannot obtain the constant C with the property that $|\epsilon(\phi(t), \psi(t))| \leq C|\psi(t)|$. Applying the mean value theorem he quotes above seems to yield 
$|\ln(\phi(t)+\psi(t)) - \ln(\phi(t)) - \frac{\psi(t)}{\phi(t)}| \leq |\psi(t)|\cdot M$
where $M = \sup_{\xi \in [-1, 1]}|\ln'(\phi(t)+\xi\psi(t)) - \ln'(\phi(t))|$.
Now from (1.3) we have that $|\ln(\phi(t)+\psi(t)) - \ln(\phi(t)) - \frac{\psi(t)}{\phi(t)}| = |\psi(t)||\epsilon(\phi(t), \psi(t))|$
and hence
$|\psi(t)||\epsilon(\phi(t), \psi(t))| \leq |\psi(t)|\cdot M$.
Now since $\psi > 0$ this gives us $|\epsilon(\phi(t), \psi(t))| \leq M$ and I cannot seem to convert this into an expression of the form $|\epsilon(\phi(t), \psi(t))| \leq C|\psi(t)|$ where C does not depend on $t$.
In my frustration, I attempted a more direct approach to the obtaining the result
$\ln(\phi + \psi) - \ln(\phi) - \frac{\psi}{\phi} = ||\psi||\alpha(\phi, \psi)$ 
where $\alpha(\phi, \psi) \to 0$ as $\psi \to 0$. 
Let, as suggested, $\alpha$ be the function $\alpha(\phi, \psi)(t) = \frac{|\psi(t)|}{||\psi||}\epsilon(\phi(t), \psi(t))$.  We have, since $\frac{|\psi(t)|}{||\psi||}$ is bounded by 1, that 
$|\alpha(\phi, \psi)(t)| \leq |\epsilon(\phi(t), \psi(t))|$. 
Now since $|\psi(t)| \leq ||\psi||$ it seems clear that as $||\psi|| \to 0$ we must have $|\psi(t)| \to 0$ independently of $t$ and thus, by the definition of $\epsilon(\phi(t), \psi(t))$ we have that  
$|\alpha(\phi, \psi)(t)| \leq |\epsilon(\phi(t), \psi(t))| \to 0$ as $||\psi|| \to 0$
as required. 
I've been playing around with this for a while now and I'm sure my error will be suitably embarrassing, but any help with either my original problem or comments regarding my alternative would be greatly appreciated. 
Thanks in advance.
 A: I have to admit, I didn't fully read through all the details, simply because everything is being done with the particular function $\ln$, which makes it hard (for me anyways) to keep track of what steps are true in general and what holds in this special case. But, as a general setup, we can state the following:

Theorem (Special case of Henri Cartan: Differential Forms, Chapter 2, Proposition 1.2.2) Let $V$ be a real Banach space, fix an interval $[a,b] \subset \mathbb{R}$, let $E = \mathcal{C^0}([a,b], V)$ (set of continuous maps, equipped with the supremum norm). Let $A$ be an open subset of $V$, and suppose we are given a $\mathcal{C}^k$ function $F: A \to \mathbb{R}$ ($k \geq 1$). Then, the set $U = \{ \varphi \in E : \text{image}(\varphi) \subset A\}$ is open in $E$. Now, define the map $f: U \to \mathbb{R}$ by
  \begin{equation}
f(\varphi) = \int_a^b F(\varphi(t)) \, dt
\end{equation}
  Then $f$ is also of class $\mathcal{C^k}$, and its derivative at a point $\varphi \in U$ (which is a bounded linear map from $E$ to $\mathbb{R}$),  evaluated at $\psi \in E$ is given by the equation
  \begin{equation}
df_{\varphi}(\psi) = \int_a^b dF_{\varphi(t)}(\psi(t)) \, dt.
\end{equation}

In the particular example you were looking at we had: $V = \mathbb{R}$, $[a,b] = [0,1]$, $A = (0, \infty), F = \ln$; so it should be easy to see how the general formula of this theorem reduces to the one presented in those notes.
To prove $U$ is open, use the fact that for a given $\varphi \in U$, its image is a compact subset of  the open set $A$; hence by the $\epsilon$-neighbourhood theorem, (which is valid in any metric space), there is an $\varepsilon > 0$ such that the $\varepsilon$ neigbourhood of $\text{image}(\varphi)$ lies in $A$. It follows that if $\lVert \psi - \varphi\rVert < \dfrac{\varepsilon}{2}$, then $\psi \in U$.
Next, the difficult part is showing $f$ is differentiable. Deducing that $f$ is $\mathcal{C^k}$ is pretty straightforward from the formula for the derivative; use induction on $k$ (for these details, I highly recommend reading Henri Cartan's book; he does an amazing job there).
Assuming $f$ is differentiable, the formula for the derivative is easy to obtain, because differentiability at $\varphi$ implies that for every $\psi \in E$, the directional derivative $(D_{\psi}f)(\varphi)$ exists with the following relationship:
\begin{align}
df_{\varphi}(\psi) &= (D_{\psi}f)(\varphi) \\
&= \dfrac{d}{dx} \bigg\rvert_{x=0} f(\varphi + x \psi) \\
&= \dfrac{d}{dx} \bigg\rvert_{x=0} \int_a^b F\left( \varphi(t) + x \cdot \psi(t) \right) \, dt \\
&= \int_a^b \dfrac{\partial}{\partial x} \bigg\rvert_{x=0} F\left( \varphi(t) + x \cdot \psi(t) \right) \, dt \\
&= \int_a^b dF_{\varphi(t)} \left( \psi(t) \right) \, dt
\end{align}
I used the typical abuse of notation which comes up in this context, and in the $4^{th}$ equal sign, I made use of the Leibniz integral rule for differentiating under the integral sign.
(If you still wish to see why $f$ is differentiable, let me know; I'll try to outline it as best I can)
