Uniqueness of the Frechet Derivative: the role of $x \in int_X(T)$ I'm currently trying to learn some functional analysis as a way to improve my ability to read economic theory papers. I've come across what I thought was a simple proof but on reflection I don't think I'm grasping it. I'm not a mathematician so I apologise if this question is rather trivial! My problem lies in the proof of the uniqueness of the Frechet derivative. 
Here is the definition that I'm using (From Efe OK's book Real Analysis with Economic Applications). 
Definition
Let $X$ and $Y$ be two normed linear spaces and $T$ a subset of $X$. For any $x \in int_X(T)$, a map $\Phi : T \rightarrow Y$ is said to be Frechet differentiable at $x$ if there is a continuous linear operator $D_{\Phi,x}\in \mathcal{B}(X,Y)$ such that 
\begin{equation}
\lim_{\omega \rightarrow x} \frac{\Phi(\omega)-\Phi(x)-D_{\Phi,x}(\omega-x)}{\left\lVert \omega-x \right\rVert} = \mathbf{0}
\end{equation}
The linear operator $D_{\Phi,x}$ is called the Frechet derivative of $\Phi$ at $x$. 
The proof proceeds by taking any two $K,L \in \mathcal{B}(X,Y)$ that satisfy the definition of the Frechet derivative, with $D_{\Phi,x} = K$ and $D_{\Phi,x} = L$. We must then have 
\begin{equation}
\lim_{\omega \rightarrow x} \frac{(K-L)(\omega-x)}{\left\lVert \omega-x \right\rVert} = \mathbf{0}
\end{equation}
The next step is where I'm confused. Since $int_X(T)$ is open, this is equivalent to saying that 
\begin{equation}
\lim_{v \rightarrow \mathbf{0}} \frac{(K-L)(v)}{\left\lVert v \right\rVert} = \mathbf{0}
\end{equation}
The rest of the proof is reasonably straightforward. The author provides a warning in the footnotes that if $x \notin int_X(T)$ the final two displayed equations are not equivalent, and the Frechet derivative in this case is not unique.  
It seems intuitively reasonable that the final two expressions are equivalent but I'm not sure how to show it. My initial thought is that, with $x$ in the boundary, it limits the directions from which one can converge to it. 
 A: It is easier to see this using an $\epsilon$-$\delta$ argument.
 $\Phi$ is differentiable at $x$ iff there exists some continuous linear $L$ such that for all $\epsilon>0$ there is some $\delta>0$ such that if $\|x-y\| < \delta$ then
$\|\Phi(y)-\Phi(x) - L(y-x) \| \le \epsilon \|y-x\|$.
Suppose $K,L$ satisfy the equation, then
$\|(K-L)(x-y)\| \le \|\Phi(y)-\Phi(x) - K(y-x) \| + \|\Phi(y)-\Phi(x) - L(y-x) \|$. Now choose $\epsilon>0$ and get some $\delta_L,\delta_K >0$
such that the above holds. Then if $\|x-y\| < \min(\delta_L,\delta_K)$ we have
$\|(K-L)(x-y)\| \le 2 \epsilon \|y-x\|$.
Since $x$ in the the interior, there is some $B(x,\eta) \subset T$ and so
for any $h \in B(0,1)$ we have
$\|(K-L) \eta h\| \le \epsilon \eta \|h\|$ or $\|(K-L) h\| \le \epsilon \|h\|$, since $(K-L)$ is linear.
In particular, $\|K-L\| \le \epsilon$. Since $\epsilon>0$ was arbitrary 
we have the desired result $K=L$.
A: What if we take it from \begin{equation}
\lim_{\omega \rightarrow x} \frac{(K-L)(\omega-x)}{\left\lVert \omega-x \right\rVert} = \mathbf{0}
\end{equation}

*

*It follows that ${(K-L)(\omega-x))} =\mathbf{0}$ for any $\omega \in T$ because

\begin{equation}
\lim_{\omega \rightarrow x} \frac{(K-L)(\omega-x)}{\left\lVert \omega-x \right\rVert} = \lim_{\omega \rightarrow x} {(K-L)(u)} = {(K-L)(u)} =\mathbf{0}
\end{equation}  Where $ u = \frac{(\omega-x)} {\left\lVert \omega-x \right\rVert}$ is a non-zero vector (of unit length).
(Since $K - L$ is continuous it is everywhere defined (even if $\Phi$ isn't) so $u$ is in its domain even if it isn't in the domain of $\Phi$ )
Hence $(K - L){\left\lVert \omega-x \right\rVert}u = {(K-L)(\omega-x))} = 0$


*Since $x \in T$ and $T$ is open then there is an open ball $B_r(x)$
around $x$.
So for any non-zero vector $v \in X$ take $w = x + r.v/2.||v||$ and then $||w - x|| < r$ i.e. $w \in B_r(x)$ and $v = \alpha (w - x) $ where $\alpha = 2.||v||/r$
Since ${(K-L)(\omega-x))} =\mathbf{0}$ it follows that ${(K-L)(v)} =\mathbf{0}$ for all $v \in X$ and therefore that $K = L$
