Understanding a "Trivial" Result In Optimal Control Theory In Lawrence C. Evans' online notes : Optimal control theory, page 33, Evans makes a very trivial looking statement,which doesn't seem trivial to me. I shall elaborate, giving necessary details, so that this reference is only for further reading.
Let $\alpha : [0, \infty) \to [-1,1]^m$ be a measurable function. Consider the differential equation:
$$
\dot x(t) = M x(t) + N\alpha(t) \\ x(0) = x_0
$$
Where $M,N$ are constant real matrices of appropriate dimension, so that $x(t)$ has codomain $\mathbb{R}^n$. Now, we can solve this equation, with the solution:
$$
x(t) = X(t)x_0 + X(t) \displaystyle\int_0^t X^{-1}(s) N\alpha(s) \operatorname{ds} 
$$
Now, fix a  time $T$, and define the following set: $$K(T,x_0) = \left\{ X(T)x_0 + X(T) \displaystyle\int_0^T X^{-1}(s) N\alpha(s) \operatorname{ds}\right\}$$ where the $\alpha$ can vary over all measurable functions possible from $[0,\infty) \to [-1,1]^m$. 
In words, we are trying to find all reachable points at time $T$. If $x_1 \in K(T,x_0)$, there exists $\alpha_{x_1}$ such that $x_1 =  X(T)x_0 + X(T) \displaystyle\int_0^T X^{-1}(s) N\alpha_{x_1}(s) \operatorname{ds}$.
One can prove that $K(T,x_0)$ is convex and closed, the former using an obvious candidate, and the latter via an application of the Banach-Alaoglu theorem.
Now, suppose  that $0 \in K(\tau, x_0)$ for some $\tau$, but it is not true that $0 \in K(t, x_0)$ for all $t < \tau$. I want to prove that $0$ is a boundary point of $K(\tau,x_0)$.
In words: "If $0$ is reachable in time $\tau$, but is not reachable in any time smaller than $\tau$, then $0$ is a boundary point of the set of all points reachable in time $\tau$".
It did not seem trivial, so I tried letting $0$ be an interior point. It is clear that in a neighbourhood of $0$, there are points which can be reached in time $< \tau$ (when we take the trajectory from $x_0$ to $0$, it enters the neighbourhood before time $\tau$, so all points falling on that trajectory are reachable in time $< \tau$, naturally).However, I am unable to show why there must be points which are not reachable in time $\tau$ in a neighbourhood of $0$. This is because I am unable to use the nature of the given set to my benefit.
Surely if the question is trivial, then I am missing something. Do guide me across this one.
 A: Sorry, I'm not an expert. However, I try to give you my insight with the hope that it helps.
In order to check whether $0$ is a boundary point of $K(\tau,x_0)$, it suffices to show that any neighbourhood of $0$ contains a point in $K(\tau,x_0)$ and a point outside.
Let's fix $\epsilon>0$ and take the open ball $B:=\{x\in\mathbb R^n \,:\, \|x\|<\epsilon\}$. Since $0\in K(\tau,x_0)$, there exists $\alpha:[0,\tau]\to[-1,1]^m$ so that
$$
0 = X(\tau)x_0 + X(\tau)\int_0^\tau X^{-1}(s)N\alpha(s)ds
$$
For any $T<\tau$ take the state
$$
x(T) = X(T)x_0 + X(T)\int_0^T X^{-1}(s)N\alpha(s)ds
$$
Then $x(T)$ is in $K(\tau,x_0)$, since it is in $K(T,x_0)$.
Let's choose $T$ so that $x(T)\in B$. That is possible, since, once fixed $x_0$, $t\mapsto x(t)$ is differentiable and thus locally Lipschitz. Hence we can bound $\|x(\tau)-x(T)\|$ as
$$\|x(\tau)-x(T)\|\le L\|\tau-T\|$$
with $L$ that depends on $\tau$. We can thus chose $T$ such that $\|x(\tau)-x(T)\|<\epsilon$. This proves that in the $\epsilon$-neighbourhood of $x(\tau)$ there exists a point in $K(\tau,x_0)$. Take now $T>\tau$ and $\alpha'(s)$ that is any properly defined non-zero function that agrees with $\alpha$ on $[0,\tau]$. then $x(T)\notin K(t,x_0)$, for all $t\le \tau$, otherwise $x(\tau)$ would be reachable in less that $\tau$ seconds, that violates the hypothesis. Using the same argument as before you can chose $T>\tau$ so that $x(T)$ is in the $\epsilon$-neighbourhood of $x(\tau)$, thus proving that in such neighbourhood there exists a point that does not belong to $K(\tau,x_0)$. At this point, the claim should follow from the arbitrariness of $\epsilon$.
Maybe this is not a rigorous proof, but I hope that it can be an idea.
A: It is easier to argue if we introduce the following set
$$
R_T=\left\{\int_0^T X^{-1}(s) N\alpha(s)\,ds\right\}.
$$
Then
$$
K(T,x_0)=X(T)(x_0+R_T)
$$
and it is quite obvious that
\begin{align}
0\not\in K(t,x_0)\quad&\Leftrightarrow\quad -x_0\not\in R_t,\\
0\in \partial K(\tau,x_0)\quad&\Leftrightarrow\quad -x_0\in \partial R_{\tau}.
\end{align}
The sets $R_t$ are convex and closed as well, but they are ordered by inclusion as
$$
t_1\le t_2\quad\Rightarrow\quad R_{t_1}\subset R_{t_2}
$$
since $\alpha=0$ is allowed. The sets are also continuous: for any $\epsilon>0$ there exists $t<\tau$ such that
$$
R_t\subset R_{\tau}\subset R_t+\epsilon B_1
$$
where $B_1$ is the unit ball. It is because the integral
$$
\int_0^t X(s)^{-1}N\alpha(s)\,ds
$$
is continuous in $t$ (actually equicontinuous for all $\alpha$ in our class).
To prove that $-x_0$ is the boundary point we can take a ball $B$ around it of an arbitrary radius $\mu>0$ and show that there are points in it that are outside of $R_{\tau}$. 
The sets $R_t$ are convex and closed. Moreover, the center point $-x_0\not\in R_t$ for $t<\tau$, so one can separate the center of the ball from $R_t$ by a hyperplane. Thus the volume of $R_t$ inside the ball is less than half of the ball volume. By continuity we can find $t<\tau$ such that
$$
R_{\tau}\subset R_t+\frac{\mu}{4}B_1
$$
and $R_t+\frac{\mu}{4}B_1$ not filling up the whole ball $B$, leaving the top quarter for points outside $R_{\tau}$.
