I am writing a paper, where I need to apply a quite simple convexity argument. But it's lengthy, and in order to save space in the paper, I would like to just cite a reference. The claim is as follows:
Let $A\in\mathbb R^n$ be convex such that $\overline B_\varepsilon(0)\subset A$. If $x_0\in\mathbb R^n$ such that $\operatorname{span}\{x_0\}\subset\overline A$, then $\operatorname{span}\{x_0\}+ B_\varepsilon(0)\subset A$.
Here,
- $\overline A$ denotes the closure of $A$
- $\overline B_\varepsilon(0)$ denotes the closure of the $\varepsilon$-ball around zero
- $\operatorname{span}\{x_0\} = \{tx_0 : t\in\mathbb R\}$.
Sketch of proof: Let $x\in\operatorname{span}\{x_0\}$ and connect every point from $\overline B_\varepsilon(0)$ with $x_\delta$, where $x_\delta\in A$ is close to $x$. The union of all these lines is in $A$ since $A$ is convex. Also, the closer $x_\delta$ is to $x$, the more of the segment $[0,x]$ is contained in this union. Letting $\delta\to 0$, it follows that $[0,x)\subset A$. So, $\operatorname{span}\{x_0\}\subset A$. Now, we can even connect every point in $\operatorname{span}\{x_0\}$ with every point in $\overline B_\varepsilon(0)$. The union of all these lines is in $A$ and should contain $B_\varepsilon(x)$ for every $x\in\operatorname{span}\{x_0\}$.
A two-sentence proof:
For any $y \in B_{\varepsilon}(0)\backslash\{0\}$ and any $t \in \Bbb R$, choose $z \in A$ such that $|z - \frac {\varepsilon + |y|}{\varepsilon - |y|}tx_0| < |y|$. We then have $$\left|\frac{\varepsilon + |y|}{2|y|}(y + tx_0 - \frac{\varepsilon - |y|}{\varepsilon + |y|}z)\right| \leq \frac{\varepsilon + |y|}{2|y|}\cdot (|y| + |tx_0 - \frac{\varepsilon - |y|}{\varepsilon + |y|}z|) < \varepsilon$$ and hence $$y + tx_0 = \frac{2|y|}{\varepsilon + |y|}\cdot \left(\frac{\varepsilon + |y|}{2|y|}(y + tx_0 - \frac{\varepsilon - |y|}{\varepsilon + |y|}z)\right) + \frac{\varepsilon - |y|}{\varepsilon + |y|} \cdot z\in A.$$
It's now up to the referee to check that everything is correct.