A result about the separation of a point and a convex closed set

25 Views Asked by At

I would like to show the following result :

If $C\subset\mathbb{R}^n$ is a closed convex subspace that does not contains the null vector, then there exists a vector $Y\neq 0_{\mathbb{R}^n}$ such that

$$ \forall X\in C : \langle X, Y\rangle\geq\langle Y, Y\rangle>0 $$

The attemt is the following :

We start by considering

$$ \lVert Y\rVert = \min\limits_{X\in C}\{ \lVert X\rVert\} $$ that is the minimum distance of $C$ to $0_{\mathbb{R}^n}$. We know that this minimum is achieved and now consider the vector $Y$ that realizes this minimal distance. Take $X\in C$ and $s\in(0,1)$ to remark that

$$ sX + (1-s)Y\in C\quad\text{and}\quad 0 < \langle Y, Y\rangle = \lVert Y\rVert^2\leq\lVert sX + (1-s)Y\rVert^2 =s^2\lVert X\rVert^2 + 2s(1-s)\langle X, Y\rangle + (1-s)^2\lVert Y\rVert^2 $$

which leads to

$$ 2s\lVert Y\rVert^2\leq s^2\lVert X\rVert^{2} + 2s(1-s)\langle X, Y\rangle +s^2\lVert Y\rVert^{2}\implies \lVert Y\rVert^{2}\leq \frac{s}{2}\lVert X\rVert^{2} + (1-s)\langle X, Y\rangle + \frac{s}{2}\lVert Y\rVert^{2} $$

and the conclusion follows by taking the limit when $s$ goes to $0$.

I would like to know if the proof is correct and if there is a better way to show this. Also, I was wondering how this result can be seen, what is the information I get by the fact that the inner product of $X$ by $Y$ is greater than the inner product of $Y$ by $Y$ ? A simple manipulation shows that :

$$ \lVert Y\rVert\leq\frac{\langle X, Y\rangle}{\lVert Y\rVert} $$

Perhaps something can be said about this but I am note so sure what.

Thank you a lot !

1

There are 1 best solutions below

1
On BEST ANSWER

Your proof is correct, and I don't think you can make it essentially simpler. As for the meaning of the result, the hyperplane $h(X)=\langle Y,X\rangle-\|Y\|^2=0$ is a supporting hyperplane of $C$ at $Y$, which separates $C$ from the origin. Clearly, if you shift your hyperplane closer to the origin: $h(X)=\langle Y,X\rangle-\|Y\|^2+\varepsilon=0$ with $\varepsilon<\|Y\|$ is a strict separating hyperplane. Another way to look at it is by saying that $Y$ is the vector with minimum projection on the direction of $Y$ among the vectors in $C$.