1

Theorem 18.8 in the book by Rockafellar establishes that any $n$-dimensional closed convex set $C$ in $R^n$ can be expressed as the intersection of the closed half spaces tangent to $C$. See here for the book page.

I'm having trouble seeing why the statement is true. For example, consider the square in $R^2$ which is the convex hull of $(1,1), (1,-1), (-1,1)$ and $(-1,-1)$. This is a closed convex set, but it does not seem to have any tangent hyperplanes (a hyperplane is tangent if it is the unique supporting hyperplane at a certain point), so how can the square be expressed as its tangent half spaces?

Is there something I'm missing here?

Edit: I realize that I might have some misunderstanding here, but I'm still not sure. On the same page of the book, tangent hyperplanes are mentioned as duals to the exposed points, that's why I thought for a hyperplane to be tangent it must only passes through one of the exposed points, which now I realize might not be the case. Does this mean that in the above example, hyperplanes containing one of the sides of the square are its tangent hyperplanes? This will rationalize the theorem for me, but can someone explain in what sense the tangent hyperplanes are dual to exposed points?

Thanks a lot!

Vokram
  • 1,587
  • Is the horizontal line ${(x,1): x\in\Bbb R}$ a tangent hyperplane (for your square) by your lights? – John Dawkins Nov 18 '15 at 19:27
  • @JohnDawkins Thanks for the comment! Now I realize that I probably misunderstood the notion of the tangent hyperplane. I just edited my question. – Vokram Nov 18 '15 at 19:33
  • Regarding uniqueness: Rockafellar is defining a supporting hyperplane at $x$ to be a tangent hyperplane, if it is the only supporting hyperplane at $x$. So by definition, points that admit more than one supporting hyperplanes, do not admit tangent planes. – Manos Jan 31 '17 at 22:29

1 Answers1

3

Tangent hyperplanes can touch a convex set at multiple points. For the square, you have 4 half-spaces whose defining hyperplanes are tangent to the sides of the square, namely $\{(x,y):y\geq 1\}$, $\{(x,y):y\leq -1\}$, $\{(x,y):x\geq 1\}$ and $\{(x,y):x\leq -1\}$. You also have 4 that touch the corners, at 45 degree angles (I won't write them down explicitly). If you intersect these 8 half-spaces, you'll get the square. It helps to draw a picture.

icurays1
  • 17,161
  • Thank you for the answer. Are those touching only the corners also considered tangent? It seems that they are not unique (i.e. you can tilt them to be touching at 30 and 60 degrees), which is a requirement for tangency? – Vokram Nov 18 '15 at 19:40
  • Supporting hyperplanes aren't unique for arbitrary convex sets, I'm not sure where you've seen that. They're unique for strictly convex sets. The theorem about half-space intersection doesn't require uniqueness anyway. – icurays1 Nov 18 '15 at 19:45
  • Hmm I realize I may be seriously confused. In my example on the square, isn't the hyperplane ${(x,y): y=1 }$ a unique supporting hyperplane to the point, say, $(0,1)$? The uniqueness requirement is mentioned in Rockafellar, on the same page of the link in my question. It's mentioned in the 4th paragraph on that page, if you wouldn't mind taking a look? Thanks a lot! – Vokram Nov 18 '15 at 19:52