Let $f : \mathbb{R}^M\to \mathbb{R} $
$f$ is convex if and only if $f =\sup\{\varphi:\mathbb{R}^M\to \mathbb{R}: \varphi \le f, \varphi \text { affine} \}$
In the proof of ($\implies$), I have to follow the following hints:
- Fix $x \in \mathbb{R}^M$ , and $t \in \mathbb{R}$ with $t < f (x)$ We claim that there exists an affine function $\varphi : \mathbb{R}^M \to \mathbb{R}$ with $\varphi \le f$ such that $\varphi(x) > t$; For this:
1a) Consider the point $P := (x, t)$. Prove that there exists a point $Q$ which minimizes the distance from $P$ to the graph of $f$
1b)Define the function $\varphi : \mathbb{R}^M \to \mathbb{R}$ as the affine function whose graph is the tangent plane to $\partial B(P, |P − Q|)$ and passing by $Q$
1c) Prove that $\varphi \le f$. Assume by contradiction that there exists a point $y \in \mathbb{R}^M$ such that $f (y) < \varphi(y)$. Let $A := (y, f (y))$. Prove that there exists a point $E$ on the graph of $f$ which is inside $B(P, r)$, $ r= |P − Q|$ contradicting the choice of $Q$
- Conclude that f is the supremum of affine functions majorized by f .
I have already proven 1a) and 2)
1a) was proven here
1b) I guess there is nothing to proof
I am having trouble proving the final part of 1c). I think it must be some continuity (because $f$ convex $\implies f $ is continuous) or some geometrical argument that I can't figure out yet
This is what I have so far in 1c)
As told in the hint, assume by contradiction that
$ \exists y \in \mathbb{R}^M$ such that $f (y) < \varphi(y) \tag{1}$
If I can prove that $\exists $ a point $E \in$ the open ball $ B(P,r)$ s.t. $E \in G(f)$ (the graph of f) this would contradict the choice of $Q$ because $E$ would be a point on G(f) at a distance from $P$ less than that of $Q$ ($Q$ is on $\partial B(P,r)$) and $Q$ was supposed to be the point that minimizes the distance from $P$ to $G(f)$. I would then have a contradiction and $\varphi \le f$ would be proven.
So it suffices to prove that $\exists E \in B(P,r)$ s.t. $E \in G(f)$. Let $A=(y,f(y))\in G(f)$. Let's analyze 3 cases:$ A \in B(P,r), A \in \partial B(P,r)$ and $A \in {\overline B}^C(P,r)$
If $A \in B(P,r)$ I am done, just take $E=A$
If $A \in \partial B(P,r)$. Clearly $A \neq Q $. So the segment $\overline{AQ}$ would have distinct endpoints on $\partial B(P,r)$.
Let $Q=(y_Q,f(y_Q))$ and recall $P=(x,t)$ with $x,y_Q \in \mathbb{R}^M$. Lets use that $f(y)<\varphi(y)$, $f(y_Q)=\varphi(y_Q)$ Since $f$ is convex: $ \forall \lambda \in [0,1]: f(z)= f(\lambda y+(1-\lambda)y_Q)\le \lambda f(y) +(1-y)f(y_Q) < \lambda \varphi(y)+(1-\lambda)\varphi(y_Q)=\varphi(\lambda y +(1-\lambda)y_Q)= \varphi(z)$
where in the last step I used the convexity of the affine function $\varphi$, which is actually an equality: So: $f(z) < \varphi (z)$, for all the $z=\lambda y+(1-\lambda)y_Q$ such that $(z, \lambda \varphi(y)+(1-\lambda)\varphi(y_Q))\in \overline{AQ}$. This proves that there are points of $f$ below the segment $\overline{AQ}$, now I just have to prove that at least one of these $(z,f(z))$, with $\lambda \in (0,1)$ is inside the ball.
Here a 2-D sketch of the situation:
This is where I am stuck,
1) Is everything OK so far ? and how do I prove that last part? I think it may be some continuity argument.
2) For the final case $A \in {\overline B}^C(P,r)$, I need to prove that the segment $\overline{AQ}$ intersects the sphere in some other point $A'$ besides $Q$, and then the thesis follows by using the previous result for the segment $\overline{A'Q}$. How do I prove that?
Both of these results seem intuitive in 2 or 3 dimensions, but since we are in n dimensions, I am not so sure how to formalize them
