Bound functin using convex function Given a continuous function $f$ on the interval $[a,b]$, I want to prove that there is a convex function $ g$ such that: 
1. $g(x) \ge f(x) , x\in [a,b]$
2. $g(a) = f(a)$
3. There is $p\in (a,b]$ such that $g(p) = f(p)$ 
Is it true?
 A: No, try $f(x) = \sqrt{x}$ on $[0,1]$. Let us assume that a $g$ exists. Then, by convexity, $g$ lies below the line through $(0,f(0))$, $(p,f(p))$ on $[0,p]$. But this contradicts $g(x) \ge f(x)$ for $x \searrow 0$.
If $f'(a) < \infty$ (as a one sided limit), such a $g$ should exist. In this case, set $g_t(x) = f(a) + t \, (x-a)$. Then, for $t$ large enough, you have $g_t(x) \ge f(x)$ for all $x \in [a,b]$ (note that this is not immediate, one has to proof it). Now, let $t_0$ be the infimum of all $t$, such that $g_t(x) \ge f(x)$ for all $x \in [a,b]$. Prove that $g_{t_0}$ has one additional touching point with $f$.
A: Consider a convex quadratic coord $h :[a, b] \rightarrow \mathbb{R}$ pinned at $P := (a, -f(a)) \in \mathbb{R}^2$ and $Q := (b, -f(b)) \in \mathbb{R}^2$, and stretch it vertically until it is below the graph of $-f$. Take $p := b$, and use a bit of algebra to verify that the choice $g := -h$ (obtained by flippling $h$ about the horizontal axis) does the job.
Alternatively, If you're a physicist, think of $f$ as a segment of a heavy chain pinned at $P$ and $Q$, and let it sag under gravity. When it comes to a stand-still, it produces a hyperbolic cosine curve $h$. Now, taking $g := -h$ trivially satisfies the conditions you seek (once again, with $p := b$).
