# Two Strictly Convex Functions with Contact of Order 1

Let $f,g: \mathbb{R}\rightarrow \mathbb{R}$ be two strictly convex functions, where $f$ is differentiable, $g$ is smooth, and $f\geq g$. Suppose that for some $x_0\in \mathbb{R}$:

$f(x_0)=g(x_0)\hspace{10mm} f'(x_0)=g'(x_0)$,$\hspace{10mm}$ and $g''(x_0)>0$

Does there exists an interval $I$ of $x_0$ such that $f$ is strongly convex on $I$?

Note: There does exist such an interval for $g$.

No, there need not be such an interval. Take $g(x)=x^2$ and $f_0(x)=2x^2$ (this is not yet $f$). Choose a sequence of disjoint intervals converging to $0$, for example $I_n=[2^{-2n},2^{1-2n}]$. Redefine $f_0$ to be linear on each $I_n$; this creates a function $f_1$ which is not strictly convex in any neighborhood of $0$, and still satisfies $f_1\ge g$. But you want $f$ to be strictly convex, so let $f(x)=f_1(x)+x^4$.
It's true that $f_1$, as written above, is not differentiable. To fix this, replace the line segment over $I_n$ with the following three-piece gadget:
Namely, the middle third of line segment is brought down a bit (it stays above the graph of $f_0$) and smooth transitions are added on left and right.
• Thanks for great example, but $f_1$ isn't differentiable, is it? I get $f'_1(1/4)=1$ from left and $f'_1(1/4)=3/2$ from right. – Ashley Aug 8 '14 at 18:35
• Okay, so you're making the transitions to the middle third of the line segment smooth and the transitions from the bottom and top thirds to $f_0$ smooth. That seems like it should work. I;m going to think about it a little more. – Ashley Aug 8 '14 at 19:03