A question regarding the convex envelope of a function I know that by definition, the convex envelope of a function $f$ ($f$ not necessarily convex), denoted $\operatorname{conv}f$, is the largest convex function majorized by $f$. That is, it is a convex function $h$ such that $h \leq f$. In fact, I believe that 
$$\operatorname{conv}f(x)= \sup\{h(x) \mid h \text{ convex}, h \leq f\}\text{.}$$ 
My question is how can we directly apply this definition in order to find the convex envelope of $f(x)= \sin x$? Wouldn't it just be all points on and below the line $y= -1$? Not exactly sure why I can't figure this out, but I would really appreciate some helpful input regarding this issue.
 A: The only convex functions majorized by $\sin$ are the functions $h(x) = c$, where $c \le -1$. Hence $(\operatorname{conv} \sin)(x) = -1$.
To see this, suppose $h \le \sin$, $h$ is convex but $h$ is not constant. Then for some $x_1<x_2$, we have $h(x_1) \neq h(x_2)$. Suppose $h(x_1) < h(x_2)$, let $l(x) = \frac{x-x_2}{x_1-x_2}h(x_1) + \frac{x-x_1}{x_2-x_1}h(x_2)$, $l$ is affine and $\lim_{x \to \infty} l(x) = \infty$. Choose $x > x_2$, and write $x_2$ as a convex combination of $x_1,x$. This gives $x_2 = \frac{x_2-x_1}{x-x_1}x+\frac{x-x_2}{x-x_1}x_1$, and hence $h(x_2) \leq \frac{x_2-x_1}{x-x_1}h(x)+\frac{x-x_2}{x-x_1}h(x_1)$. Rearranging gives $l(x) \le h(x)$, which is a contradiction, as $h$ is bounded above by $1$ (since it is majorized by $\sin$). Hence $h$ is constant. Since $h \le \sin$, and $\sin (-\frac{\pi}{2}) = -1$, it follows that the constant is $\le -1$.
The following is a change from my previous answer, thanks to @byk7 (see the comments below) for catching a problem with my previous answer.
Here is another way of seeing this. Given a convex set $C \subset \mathbb{R}^n \times \mathbb{R}$, define $L_C(x) = \inf \{ \mu | (x,\mu)  \in C \}$. It is straightforward to see (Theorem 5.3 Rockafellar) that $L_C$ is a convex function.
Now note that $\operatorname{conv} \sin=L_{\operatorname{co} ( \operatorname{epi} \sin)}$.
Since $\operatorname{co} ( \operatorname{epi} \sin) = \mathbb{R} \times [-1,\infty)$, we have $h(x) = \inf \{ \alpha | (\alpha,x) \in \operatorname{co} ( \operatorname{epi} \sin) \} = -1$.
