I want to solve the following problem but I stuck at the last part:
Let $f(x) = (1-x^2)^2$ for $x \in \mathbb{R}$. Determine the convex envelope $Cf$ of $f$.
First I guessed, that the function is
\begin{equation*}
F(x) = \begin{cases}
(1-x^2)^2 &, \mbox{ for } |x| \geq 1, \\
0 &, \mbox{ for } |x| < 1
\end{cases}.
\end{equation*}
I have already proofen that $F$ is convex.
Now I want to show that for any given convex function $g$ with $g \leq f$, we must have $g \leq F$ such that $F$ is the convex envelope of $f$.
For $|x| \geq 1$ it is clear. But if $|x| < 1$ I can't figure out how $g \leq 0 = F(x)$ for $|x| < 1$.
I wanted to prove it by contraposition, assuming there exists a $|x| < 1$ such that $g(x) > 0$ and than using $g \leq f$ that $g$ must be $f$ or $g \leq 0$.
For $x_1 = -x_2$ I get for $\lambda \in [0,1]$
\begin{equation*}
\lambda (g(x_2) – g(x_1)) + g(x_1) \leq \lambda (f(x_2) – f(x_1)) + f(x_1) = f(x_1)
\end{equation*}
which yields
\begin{equation*}
g(x_1) \leq f(x_1)
\end{equation*}
and
\begin{equation*}
g(x_2) \leq f(x_1)
\end{equation*}
I don't know if this is useful.
I failed showing this property. Am I missing something or is the function which I choose not the right one?
I am thankful for every hint I can get.
Best Answer
Every $x \in [-1, 1]$ is a convex combination of $-1$ and $1$: $$ x = \frac{1-x}{2}\cdot(-1) + \frac{x+1}{2} \cdot 1 $$ Therefore, if $g$ is convex with $g \le f$ then $$ g(x) \le \frac{1-x}{2}\cdot g(-1) + \frac{x+1}{2} \cdot g(1) \\ \le \frac{1-x}{2}\cdot f(-1) + \frac{x+1}{2} \cdot f(1) = 0 $$
for $-1 \le x \le 1$. In other words: The graph of the convex envelope of $f$ lies below the secant joining $(-1, f(-1)) = (-1, 0)$ and $(1, f(1)) = (1, 0)$.
So your solution is correct: $F$ is a convex function with $F \le f$, and every convex function $g \le f$ satisfies $g \le F$. Therefore $F$ is the convex envelope of $f$.