Let $f\colon [0, \infty)\to\mathbb{R}$ be nonnegative, nondecreasing, and concave. Prove the following claim or give a counter example: There is a sequence of functions $f_n\colon [0, \infty)\to\mathbb{R}$ with following properties:
- $f_n$ is nonnegative, nondecreasing, and concave for all $n\in\mathbb N$.
- $f_n$ is smooth, i.e., infinitely differentiable on $(0,\infty)$, for all $n\in\mathbb N$.
- $\lim_{n\to\infty} f_n(x) = f(x)$ for all $x\in[0,\infty)$.
My first approach was to use mollifiers. But I get problems close to $x=0$:
Let $\varphi\colon\mathbb{R} \to \mathbb{R}$ be a positive mollifier, e.g.,
$$
\varphi(x) := \begin{cases}
C \exp(-\frac{1}{1-x^2}) & \text{ if } x^2 < 1,\\
0 & \text{ if } x^2 \geq 1,
\end{cases}
$$
where $C\in(0,\infty)$ is a normalization constant such that $\int_{-\infty}^\infty \varphi(x) d x = 1$. Define
$$
f_n(x) := \int_{-\infty}^\infty n \varphi(ny) \tilde f(x-y) d y,
$$
where $\tilde f\colon\mathbb{R} \to \mathbb{R}$ is an extension of $f$ to $\mathbb R$. If we could find an extension $\tilde f$ that is nonnegative, nondecreasing, and concave on $[-1/n, \infty)$, we would be fine. But if $f(0) = 0$ and $f$ is strictly increasing, this is not possible. If we use $\tilde f(x) = f(\max(0, x))$, then we can only guarantee concavity on $[1/n, \infty)$ but not in $(0, 1/n)$.
Any ideas?
Best Answer
Replace the convolution on $\mathbb{R}$ by a convolution on the group $\mathbb{R}_+^*$, endowed with the invariant measure $dx/x$, namely set $$f_n(x) := \int_0^\infty \varphi_n(y) f(x/y) \frac{dy}{y},$$ where $(\varphi_n)_{n \ge 1}$ is a sequence of non-negative $\mathcal{C}^\infty$ functions on $\mathbb{R}_+^*$ with support contained in $[e^{-1/n},e^{1/n}]$ such that for every $n \ge 1$, $$\int_0^\infty \varphi_n(y) \frac{dy}{y} = 1.$$