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:


1. $f\_n$ is nonnegative, nondecreasing, and concave for all $n\in\mathbb N$.
2. $f\_n$ is smooth, i.e., infinitely differentiable on $(0,\infty)$, for all $n\in\mathbb N$.
3. $\lim\_{n\to\infty} f\_n(x) = f(x)$ for all $x\in[0,\infty)$.




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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 $(0, 1/n)$.




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Any ideas?



Smooth approximation of nonnegative, nondecreasing, concave functions
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 $n \ge 1$,
$$\int\_0^\infty \varphi\_n(y) \frac{dy}{y} = 1.$$


