Let $f$ be your two-piece linear function. Let $\varphi\in C^\infty_0((-\epsilon,\epsilon))$ for some small $\epsilon$, such that

• $\varphi$ is even
• the integral $\int \varphi = 1$
• $x\varphi' \leq 0$

Then you can check that the convolution $\varphi*f$ is increasing, smooth, and agrees with $f$ outside $(-2\epsilon,2\epsilon)$.

Taking appropriately rescaled versions of $\varphi$ you get uniform approximations.

Let $f$ be your two-piece linear function. Let $\varphi\in C^\infty_0((-\epsilon,\epsilon))$ for some small $\epsilon$, such that

• $\varphi$ is even
• the integral $\int \varphi = 1$
• $x\varphi' \leq 0$

Then you can check that the convolution $\varphi*f$ is increasing, smooth, and agrees with $f$ outside $(-2\epsilon,2\epsilon)$.

Taking appropriately rescaled versions of $\varphi$ you get uniform approximations.

Sketch of Proof:

1. Observe that since $\varphi$ is even $\int \varphi(x) x \mathrm{d}x = 0$, this guarantees that if $f$ is linear in $(x-\epsilon,x+\epsilon)$, $\varphi* f(x) = f(x)$ (write $f(y) = (f(y) - f(x)) + f(x)$.)
2. Let $g(x) = \varphi*f(x)$, we have $g' = \varphi'*f$. In particular $g'(x) = \int_0^\epsilon \varphi'(y) ( f(x-y) - f(x+y)) \mathrm{d}y$. So using that $x\varphi'(x) \leq 0$ and $f$ is increasing you get that $g'(x) \geq 0$.
3. It is a standard fact that $\varphi*f$ is smooth.
4. Let $\varphi_\delta(x) = \frac1\delta \varphi(x / \delta)$. Observe that $$ \varphi_\delta* f(x) - f(x) = \int \varphi_\delta(y) (f(x - y) - f(x)) ~\mathrm{d}y $$ and observe that the difference is always zero when $|x| > 2\delta\epsilon$. You have that $$ |\varphi_\delta*f(x) - f(x)| \leq \sup_{x,y\in (-2\delta\epsilon,2\delta\epsilon)} |f(x) - f(y)| $$ which can be easily controlled by uniform continuity (of continuous functions on compact sets).