Existence of a certain decreasing function I'm in the following strange situation:
$\alpha:[0,\infty)\to\mathbb{R}_{>0}$ is a continuous, positive, decreasing function with $\int\alpha=\infty$. Let $f$ be the function $x \mapsto e^{-2x}.$
Define $k(x)=\max(f,\alpha-f)$, which is a continuous, positive function with $\int k=\infty$.
Question: Does there exist a continuous, decreasing, positive, non-integrable function $K$ bounded above by $k$?
After looking at some 'bad' examples of $\alpha$, it seems that $K(x):=\inf_{y\leq x}k(y)$ might work.
I'm hoping to find an idea for a proof or a way to construct a counter-example.
Thanks for reading! 
 A: Your choice $$K(x) := \inf_{y \le x} k(y)$$ is not bad: $K$ is a positive, monotonically decreasing function with $K(x) \le k(x)$. Since $$k(x) \ge \frac{f(x)}{2}+ \frac{\alpha(x)-f(x)}{2} = \frac{\alpha(x)}{2}$$
by definition and $\alpha$ is decreasing, we find that 
$$K(x) \ge \frac{\alpha(x)}{2}.$$
This implies already that $\int_0^\infty K(x) \, \mathrm{d} x = \infty$.
However, $K$ is not necessarily a continuous function. To repair this defect, we smooth $K$ as follows. Set
$$H(x) = \int_0^1 K(y+x) \, \mathrm{d} y.$$
Then $H$ is positive, $H(x) \le \int_0^1 K(x) \, \mathrm{d} y = K(x) \le k(x)$ because $K$ is decreasing and $$H(x) \ge \int_0^1 \frac{\alpha(x+y)}{2} \, \mathrm{d} y \ge \frac{\alpha(x+1)}{2}.$$
Since $\alpha$ is obviously integrable on $[0,1]$, the last inequality implies that $\int_0^\infty H(x) \, \mathrm{d}x = \infty$. Moreover, we have for $x > x'$
$$ 0 \le H(x') - H(x) = - \int_{x'}^x K(y) \, \mathrm{d} y + \int_{x'+1}^{x+1} K(y) \, \mathrm{d} y \le 2 k(1) (x-x').$$
Hence $H$ is also Lipschitz-continuous.
