Sum of slowly varying functions

Question

We call a function $L:(0,\infty)\rightarrow(0,\infty)$ slowly varying if for each $c>0$ one has $\displaystyle \lim_{x\to\infty}\frac{L(cx)}{L(x)}=1$.

Can somebody give me a hint why the sum of such functions is also slowly varying?

Answer 1

Suppose that $L$ and $K$ are slowly varying and that $c > 0$.

Fix $\epsilon > 0$. There exists a point $x_0$ with the property that $x \ge x_0$ implies $$\left| \frac{L(cx)}{L(x)} - 1 \right| < \epsilon \quad \text{and} \quad \left| \frac{K(cx)}{K(x)} - 1 \right| < \epsilon.$$ Thus $x \ge x_0$ implies $$|L(cx) - L(x)| < \epsilon L(x) \quad \text{and} \quad |K(cx) - K(x)| < \epsilon K(x).$$ Now apply the triangle inequality: if both inequalities above hold then $$|(L(cx) + K(cx)) - (L(x) + K(x))| \le |L(cx) - L(x)| + |K(cx) - K(x)| < \epsilon (L(x) + K(x))$$ so that $x \ge x_0$ implies $$\left| \frac{L(cx) + K(cx)}{L(x) + K(x)} - 1 \right| < \epsilon.$$The conclusion is that $L + K$ is slowly varying.