\frac{u^{\lambda}(-x)}{u^{\lambda}(0)} = 1- \frac{u^{\lambda}(0)-u^{\lambda}(-x)}{u^{\lambda}(0)} \to 1-\kappa K(-x),$$ as $\lambda \to 0^+$.

Instead of $K$, let us consider the symmetrized $\lambda$-potential kernel $$H^{\lambda}(x) = K^{\lambda}(x) + K^{\lambda}(-x).$$ By [@Bertoin96 Theorem II.19], $$H^{\lambda}(x) = \frac{1}{\pi}\int_{{\mathbb{R}}} \big(1-\cos xs\big){\operatorname{Re}}\Bigg[ \frac{1}{\lambda+\psi(s)} \Bigg]\,ds.$$ Under one can show that *the symmetrized potential kernel* $$H(x) = \lim_{\lambda \to 0^+} H^{\lambda}(x), \quad x \in {\mathbb{R}},$$ is well-defined and $$H(x) = \frac{1}{\pi}\int_{{\mathbb{R}}} \big(1-\cos xs\big){\operatorname{Re}}\Bigg[ \frac{1}{\psi(s)} \Bigg]\,ds, \quad x \in {\mathbb{R}}.$$ Proceeding as in the proof of [@GR17 Proposition 2.2] one may prove that $H$ is subadditive on ${\mathbb{R}}$.

We will often assume the so-called global weak lower scaling condition on the real part of the characteristic exponent with the scaling index strictly bigger than $1$. We will say that a function $f$ satisfies *the global weak lower scaling condition*, if there are $\alpha \in {\mathbb{R}}$ and $c \in (0,1]$ such that for all $u>0$ and $\lambda {\geqslant}1$, $$f(\lambda u) {\geqslant}c\lambda^{\alpha} f(u), \quad u>0.$$ We will write shortly $f \in {{\rm WLSC}(\alpha,c)}$. The pair $(\alpha,c)$ will be refered to as *scaling characteristics* or simply *scalings*. Note that by [@BGR14 Lemma 11], $f \in {{\rm WLSC}(\alpha,c)}$ for some $\alpha \in {\mathbb{R}}$ and $c \in (0,1]$ if and only if the function $$(0,\infty) 