I'm learning about functions of bounded variations and need help to understand the proof of this lemma:
Lemma. If $f : [a,b] \rightarrow \mathbb{R}$ is of bounded variation, then f is also bounded and satisfies $\left\lVert f \right\rVert_{\infty} \leq \left|f(a) \right| + V_{a}^{b} f$.
Proof. Let $a \leq x \leq b$, and set $P = \{a, x, b\}$. Then $\left| f(x) - f(a) \right| \leq V(f, P) \leq V_{a}^{b} f$. Consequently, $\underbrace{\left| f(x) \right| \leq \left| f(a) \right| + V_{a}^{b} f}_{(1)}$.
In inequality (1) I don't understand how we're able to break the absolute value.
Also, I'm a little confused by the symbol $\left\lVert f \right\rVert_{\infty}$. Can someone explain it?
The symbol $\|f\|_{\infty}$ is the $L^{\infty}$ norm of the function, also called the essential supremum. It is the largest (more precisely, the supremum of) number $\alpha$ for which $\{x : |f(x)| > \alpha\}$ has positive measure. For a continuous function on a compact set, it coincides with the maximum.
As far as the inequality, it appears to be a typo and one $x$ ought to be an $a$. It follows from the (reverse) triangle inequality:
$$|f(x)| - |f(a)| \le |f(x) - f(a)| \le ...$$
One way to see the first inequality is to apply the usual triangle inequality to $f(x) = f(x) - f(a) + f(a)$.
The correct statement of the lemma should actually be
$$\|f\|_{\infty} \le |f(a)| + V_{a}^b f$$