Are integral functions of continuous functions of bounded variation?

Question

Given a continuous function $f$, can we say that the integral function $F(t)=\int_{t_0}^t f(s) ds$ is of bounded variation?

Answer 1

Yes.

• If $f$ is continuous on $[a,b]$: Then $f$ is bounded, i.e., there is $M>0$ s.t. $|f(x)|\leq M$ for all $x\in [a,b]$. Therefore $$|F(t)-F(s)|=\leq \int_s^t |f(x)|\,\mathrm d x\leq M|s-t|,$$ for all $s,t\in [a,b]$. Therefore, it has bounded variation on $[a,b]$.

• If $f$ is locally $L^1(a,b)$ only :

Set$$f^{+}(x)=\max\{f(x),0\}\quad \text{and}\quad f^-(x)=-\min\{f(x),0\},$$ then $f^+$ and $f^-$ are positive. Therefore $$t\mapsto \int_0^tf^+(x)\,\mathrm d x\quad \text{and}\quad t\mapsto \int_0^tf^-(x)\,\mathrm d x$$ are increasing. Since $$F(t)=\int_0^t f^+(x)\,\mathrm d x-\int_0^t f^-(x)\,\mathrm d x,$$ (becaue $f(x)=f^+(x)-f^-(x)$), you can write $F$ as a difference of two increasing function. And thus, it has bounded variation.