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Suppose that $f$ is a continuous function on $\mathbb{R}$. I want to estimate the definite integral $$ I:= \int_{0}^a [f(x)-f(0)]dx $$ by the upper bound $M = \sup_{x\in[0,a]}|f(x)|$ and the variation $V_f(x,\delta):=\sup_{y\in [x,x+\delta]}|f(x)-f(y)|$

I have a small $\delta>0$ and eventually make $\delta \rightarrow 0$. And fix a positive constant integer $k$. Then the part $$I_1:=\int_0^{k\delta} [f(x)-f(0)]dx$$ can be precisely estimated by $$ |I_1|\leq \sum_{i=0}^{k-1}\int_{i\delta}^{(i+1)\delta} |f(x)-f(0)| \leq \delta \sum_{i=0}^{k}\sum_{j=0}^{i} V_f(j\delta,\delta)$$ due to the continuity of $f$. Although we can also estimate it by upper bound $|I_1|\leq 2M \delta$, it is clearly not the best choice.

However, if I have a constant $0<c<a$ and don't do anything to it later, then the variation is not the best choice for the part $$ I_2 := \int_{c}^a [f(x)-f(0)]dx$$ because $x$ is far away from $0$. So, it may be the best choice to estimate by the upper bound. $$ I_2 \leq 2(a-c)M. $$

The question is what is the best estimation for the rest interval $$ I_3 := \int_{k\delta}^{c}[f(x)-f(0)] ?$$ It seems either the variation or the upper bound is not the best choice. That is, they both overestimate $|I_3|$. Of course, if $f$ satisfies some Holder continuity, $|I_3|$ can be estimated well. Here, nevertheless, I wonder what estimation we can do for the most general continuous functions.

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    $\begingroup$ Your bound on $|I_1|$ is incorrect. Consider e.g. the case when $f$ is nondecreasing, $f(0)=0$, and $f=1$ on $[\delta,\infty)$. $\endgroup$ Commented Jun 8, 2023 at 3:31
  • $\begingroup$ @IosifPinelis I'm sorry. I have missed a sum sign. $\endgroup$
    – Watheophy
    Commented Jun 8, 2023 at 6:06

1 Answer 1

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Without any further conditions on $f$ besides continuity and bounded variation, the best you can do is to set the upper limit of $I_1$ to be that $y$ such that $\text{Var}_{[0, y]} f = 2M$ (or $a$ if the variation is always less than $2M$), and take $c = y$, so that $I_3$ vanishes.

This gives us the upper bound

$$I \leq 2(a-c)M + \int_{0}^c \text{Var}_{[0, s]} f \, ds,$$

and this is sharp, as displayed by say the function $f$ on $[0, a]$ given by

$f(0) = -C;$

$f(x) = C$ for $x \in [\frac{a}{2}, a]$,

and linearly interpolated between $0$ and $\frac{a}{2}$ otherwise, where $C$ is a positive constant.

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