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Nate River
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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

$f(0) = -C;$ $f = C$ on

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

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

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

$f(0) = -C;$ $f = C$ on $[\frac{a}{2}, a]$

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

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.

added 39 characters in body
Source Link
Nate River
  • 6.2k
  • 2
  • 23
  • 99

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

$$I \leq 2(a-c+1)M,$$$$I \leq 2(a-c)M + \int_{0}^c \text{Var}_{[0, s]} f \, ds,$$

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

$f(0) = -C;$

$ f(\frac{a}{2}) = f(a) = C$, $f = C$ on $[\frac{a}{2}, a]$

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

Without any further conditions on $f$ besides continuity, 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 $c = y$.

This gives us the upper bound

$$I \leq 2(a-c+1)M,$$

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

$f(0) = -C;$

$ f(\frac{a}{2}) = f(a) = C$,

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

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

$f(0) = -C;$ $f = C$ on $[\frac{a}{2}, a]$

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

Source Link
Nate River
  • 6.2k
  • 2
  • 23
  • 99

Without any further conditions on $f$ besides continuity, 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 $c = y$.

This gives us the upper bound

$$I \leq 2(a-c+1)M,$$

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

$f(0) = -C;$

$ f(\frac{a}{2}) = f(a) = C$,

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