A Lipschitz Bound For Accumulation Functions

From a Bounded Rate to a Bound on Change

An accumulation function records the net amount contributed by a rate over an interval. If the rate never has magnitude greater than $M$, then over a short interval it cannot contribute more than $M$ times the interval's length in magnitude. This observation gives a bound not only on accumulation from a chosen basepoint, but on the difference between the function's values at any two inputs.

The resulting estimate is called a Lipschitz bound. It is useful because it controls changes uniformly: the same constant works for every pair of inputs in the interval. The estimate follows directly from the interval-increment identity and a basic comparison property of definite integrals.

Accumulation Functions and Lipschitz Bounds

Let $J$ be an interval, let $c\in J$ be a fixed basepoint, and suppose $f$ is Riemann integrable on every compact subinterval of $J$. Define the accumulation function $A$ by integrating $f$ from $c$ to the input $x$. The limits are oriented: reversing their order changes the sign of the integral.

A(x)=\int_c^x f(t)\,dt,\qquad x\in J.

The bound on the rate must apply throughout the interval where the estimate is being used. In particular, suppose a nonnegative constant $M$ satisfies $|f(t)|\le M$ for every $t\in J$.

Definition: A function $A$ is Lipschitz on $J$ with Lipschitz constant $M$ if, for every $x,y\in J$, $$ |A(y)-A(x)|\le M|y-x|. $$ A Lipschitz constant is any constant for which this inequality holds; it need not be the smallest possible one.

The definition compares two function values, so it does not depend on which point was chosen as the basepoint. The accumulation function has this property whenever its integrand has the stated uniform bound.

The Lipschitz Estimate

For any $x,y\in J$, the interval-increment identity says that the difference between the two accumulated values is the integral of the rate between those inputs. This identity holds regardless of the location of the basepoint: the contributions from the basepoint to $x$ and from the basepoint to $y$ differ only on the interval between $x$ and $y$.

Theorem: Suppose $f$ is Riemann integrable on every compact subinterval of $J$, and $|f(t)|\le M$ for every $t\in J$, where $M\ge 0$. If $$ A(x)=\int_c^x f(t)\,dt $$ for a fixed $c\in J$, then for all $x,y\in J$, $$ |A(y)-A(x)|\le M|y-x|. $$ Thus $A$ is Lipschitz on $J$ with Lipschitz constant $M$.

Why the Estimate Holds

It is enough to compare two inputs in increasing order. If $x\le y$, the interval-increment identity gives $A(y)-A(x)=\int_x^y f(t)\,dt$. Since $-M\le f(t)\le M$ throughout this interval, the comparison property of definite integrals bounds the integral between the integrals of the two constant functions $-M$ and $M$.

Use the interval-increment identity.
For $x\le y$, $A(y)-A(x)=\int_x^y f(t)\,dt$.

Compare the integrand with constant bounds.
From $-M\le f(t)\le M$, integration over $[x,y]$ gives $-M(y-x)\le\int_x^y f(t)\,dt\le M(y-x)$.

Take the absolute value.
The two-sided inequality implies $|A(y)-A(x)|\le M(y-x)$, and $y-x=|y-x|$.

Cover either ordering.
If $y<x$, interchange the names of the two inputs. Both absolute values are unchanged, so the same estimate follows.

This proves the result. In the increasing-order case, the key comparison can also be written directly as

-M(y-x)\le A(y)-A(x)\le M(y-x),

which places the signed change between two equal bounds of opposite sign. Taking its magnitude gives the claimed estimate. The proof requires integrability on the interval between the inputs and a bound on the integrand there; it does not require that the rate be positive.

Worked Examples

Worked Example: A Constant Rate Attains the Bound

Let $f(t)=3$ on $J=[0,5]$, choose basepoint $c=2$, and define $A(x)=\int_2^x 3\,dt$. Here $|f(t)|=3$ at every point, so we can take $M=3$. Evaluating the integral gives

$$ A(x)=3x-3(2)=3x-6. $$

This expression is consistent with the defining integral: its derivative is $3=f(x)$, and $A(2)=3(2)-6=0$, as required by the basepoint. Take $x=1$ and $y=4$. Then

A(1)=3(1)-6=-3,\qquad A(4)=3(4)-6=6, $$ $$ |A(4)-A(1)|=|6-(-3)|=9,\qquad M|4-1|=3\cdot 3=9.

The estimate holds with equality. This shows that the constant $M$ cannot in general be reduced when all that is known is $|f|\le M$.

Worked Example: A Polynomial Rate

Let $f(t)=t^2$ on $J=[0,2]$, with basepoint $c=0$. Since $0\le t^2\le4$ on this interval, $|f(t)|\le4$, so use $M=4$. The accumulation function is

A(x)=\int_0^x t^2\,dt=\frac{x^3}{3}.

The antiderivative $\frac{t^3}{3}$ differentiates to $t^2$, and its value at $0$ is $0$, verifying the integral evaluation. Compare $x=1$ and $y=\frac32$:

A(1)=\frac{1}{3},\qquad A\left(\frac32\right)=\frac{(3/2)^3}{3}=\frac{27}{24}=\frac98, $$ $$ \left|A\left(\frac32\right)-A(1)\right| =\left|\frac98-\frac13\right| =\left|\frac{27}{24}-\frac{8}{24}\right| =\frac{19}{24}.

The Lipschitz bound for this input change is

M\left|\frac32-1\right|=4\cdot\frac12=2=\frac{48}{24}.

Since $\frac{19}{24}\le\frac{48}{24}$, the estimate holds. Here it is not exact, because the rate's magnitude is smaller than $4$ on much of the interval between the inputs.

Worked Example: A Rate That Changes Sign

Let $f(t)=1-2t$ on $J=[0,1]$, with $c=0$. The rate decreases from $1$ to $-1$, so $|1-2t|\le1$ throughout $J$, and $M=1$ is a valid bound. Integrating gives

A(x)=\int_0^x(1-2t)\,dt=x-x^2.

The antiderivative $t-t^2$ differentiates to $1-2t$, and it is zero at the basepoint, confirming the evaluation. For $x=\frac15$ and $y=\frac{7}{10}$,

A\left(\frac15\right)=\frac15-\frac1{25}=\frac4{25}, $$ $$ A\left(\frac{7}{10}\right)=\frac7{10}-\frac{49}{100} =\frac{70}{100}-\frac{49}{100} =\frac{21}{100}.

Thus the magnitude of the change is

\left|\frac{21}{100}-\frac4{25}\right| =\left|\frac{21}{100}-\frac{16}{100}\right| =\frac5{100}=\frac1{20}.

The bound is $M|y-x|=1\left|\frac{7}{10}-\frac15\right|=\frac12$. Therefore $\frac1{20}\le\frac12$. The rate changes sign between the inputs, but the estimate remains valid: it controls the net change in accumulation using the magnitude bound, without requiring a fixed sign.

What the Bound Tells You

The inequality states that if the input changes by a distance $d$, the accumulated value changes by no more than $Md$. In this sense, $M$ limits the sensitivity of the accumulation function to changes in its input. For a rate measured in accumulated units per input unit, the product $M|y-x|$ has the units of accumulated value, as it should.

A Lipschitz bound also gives a direct continuity estimate. Given any $\varepsilon>0$, if $|y-x|<\varepsilon/M$ when $M>0$, then $|A(y)-A(x)|<\varepsilon$. If $M=0$, the rate is zero throughout $J$ and the estimate says $A(y)=A(x)$ for every pair of inputs. Thus a bounded rate gives uniform control over changes, rather than a bound that depends on which input is being considered.

The estimate concerns differences in accumulation values, not the values themselves. A bound on the rate over an interval can also bound accumulation from a particular basepoint, but that is a different comparison: it uses the distance from the basepoint. The Lipschitz estimate instead uses the distance between the two inputs being compared.

Common Pitfall: Comparing to the Basepoint Instead

A tempting but unhelpful approach is to bound the two accumulated values separately and then add the bounds. For $f(t)=1$ on $[0,5]$ with basepoint $0$, $A(x)=x$. At inputs $4$ and $5$, the separate bounds give $|A(5)|+|A(4)|=5+4=9$. That is a valid, but loose, upper bound for $|A(5)-A(4)|$; it does not establish the sharper estimate $1$ merely by replacing the sum $5+4$ with the difference $5-4$.

The correction is to use the interval-increment identity on the interval between the inputs:

A(5)-A(4)=\int_4^5 1\,dt=1, \qquad |A(5)-A(4)|=1\le 1\cdot|5-4|.

In general, apply the rate bound over the interval between $x$ and $y$, not two separate bounds measured from the basepoint. Also keep the absolute value: the integral may be negative, but the theorem bounds the magnitude of the change.

Check Your Understanding

Use the definition and the integral comparison in each question. State the relevant rate bound when applying the estimate.

If $|f(t)|\le 5$ on an interval and $A(x)=\int_c^x f(t)\,dt$, what bound does the theorem give for $|A(7)-A(2)|$?
Why does a sign change in $f$ not invalidate the Lipschitz estimate?
For $f(t)=-2$ on an interval, what is a Lipschitz constant for its accumulation function? Is the estimate attained for distinct inputs?
Which interval must satisfy the bound $|f(t)|\le M$ to estimate $|A(y)-A(x)|$?
Explain why a bound on $|A(x)|+|A(y)|$ from the basepoint does not by itself give the Lipschitz estimate in terms of $|y-x|$.