One-Sided Limits And Jump Sizes For BV Functions

Why One-Sided Limits Are Forced by Bounded Variation

A function of bounded variation can have discontinuities, but it cannot oscillate by a fixed amount infinitely often while remaining on a compact interval. Near an interior point, that restriction forces the function values on either side to settle toward finite limits. Those limits describe the jump across the point; the value assigned at the point itself is a separate matter.

Let \(a<b\), and let \(f:[a,b]\to\mathbb{R}\) be real-valued. For \(a\le u\le v\le b\), the variation of \(f\) on \([u,v]\) is

V_u^v(f)=\sup\left\{\sum_{j=1}^{n}|f(x_j)-f(x_{j-1})|:\ u=x_0<x_1<\cdots<x_n=v\right\},

where \(V_u^u(f)=0\). The function \(f\) is of bounded variation, written \(f\in BV[a,b]\), if \(V_a^b(f)<\infty\). Variation is computed from all finite partitions; it is not an integral or a measure. Throughout this tutorial, all limits of function values are ordinary pointwise one-sided limits, not almost-everywhere limits.

Lemma (Additivity of variation): If \(f\in BV[a,b]\) and \(a\le u\le w\le v\le b\), then \(V_u^v(f)=V_u^w(f)+V_w^v(f)\).

Proof. First suppose \(u<w<v\). Take any partition of \([u,v]\) and insert \(w\) if it is not already present. Inserting a point cannot decrease the partition sum, since \(|f(y)-f(x)|\le |f(w)-f(x)|+|f(y)-f(w)|\) whenever \(x<w<y\). The refined sum is the sum of a partition sum on \([u,w]\) and one on \([w,v]\), so it is at most \(V_u^w(f)+V_w^v(f)\). Taking the supremum over partitions of \([u,v]\) gives \(V_u^v(f)\le V_u^w(f)+V_w^v(f)\).

For the reverse inequality, both subinterval variations are finite because \(f\in BV[a,b]\). Given \(\varepsilon>0\), choose a partition of \([u,w]\) with sum greater than \(V_u^w(f)-\varepsilon/2\), and a partition of \([w,v]\) with sum greater than \(V_w^v(f)-\varepsilon/2\). Such partitions exist by the definition of a finite supremum. Concatenating them at \(w\) gives a partition of \([u,v]\), so \(V_u^v(f)>V_u^w(f)+V_w^v(f)-\varepsilon\). Since this holds for every \(\varepsilon>0\), the reverse inequality follows. If \(w=u\) or \(w=v\), the identity follows from \(V_u^u(f)=0\). \(\square\)

Existence of the One-Sided Limits

Theorem: If \(f\in BV[a,b]\) and \(c\in(a,b)\), then the finite limits \(f(c-)=\lim_{x\uparrow c}f(x)\) and \(f(c+)=\lim_{x\downarrow c}f(x)\) exist.

Proof. Define the variation function \(W(t)=V_a^t(f)\) for \(t\in[a,b]\). Additivity shows that \(W\) is nondecreasing and that, whenever \(a\le x\le y\le b\),

|f(y)-f(x)|\le V_x^y(f)=W(y)-W(x).

Since \(0\le W(t)\le V_a^b(f)<\infty\), the number \(L_-=\sup_{a\le t<c}W(t)\) is finite. Choose a sequence \(r_n\in(a,c)\) increasing to \(c\). Then \(W(r_n)\to L_-\): for every \(\varepsilon>0\), the definition of the supremum gives some \(t<c\) with \(W(t)>L_- -\varepsilon\), and eventually \(r_n>t\), so \(L_- -\varepsilon<W(r_n)\le L_-\). For \(m>n\), the displayed variation bound gives \(|f(r_m)-f(r_n)|\le W(r_m)-W(r_n)\). Thus \((f(r_n))\) is Cauchy and has a finite real limit, say \(\ell_-\).

To check that this sequence limit is the full left-hand limit, fix \(\varepsilon>0\). Choose \(r<c\) such that \(L_- - W(r)<\varepsilon/3\). Choose \(n\) large enough that \(r_n>r\) and \(|f(r_n)-\ell_-|<\varepsilon/3\). For every \(x\in(r,c)\), both \(x\) and \(r_n\) lie to the right of \(r\), and additivity and monotonicity of \(W\) give \(|f(x)-f(r_n)|\le L_- - W(r)<\varepsilon/3\). Therefore \(|f(x)-\ell_-|<2\varepsilon/3<\varepsilon\), proving \(f(x)\to\ell_-\) as \(x\uparrow c\).

For the right-hand limit, set \(L_+=\inf_{c<t\le b}W(t)\), which is finite. Choose \(s_n\in(c,b)\) decreasing to \(c\). Then \(W(s_n)\to L_+\): given \(\varepsilon>0\), choose \(t>c\) with \(W(t)<L_++\varepsilon\); eventually \(s_n<t\), and hence \(L_+\le W(s_n)\le W(t)<L_++\varepsilon\). If \(m>n\), then \(s_m<s_n\), so \(|f(s_m)-f(s_n)|\le W(s_n)-W(s_m)\). It follows that \((f(s_n))\) is Cauchy and converges to a finite real number \(\ell_+\).

Given \(\varepsilon>0\), choose \(r>c\) such that \(W(r)-L_+<\varepsilon/3\), and then choose \(n\) with \(s_n<r\) and \(|f(s_n)-\ell_+|<\varepsilon/3\). For each \(x\in(c,r)\), the variation bound yields \(|f(x)-f(s_n)|\le W(r)-L_+<\varepsilon/3\). Thus \(|f(x)-\ell_+|<\varepsilon\), proving \(f(x)\to\ell_+\) as \(x\downarrow c\). Set \(f(c-)=\ell_-\) and \(f(c+)=\ell_+\). \(\square\)

Definition: For \(c\in(a,b)\), the left trace \(f(c-)\) and right trace \(f(c+)\) are the one-sided limits just proved to exist. The signed jump of \(f\) at \(c\) is \(\Delta f(c)=f(c+)-f(c-)\); its magnitude is \(|\Delta f(c)|\). These quantities depend on the limiting values on either side, not on \(f(c)\).

The sign records the direction of the change from left to right: a positive jump is upward, and a negative jump is downward. A zero jump means the two one-sided limits agree. It does not, by itself, assert that \(f(c)\) equals that common limit.

Worked Step-Function Calculations

Worked Example: A Single Downward Step

On \([0,2]\), define \(f(x)=2\) for \(0\le x<1\) and \(f(x)=-3\) for \(1\le x\le2\). Find the one-sided limits and jump at \(c=1\), and compute \(V_0^2(f)\).

For every \(x<1\), \(f(x)=2\), so \(f(1-)=2\). For every \(x>1\), \(f(x)=-3\), so \(f(1+)=-3\). Therefore

\Delta f(1)=f(1+)-f(1-)=-3-2=-5,\qquad |\Delta f(1)|=5.

In any increasing partition of \([0,2]\), all sampled values are \(2\) until the partition reaches a point in \([1,2]\), after which all sampled values are \(-3\). Thus its sum of absolute increments is either \(0\) or a single change of size \(|-3-2|=5\). The partition \(0,1,2\) has sum \(|2-(-3)|+|-3-(-3)|=5+0=5\), so \(V_0^2(f)=5\). Check: the computed variation equals the magnitude of the only change between the constant levels, and the jump’s sign is negative because the level decreases from \(2\) to \(-3\).

Worked Example: An Isolated Point Value Is Not a Jump

On \([-1,1]\), let \(g(0)=4\) and \(g(x)=0\) for every \(x\ne0\). Compute the traces, signed jump, and total variation.

For all \(x<0\), \(g(x)=0\), and for all \(x>0\), \(g(x)=0\). Hence \(g(0-)=0\) and \(g(0+)=0\), giving \(\Delta g(0)=0-0=0\). The value \(g(0)=4\) does not enter this subtraction.

A partition that omits \(0\) has sum \(0\), since all its sampled values are zero. A partition that includes \(0\) has at most one change from \(0\) to \(4\) and at most one change back from \(4\) to \(0\); its sum is at most \(4+4=8\). The partition \(-1,0,1\) attains this bound: \(|g(0)-g(-1)|+|g(1)-g(0)|=|4-0|+|0-4|=4+4=8\). Therefore \(V_{-1}^{1}(g)=8\). Check: the traces agree, so the jump is zero, while the variation is positive because the assigned value at the point differs from both surrounding levels.

Worked Example: Several Step Changes

On \([-2,3]\), define \(h\) by the levels \(-1\) on \([-2,-1)\), \(2\) on \([-1,1)\), \(-2\) on \([1,2)\), and \(1\) on \([2,3]\). Find the traces and jumps at \(c=1\), and compute the total variation.

Immediately to the left of \(1\), the function equals \(2\); immediately to the right, it equals \(-2\). Thus \(h(1-)=2\), \(h(1+)=-2\), and \(\Delta h(1)=-2-2=-4\), with magnitude \(4\). The function’s value at \(1\) is \(-2\), as specified by the interval \([1,2)\); that agrees with the right trace but is not needed to compute the jump.

The three successive changes of level occur at \(-1\), \(1\), and \(2\). Their absolute sizes, computed from the displayed levels, are \(|2-(-1)|=3\), \(|-2-2|=4\), and \(|1-(-2)|=3\). Any partition sum is at most their total \(3+4+3=10\): if a partition skips one or more breakpoints, the triangle inequality bounds the direct change across those levels by the sum of the skipped successive changes. The partition \(-2,-1,1,2,3\) attains \(3+4+3+0=10\). Hence \(V_{-2}^{3}(h)=10\). Check: this partition’s successive values are \(-1,2,-2,1,1\), so its sum is \(3+4+3+0=10\), as claimed.

A Bound on Jump Magnitude and a Common Pitfall

Proposition: If \(f\in BV[a,b]\) and \(c\in(a,b)\), then \(|\Delta f(c)|\le V_a^b(f)\).

Proof. Choose sequences \(x_n<c<y_n\) with \(x_n\uparrow c\) and \(y_n\downarrow c\). By the definition of variation, \(|f(y_n)-f(x_n)|\le V_a^b(f)\) for every \(n\): the two points can be included in a partition of \([a,b]\), whose sum includes this increment and has only nonnegative terms. By the one-sided limit theorem, \(f(y_n)-f(x_n)\to f(c+)-f(c-)=\Delta f(c)\). Continuity of absolute value gives \(|f(y_n)-f(x_n)|\to|\Delta f(c)|\). Passing to the limit in the inequalities proves the result. \(\square\)

A common mistake is to call \(f(c)-f(c-)\) the jump. For the isolated-point example, that incorrect calculation gives \(4-0=4\), even though the signed jump is \(g(0+)-g(0-)=0-0=0\). The correction is to take both traces first and subtract right trace minus left trace. The point value may equal one trace, both traces, or neither; it is not part of the definition of the jump.

One-sided limits are useful because they separate the behavior immediately on either side of a point from an arbitrary value assigned at the point. For step functions, this makes jumps direct to compute; for general bounded-variation functions, the variation-function argument guarantees that the same traces exist even when there is no finite step-by-step description.

Check Your Understanding

For each question, distinguish the one-sided traces from the value at the point and justify any variation calculation from partitions.

On \([-1,2]\), let \(q(x)=1\) for \(x<0\) and \(q(x)=5\) for \(x\ge0\). Find \(q(0-)\), \(q(0+)\), \(\Delta q(0)\), and \(V_{-1}^{2}(q)\).
Suppose \(r\in BV[a,b]\) and \(c\in(a,b)\). In the existence proof, why does boundedness of \(W(t)=V_a^t(r)\) ensure that both one-sided monotone limits of \(W\) are finite?
Define \(s(0)=3\) and \(s(x)=-1\) for \(x\ne0\) on \([-1,1]\). Compute both traces and the signed jump at \(0\). Explain why the value \(s(0)\) does not determine that jump.
For the function in the preceding question, compute \(V_{-1}^{1}(s)\) and exhibit a partition attaining it.
Explain why the definition of a signed jump uses \(f(c+)-f(c-)\), rather than \(f(c)-f(c-)\).