# Re: analysis question (bounded variation)

jraul wrote:

I have a question on Theorem (2) below, but it quotes
theorem (1) which I give now:

1. Let f be defined on [a,b]. Then f is of bounded
variation on [a,b] if and only if f can be expressed
as the difference of two increasing functions.

2. Let f be of bounded variation on [a,b]. If x in (a,b],
let V(x) = V_f(a,x) and put V(a) = 0. Then every point
of continuity of f is also a point of continuity of V.
The converse is also true.

I have trouble with the beginning of the proof, but
accepting that I was able to understand the rest.
My questions are in brackets []:

Since V is monotonic, the right and lefthand limits
V(x+) and V(x-) exist for each point x in (a,b).
Because of Theorem 1, the same is true of f(x+)
and f(x-).

[Is this understanding correct? V monotonic so it
can only have jump discontinuities, so V(x+) and
V(x-) exist. By Theorem 1, because f is of bounded
variation, it is the difference of two monotonic functions,
so it too can only have jump discontinuities.]

The proof continues: If a < x < y <= b, then we have:

0 <= |f(y) - f(x)| <= V(y) - V(x)

Letting y-->x, we have

0 <= |f(x+) - f(x)| <= V(x+) - V(x)

Likewise,

0 <= |f(x) - f(x-)| <= V(x) - V(x-)

These inequalities imply that a point of continuity
of V is also a point of continuity of f.

[Is this because, as we take the limits, V(x+), V(x-),
we make |y-x| < delta, and since V is assumed continuous,
we have V(x+) - V(x) < epsilon and V(x) - V(x-) < epsilon.
Hence f is continuous as x, too?]

After looking at the result you want to prove, it occurred
to me that this is a good example where a more precise
result clarifies things. As I got into this, I noticed
that continuity is not really the main idea, at least
not in the calculations I wound up doing.

To review, there are 5 relevant values one can associate
with a function f(x) at a given point x = c:

The left lower and left upper extreme limits,
lim-inf f(c-) and lim-sup f(c-).

The value of f at x = c, f(c).

The right lower and right upper extreme limits,
lim-inf f(c+) and lim-sup f(c+).

There are various "measures of discontinuity" that can
be assigned to f(x) at x = c, such as the maximum of
the two pairs of unilateral extreme-limit differences,
the maximum of the differences between any two of the
four extreme limits, the maximum of the differences
between the value of the function and any of the four
extreme limits, etc.

In the case where f is monotone, these reduce to
just three: f(c-), f(c), and f(c+). Moreover, f(c)
lies between f(c-) and f(c+).

If f has bounded variation, then we still have just
these three, although f(c) may no longer lie between
f(c-) and f(c+). Incidentally, the reason BV functions
have left and right limits is because the left and
right limit operations commute with subtraction
(indeed, the commute with all four basic arithmetic
operations), and every BV function can be written
as a difference of monotone functions. Note that
the unilateral extreme limits (or even the bilateral
lim-inf and lim-sup's at a point) only satisfy one
inequality (which varies according to whether it's
the lim-inf or the lim-sup), and not equality, for
subtraction of functions. Thus it's really a pair of
fortunate events that allows us go from monotone functions
to BV functions regarding the existence of unilateral
limits: BV functions are differences of monotone functions,
and we're dealing with actual limits (rather than just
with lim-inf's or lim-sup's).

Now for a more precise version of your theorem.

THEOREM: Assume f belongs to BV[a,b].

(1) For each c such that a <= c < b, we have

V(c+) - V(c) <= | f(c+) - f(c) |

(2) For each c such that a < c <= b, we have

V(c) - V(c-) <= | f(c-) - f(c) |

COROLLARY 1: If f belongs to BV[a,b], a <= c < b,
and f is right continuous at c, then
V is right continuous at c.

COROLLARY 2: If f belongs to BV[a,b], a < c <= b,
and f is left continuous at c, then
V is left continuous at c.

COROLLARY 3: If f belongs to BV[a,b], a < c < b,
and f is continuous at c, then V is
continuous at c.

Roughly speaking, V is no more discontinuous at x = c
than f is. We can't improve either of the inequalities
to a strict inequality, as appropriate step functions
will show.

*****************************************************

PROOF OF (1)

Let epsilon > 0 be given.

Using the definition of f(c+), choose delta > 0 so that
c < x < c + delta ==> |f(c+) - f(x)| < epsilon.

Now let c = x_0 < x_1 < x_2 < ... < x_n = b be a
partition of [c,b] such that c < x_1 < c + delta and
Var(f, [c,b]) < epsilon + SUM(k=1 to n) |f(x_k) - f(x_(k-1))|.

This can be done, since the definition of variation
as a supremum over such partitions of such sums
implies that we can get within epsilon by using
some partition, and refining (if necessary) any such
partition so that it includes an additional 2'nd
point x_1 satisfying c < x_1 < c + delta will only
get us closer to Var(f, [c,b]). {At some earlier time
you should have proved, or you should now be able
to use, the fact that finer partitions can't result
in strictly smaller "variation sums".}

Now pick any x' such that c < x' < x_1. Note that we
have c < x' < x_1 < c + delta, and so both x' and x_1
belong to the right delta-neighborhood of c.

The purpose of x' is to allow us a vehicle that we
can use to get V(c+) and f(c+) to show up.

Then we get

V(x') - V(c) = Var(f, [c,b]) - Var(f, [x',b])

intervals, which should have been previously proved.}

This in turn is less than

[ epsilon + SUM(k=1 to n) |f(x_k) - f(x_(k-1))| ]

minus

[ |f(x_1) - f(x')| + SUM(k=2 to n) |f(x_k) - f(x_(k-1))| ].

{This comes from the fact that we're replacing A - B
with A' - B', where A' > A and B' <= B. That is, we're
subtracting less from something more. The reason A' > A
is because this is how the partition was chosen to begin
with. The reason B' <= B is because the actual variation
on an interval is never less than the variation relative
to some specific partition of that interval.}

By additively cancelling n-1 terms, the previous
expression is equal to

epsilon + |f(x_1) - f(x_0)| - |f(x_1) - f(x')|.

Using x_0 = c and using a form of the triangle inequality
(the form that says | |r| - |s| | <= |r - s|), we get
the previous expression to be less than or equal to

| f(x_1) - f(c) - f(x_1) + f(x') | + epsilon

= | f(x') - f(c) | + epsilon

= | f(x') - f(c+) + f(c+) - f(c) | + epsilon

<= |f(x') - f(c+)| + |f(c+) - f(c)| + epsilon

Therefore, we have

V(x') - V(c) < |f(x') - f(c+)| + |f(c+) - f(c)| + epsilon

Since x' lies between c and c + delta, it follows from
the way delta was chosen (at the very beginning of
this proof) that

V(x') - V(c) < epsilon + |f(c+) - f(c)| + epsilon

V(x') - V(c) < |f(c+) - f(c)| + 2*epsilon

Using the fact that V(x') can be made arbitrarily
close to V(c+) {to fill in all the details, another
epsilon-delta argument is needed at this point},
we then get

V(c+) - V(c) <= | f(c+) - f(c) | + 2*epsilon

{Note that the strict inequality turned into a
non-strict inequality.}

Since the previous inequality holds for all epsilon > 0,
we finally have the desired result:

V(c+) - V(c) <= | f(c+) - f(c) |

*****************************************************

PROOF OF (2)

The proof is very similar, so I'll just outline the
changes needed.

Choose delta so that

c - delta < x < c ==> |f(c-) - f(x)| < epsilon.

Choose a = x_0 < x_1 < x_2 < ... < x_n = c
so that c - delta < x_(n-1) < c and
Var(f, [a,c]) < epsilon + SUM(k=1 to n) |f(x_k) - f(x_(k-1))|

Choose x' so that c - delta < x_(n-1) < x' < c.

Begin with V(c) - V(x') = Var(f, [a,c]) - Var(f, [a,x'])
and estimate as before (cancel n-1 terms, put in x_n = c,
use triangle inequality, etc.).

You'll eventually get

V(c) - V(x') < | f(c) - f(c-) | + 2*epsilon.

Hence, by an epsilon-delta argument with x', we get

V(c) - V(c-) <= | f(c) - f(c-) | + 2*epsilon.

Hence, since epsilon > 0 was arbitrary, we get

V(c) - V(c-) <= | f(c) - f(c-) |.

*****************************************************

Dave L. Renfro

.

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