# Re: analysis question (bounded variation)

*From*: "Dave L. Renfro" <renfr1dl@xxxxxxxxx>*Date*: 11 Jan 2007 09:31:13 -0800

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])

{This comes from additivity of 'Var' over adjacent

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

.

**References**:**analysis question (bounded variation)***From:*jraul

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