∑ ∑
|F (si+1)− F (si)|p ≤ (KT )p−1|F (si+1 )− F (si)|
P P
≤ ∑ (KT )p−1K (s − s) = (KT )p
P i+1 i
It follows that F^{P}is in V^{p}
([0,T ])
. ■
Note that this actually shows that any such piecewise linear function is in V^{p} for every
p ≥ 1.
Next is a fundamental approximation lemma which says that when you replace a function
in V^{p} with its piecewise linear approximation the p variation gets smaller.
Lemma 10.2.3Let F ∈ V^{p}
([0,T ])
and let P be a dissection. Then
∥∥ P ∥∥
F p,[0,T] ≤ ∥F∥p,[0,T]
Proof: Let P_{ε} be a dissection of
[0,T]
such that
∑ | | ∥ ∥
|FP (ui+1)− F P (ui)|p > ∥FP ∥p − ε (10.1)
ui∈Pε p,[0,T ]
(10.1)
The idea is to show that, deleting some points of P_{ε} you can assume every grid point of P_{ε} is
also a grid point of P while preserving the above inequality. In other words, we can remove
the grid points of P_{ε} which are not already in P and end up making the left side at least as
large. This might not be too surprising because the variation of F^{P} is determined by the
F
(ti)
where t_{i}∈P. Let the mesh points of P_{ε} be denoted by
{ui}
and those of P are
denoted by
{tj}
.
Let u_{i+1} = u be the first point of P_{ε} which is not in P. Then u is not equal to any of the
t_{j} and so u ∈
(tj,tj+1)
for some j. Now let v be the very next point of P∪P_{ε} after u. The
possibilities are
ui = tk,k ≤ j,ui+1 ∈ (tj,tj+1), v < ui+2, so v = tj+1 (**)
(**)
PICT
Consider case * In this case, you have u_{i+2} comes before t_{j+1} and so F^{P} is linear on
[tj,v]
. Therefore, the map
| | | |
s → |FP (s)− FP (tk)|p + |F P (ui+2)− F P (s)|p
is convex and so its maximum occurs at an end point. You get for the maximum
| |
||F P (tj)− FP (tk)||p + ||F P − F P||p if end point is s = tj
| ui+2 tj |p
|FP (ui+2) − F P (tk)| if end point is s = v
In the sum of 10.1 you have some terms involving u_{i+1} they are, in this case
| |
||FP (ui+1) − F Pt||p + ||FPu − F P (ui+1)||p
k i+2
Here u_{i+1} is like s in what was just shown. This was just shown to be no larger than either
| |
|F P (tj)− FP (tk)|
^{p} +
| |
|FP (ui+2)− F P (tj)|
^{p} or
| |
|F P (ui+2)− FP (tk)|
^{p} so we could replace
those terms in the sum with one of these and F^{P}
(ui+1)
has been eliminated while the
inequality of 10.1 has been preserved.
Consider case ** You have in the sum
|| P P ||p || P P ||p
F (ui+1)− F (tk) + F (ui+2)− F (ui+1)
because u_{i} = t_{k}. As before, for s ∈
[tj,v]
=
[tj,tj+1]
| P P |p | P P |p
s → |F (s)− F (tk)| + |F (ui+2)− F (s)|
is convex and so it is no larger than either
| |p | |p
|F P (tj)− FP (tk)| + |FP (ui+2)− F P (tj)|
or when maximum occurs at v = t_{j+1},
||FP (t )− FP (t)||p + ||F P (u )− F P (t )||p
j+1 j i+2 j+1
Now in the sum, we still have
| |p | |p
|F P (ui+1)− FP (tk)| + |FP (ui+2)− F P (ui+1)|
where u_{i+1} is like s. Thus this sum of the two terms is dominated by either of
| |p | |p
|F P (tj)− FP (tk)| + |FP (ui+2)− F P (tj)|
or
||FP (t )− FP (t)||p + ||F P (u )− F P (t )||p
j+1 j i+2 j+1
and so if replaced with one of these, the inequality of 10.1 is preserved without using
u_{i+1}.
Thus this eliminates u_{i+1} = u from P_{ε} and preserves the inequality of 10.1. Repeat the
process until all the mesh points in P_{ε} are in P. Now using this possibly modified P_{ε}, each
mesh point is in P and so
Since ε is arbitrary, this establishes the desired inequality. ■
Theorem 10.2.4Let F ∈ V^{p}
([0,T ])
,p ≥ 1. Then
lim_{|P|
→0}
∥ ∥
∥F P − F ∥
_{q,[0,T]
} = 0 for every q > p.
Proof:First of all, note that F^{P} converges uniformly to F. To see this, let ε > 0 be given
and let δ work for uniform continuity of F on
[0,T]
with ε∕3. Then let
|P |
< δ. Let
t ∈
[0,T]
. Say t ∈
[ti,ti+1]
where the end points are successive points of P. Then
||FP (t)− F (t)|| ≤ ||FP (t)− F P (ti)||+ ||FP (ti)− F (ti)||+ |F (t) − F (ti)|
|| P P ||
= |F (t)− F (ti)| + |F (ti)− F (t)|
< |FP (t)− F (ti)|+ ε
3
Now there is a formula for F^{P} on this interval given by
--1---
ti+1−ti
(F (ti))
(ti+1 − t)
+F
(ti+1)
(t− ti)
and so
| |
= ||--1----[(F (ti))(ti+1 − t)+ F (ti+1)(t− ti)]− F (ti)||+ ε
||ti+1 − ti | | 3
||---1----- || ε
= |(ti+1 − ti) [(F (ti+1) − F (ti))(t− ti)]|+ 3
ε ε ε
≤ |F (ti+1)− F (ti)|+ 3 < 3 + 3 < ε (10.2)
Therefore, the uniform convergence certainly holds. Note that one only needs to know that
the function is continuous to show this. Then
∑
∥∥FP − F∥∥q = ||FP (ti+1)− F (ti+1)− (FP (ti)− F (ti))||q
q,[0,T] ti∈P
( )
≤ max ||FP (ti+1)− F (ti+1)− (FP (ti)− F (ti))||q−p ⋅
t∑i∈P
||F P (ti+1 )− F (ti+1) − (F P (ti) − F (ti))||p
ti∈P
(|| P ( P )||q−p)∥∥ P ∥∥p
≤ mtai∈xP F (ti+1)− F (ti+1)− F (ti)− F (ti) F − F p,[0,T]
((| P | | P |)q−p)( ∥ P∥ )p
≤ mtai∈xP |F (ti+1)− F (ti+1)|+ |F (ti)− F (ti)| ∥F ∥p,[0,T] + ∥F ∥p,[0,T]
q−p | P |q−p( )p
≤ 2 tm∈a[0x,T ]|F (t) − F (t)| 2 ∥F∥p,[0,T]