Advanced topics in analysis exercise sheet 4
1. Let f : [0, 1] → [0, 1] be the Cantor function. Find all α such that
f ∈ C α ([0, 1]).
Solution
Let µ be the Cantor measure. Consider the interval [x, y] we have that
f ([x, y]) = µ([y − x]). Suppose that 3−(n+1) ≤ y − x < 3−n for some
n ∈ N, we then have that [x, y] can intersect at most one nth level
triadic interval contained in the Cantor set. Thus
log 2
|f (y) − f (x)| = f (y) − f (x) = µ([y − x]) ≤ 2−n ≤ 3(|y − x|) log 3 .
So f ∈ C α for any 0 < α <
log 2
log 3 .
On the otherhand for any n ∈ N
f (3−n ) − f (0) = 2−n
and so f ∈
/ C α for any α >
log 2
log 3 .
2. Let f : [0, 1] → R be a Lipschitz function which is also singular. Show
that f is constant (try to do this without using that Lipschitz implies
absolutely continuous).
Solution: Since f is singular let the set
E = {x : f 0 (x) = 0}
has measure 1. If for 0 > 0 we let
I0 = {[x, y] ∈ [0, 1] : |f (x) − f (y)| ≤ |x − y|}
then note that for any > 0, E = ∪0 < I0 is a Vitali cover of E. Thus
for any δ, P
> 0 we can find disjoint intervals [x1 , y1 ], . . . , [xn , yn ] ∈ EP
such that ∞
n=1 yn − xn ≥ 1 − δ. We let y0 = 0, xn+1 = 1 and let C
be the Lipschitz constant for f . We have that
∞
X
n+1
X
|f (0) − f (1)| ≤ f (xk ) − f (yk−1 ) +
|f (yi ) − f (xi )|
k=1
n+1
X
≤ C
|yk+1 − xk | + 0
n=1
∞
X
|yi − xi |
n=1
k=1
≤ Cδ + 0 .
By taking , δ to be small we can see that f (0) = f (1). However the
same argument holds for any x ∈ [0, 1], so f (x) = f (0) for all x ∈ [0, 1].
√
3. Let f : [0, 1] → [0, 1] be defined by f (x) = x. Show that f is
absolutely continuous.
1
Solution
We give a direct proof, you could also write the function as √
an integral
to deduce the result. Let > 0 and choose δ > 0 such that 2δ ≤ /2.
Note
that f restricted to [δ, 1] is Lipschitz continuous with constant
√
12δ.
P∞ Now let I1 = [x1 , y1 ], . . . , In = [xn , yn ] be disjoint intervals with
n=1 |In | ≤ δ. We can suppose without loss of generality that there
exists k such that y1 ≤ . . . ≤ yk ≤ 2δ and xk+1 , . . . , xn ≥ δ. Thus
n
X
f (yi ) − f (xi ) ≤ f (yk ) +
i=1
n
X
f (yk+1 ) − f (xk+1 ).
i=k+1
Using the Lipschitz property this gives that
n
X
√
f (yi ) − f (xi ) ≤ /2 + δ 1δ ≤ 2.
i=1
4. Let f : [0, 1] → R be such that there exist closed intervals I1 , . . . , In
such that [0, 1] = ∪ni=1 Ii and f is monotone on each interval. Show
that f is a function of bounded variation.
Solution
First note that by taking intersions of the {In } we can find intervals
J1 ≤ . . . ≤ Jm such that ∪m
i=1 Ji = [0, 1]. We konw that for any
1 ≤ i ≤ m, f : Ji → R is of Bounded variation with variation Vi .
However for any partiiton P of [0, 1] we know that (by adding the
endpoints of the intervals Ji into the partition)
V (P, f ) ≤
m
X
Vi .
i=1
Thus V (f ) ≤
Pm
i=1 Vi
< ∞.
5. Define f : [0, 1] by f (x) = x sin(1/x) if x 6= 0 and f (0) = 0. Show that
f is not of bounded variation.
Solution
Consider the points xi = (iπ)−1 and yi = (π(i + 1/2))−1 . We have
that f (xi ) = 0 and |f (yi )| = yi . We also have that for any n ∈
N, [0, yn ], [yn , xn ], [xn , yn−1 ], [yn−1 , xn−1 ], . . . , [y − 1, x1 ], [x1 , 1] form a
partition of [0, 1]. We have that
V (P, f ) ≤ 2
n
n
X
1
2X 1
=
1.
yi
π
k
+
2
k=1
k=1
Since this tends to ∞ as n → ∞ we have that f is not of bounded
variation.
2