Solutions of Twelfth Homework
Solution of 10.1 Ex. 3: Let P be a partition of R. For a rectangle
Ri in the partition we have
|f (x) − f (y)| ≤ K|g(x) − g(y)| ≤ K| sup g(x) − inf g(y)|
y∈Ri
x∈Ri
for each x, y ∈ Ri . Therefore, we have
| sup f (x) − inf f (y)| ≤ K| sup g(x) − inf g(y)|.
x∈Ri
y∈Ri
x∈Ri
y∈Ri
This in turn implies that
|U (f, P ) − L(f, P )| ≤ K|U (g, P ) − L(g, P )|.
If g is integrable on R, for any > 0, there exists P such that U (g, P )−
L(g, P ) < . Therefore, U (f, P ) − L(f, P ) < K. Since is arbitrary,
we see that f is integrable on R.
Solution of 10.1 Ex. 4: By the triangle inequality
|a| = |(a − b) + b| ≤ |a − b| + |b|
for any two numbers a, b ∈ R. Therefore, we have |a| − |b| ≤ |a − b|. By
switching the roles of a and b, we also have |b| − |a| ≤ |a − b|. Therefore
we have
||a| − |b|| ≤ |a − b|.
This implies that
||f (x)| − |f (y)|| ≤ |f (x) − f (y)|
for all x, y ∈ R. By the Ex. 3, it follows that if f is integrable on R,
|f | is also integrable on R.
Solution of 10.1 Ex. 5: Since f is bounded on R, there exists M
such that |f (x)| ≤ M . Therefore, we have
|f 2 (x) − f 2 (y)| ≤ |f (x) − f (y)||f (x) + f (y)|
≤ |f (x) − f (y)|(|f (x)| + |f (y)|) ≤ 2M |f (x) − f (y)|
for all x, y ∈ R. By Ex. 3, we see that f 2 is integrable on R if f is
integrable on R.
Solution of 10.1 Ex. 6: If f and g are integrable on R, the functions f + g and f − g are also integrable on R. By Ex. 5, this implies
that (f + g)2 and (f − g)2 are integrable on R. On the other hand, it
follows that
1
f g = ((f + g)2 − (f − g)2 )
4
1
2
is integrable on R.
Solution of 10.1 Ex. 7: Clearly, U (k, P ) = kV (R) and L(k, P ) =
kV (R) for any partition P . Therefore k is integrable on R and
Z
kdV (x) = kV (R).
R