By page 121 we know that
must be bounded, say by
. We need to show that given
we can find some
such that
So, by Theorem 6.12 (c) we have
and
.
Hence,
but since we can choose any
and
is fixed we can choose
which yields
So, given
we can always choose a
such that
as desired.
Considered the function which is defined to be
on the last
of the interval [0,1] and zero at those
where
. This function is well defined, since we know that
.
More specifically the function has value
on the open interval from
First we evaluate the integral of the function itself. Consider a partitioning of the interval
at each
for some
Then, the lower and upper sums corresponding to the intervals of the partition from
to
are the same, since the function is constant valued on these intervals. Moreover, as
the value of the upper and lower sums both approach
.
Thus we can express the value of the integral as the sum of the series
but we recognize this sum as just a constant multiple of the alternating harmonic series. Hence, the integral converges.
Now we examine the integral of the absolute value of the function. We argue similarly to the above, again partitioning the function at
as defined above. The difference is that now, as we let
the upper and lower sums both go to
and so the integral does not exist, as this is the harmonic series, which does not converge.
In the above proof of divergence the important point is that the lower sums diverge. The fact that the upper sums diverge is an immediate consequence of this.
So, we have demonstrated a function whose integral converges, but does not converge absolutely as desired.
We begin by showing (
converges if
converges.
So, we assume to start that
converges. Now consider the partition
. Since
decreases monotonically it must be that
and similarly that
. Thus, the integral which we are trying to evaluate is bounded above by
and below by
.
Now we observe that
may be written as a sum over the domain as
We know moreover that each of these integrals exist, by Theorem 6.9. Also, since
is always positive each such integral must be positive. Therefore, the integral may be expressed as a sum of a nonnegative series which is bounded above. Hence, by Theorem 3.24 the integral exists.
Now we prove (
) that if
converges then
converges.
So assume now that
converges. Then we can prove that the summation
satisfies the Cauchy criterion. We established above
is bounded above by
and below by
. This implies that given a sum
it is bounded above by the integral
. Moreover, since the integral
exists and
is nonnegative we know that it has the property given
such that
. For otherwise the integral would not exist and instead tend to infinity.
So now we can apply the Cauchy criterion for series. Since an upper bound of the series has the property that given
such that
. So must the series itself have this property.
Thus, the sum converges as desired.
We will prove that If
and
then
and that equality holds if and only if
\begin{proof}
We begin by proving the special case of equality
Assume that
.
(Similarly we can show that
.)
Thus,
and we see moreover that
since in this case we have
Also, if it is not the case that
then it is easy to see that
as for a sum of quotients by
and
to not contain
,
we must have the numerators equal.
Now we show that as we vary
we must always have
. For, compute the derivative of
with respect to
, and the derivative of
with respect to
. We get
and
respectively. If we have
then these are equal as demonstrated above (we showed that
in that case). In the case that
is larger than this value then
and in the case that
is less than this value then
.
This argument can be repeated in an analogous manner for variations in
, and given any
and
we can find values for which
.
Thus, we observe that
as desired\end{proof}
If
,
,
,
, and
then
\begin{proof}
If
and
then
and
are in
by Theorem 6.11. Also, we have
so we get
as desired.\end{proof}
We prove H\"older's inequality
\begin{proof}
If
and
are complex valued then we get
If
and
then applying the previous part to the functions
and
where
and
gives what we wanted to show.
However, if one of the above is zero (say without loss of generality
then we just have
for
. Taking the limit
we observe that the inequality is still true.
\end{proof}
\begin{enumerate}
We take the expression
and express it as a sum of integrals on the intervals
to get
but since each such interval
is the same, we just write
(1)
Now we exploit the Fundamental Theorem of Calculus, computing
So, the summation in Equation 1 can, more explicitly be written as
However, grouping common denominators, we observe that the sum partially telescopes to yield more simply
Having now proved Part a it suffices to show that
By the Fundamental Theorem of Calculus we have
So
\begin{eqnarray}
\int_1^\infty \frac{x}{x^{s+1}} dx&=&\frac{1}{s-1}\\
\Rightarrow s \int_1^\infty \frac{x}{x^{s+1}} dx&=&\frac{s}{s-1}\\
\Rightarrow s \int_1^\infty \left( \frac{x-[x]}{x^{s+1}} + \frac{[x]}{x^{s+1}} \right) dx&=&\frac{s}{s-1}\\
\Rightarrow s \int_1^\infty \left( \frac{x-[x]}{x^{s+1}} + \frac{[x]}{x^{s+1}} \right) dx&=&\frac{s}{s-1}\\
\Rightarrow s \int_1^\infty \frac{[x]}{x^{s+1}} dx &=&\frac{s}{s-1} - s \int_1^\infty \frac{x-[x]}{x^{s+1}} dx\\
\end{eqnarray*}
as desired\
end part b
It remains now to show that the integral in Part \ref{2} converges.
Since for
we know that
converges if and only if
converges.
However,
converges by the integral test (Problem 8) since we have already shown that the sequence
is convergent for