Solutions To Mathematics Textbooks/Principles of Mathematical Analysis (3rd edition) (ISBN 0070856133)/Chapter 6

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Chapter 6[edit]



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 nonegative 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.





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