Exercises: Chapter 3, Section 6

  1. Use Theorem 3-14 to prove Theorem 3-13 without the assumption that ch3f1.png .

    Let ch3f2.png Then ch3f3.png is open and Theorem 3-13 applies with ch3f4.png in place of the ch3f5.png in its statement. Let ch3f6.png be a partition of unity subordinate to an admissible cover ch3f7.png of ch3f8.png . Then ch3f9.png is a partion of unity subordinate to the cover ch3f10.png . Now ch3f11.png is absolutely convergent, and so ch3f12.png also converges since the terms are identical. So, ch3f13.png . By Theorem 3-14, we know that ch3f14.png . Combining results, we get Theorem 3-13.

  2. If ch3f15.png and ch3f16.png , prove that in some open set containing ch3f17.png we can write ch3f18.png , where ch3f19.png is of the form ch3f20.png , and ch3f21.png is a linear transformation. Show that we can write ch3f22.png if and only if ch3f23.png is a diagonal matrix.

    We use the same idea as in the proof of Theorem 3-13. Let ch3f24.png be a point where ch3f25.png . Let ch3f26.png , and ch3f27.png . Then ch3f28.png . Define for ch3f29.png , ch3f30.png . Then ch3f31.png . So we can define on successively smaller open neighborhoods of ch3f32.png , inverses ch3f33.png of ch3f34.png and ch3f35.png . One then can verify that ch3f36.png . Combining results gives

    ch3f37.png

    and so ch3f38.png .

    Now, if ch3f39.png is a diagonal matrix, then replace ch3f40.png with ch3f41.png . for ch3f42.png and ch3f43.png . Then the ch3f44.png have the same form as the ch3f45.png and ch3f46.png .

    On the other hand, the converse is false. For example, consider the function ch3f47.png . Since ch3f48.png is linear, ch3f49.png ; so ch3f50.png is not a diagonal matrix.

  3. Define ch3f51.png by ch3f52.png .
    1. Show that ch3f53.png is 1-1, compute ch3f54.png , and show that ch3f55.png for all ch3f56.png . Show that ch3f57.png is the set ch3f58.png of Problem 2-23.

      Since ch3f59.png , to show that the function ch3f60.png is 1-1, it suffices to show that ch3f61.png and ch3f62.png imply ch3f63.png . Suppose ch3f64.png . Then ch3f65.png implies that ch3f66.png (or ch3f67.png ). If ch3f68.png , it follows that ch3f69.png . But then ch3f70.png and ch3f71.png has the same value, contrary to hypothesis. So, ch3f72.png is 1-1.

      One has

      ch3f73.png

      So, ch3f74.png for all ch3f75.png in the domain of ch3f76.png .

      Suppose ch3f77.png , i.e. ch3f78.png and ch3f79.png . If ch3f80.png , then ch3f81.png implies ch3f82.png and so ch3f83.png . But then ch3f84.png contrary to hypothesis. On the other hand, if ch3f85.png , then let ch3f86.png and let ch3f87.png be the angle between the positive ch3f88.png -axis and the ray from (0,0) through ch3f89.png . Then ch3f90.png .

    2. If ch3f91.png , show that ch3f92.png , where

      ch3f93.png

      ch3f94.png

      (Here ch3f95.png denotes the inverse of the function ch3f96.png .) Find P'(x,y). The function ch3f97.png is called the polar coordinate system on ch3f98.png .

      The formulas for ch3f99.png and ch3f100.png follow from the last paragraph of the solution of part (a). One has ch3f101.png . This is trivial from the formulas except in case ch3f102.png . Clearly, ch3f103.png . Further, L'H@ocirc;pital's Rule allows one to calculate ch3f104.png when ch3f105.png by checking separately for the limit from the left and the limit from the right. For example, ch3f106.png .

    3. Let ch3f107.png be the region between the circles of radii ch3f108.png and ch3f109.png and the half-lines through 0 which make angles of ch3f110.png and ch3f111.png with the ch3f112.png -axis. If ch3f113.png is integrable and ch3f114.png , show that

      ch3f115.png

      If ch3f116.png , show that

      ch3f117.png

      Assume that ch3f118.png and ch3f119.png . Apply Theorem 3-13 to the map ch3f120.png by ch3f121.png . One has ch3f122.png and ch3f123.png . So the first identity holds. The second identity is a special case of the first.

    4. If ch3f124.png , show that

      ch3f125.png

      and

      ch3f126.png

      For the first assertion, apply part (c) with ch3f127.png . Then ch3f128.png . Applying (c) gives ch3f129.png .

      The second assertion follows from Fubini's Theorem.

    5. Prove that

      ch3f130.png

      and conclude that

      ch3f131.png

      One has ch3f132.png and the integrands are everywhere positive. So

      ch3f133.png

      Since part (d) implies that ch3f134.png , the squeeze principle implies that ch3f135.png also.

      But using part (d) again, we get ch3f136.png also exists and is ch3f137.png (since the square root function is continuous).