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


    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


      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



      (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


      If ch3f116.png , show that


      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




      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


      and conclude that


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


      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).