Spivak’s Calculus On Manifolds: Solutions
                        Manual
                       Thomas Hughes
                        August 2017
                                 Chapter 1
                                 Functions on Euclidean
                                 Space
                                                         P
                                   1.1 Prove that |x| ≤     n  |x |
                                                            i=1  i
                                       Proof. If {e ,e ,...,e } is the usual basis on Rn, then we can write
                                                   1   2      n
                                                              x=x e +x e +...+x e
                                                                     1 1    2 2           n n
                                       and thus
                                                          n           n           n             n
                                                                 
                                                          X           X            X             X
                                                   |x| =     x e  ≤     |x e | =     |x ||e | =    |x |
                                                               i i        i i          i  i          i
                                                         i=1         i=1         i=1           i=1
                                   1.2 When does equality hold in Theorem 1-1(3)?
                                       Proof. Notice in the proof that we get
                                                         n          n            n
                                                   2    X 2 X 2 X                             2      2
                                            |x +y| =       (x ) +      (y ) +2      x y ≤|x| +|y| +2|x||y|
                                                             i           i            i i
                                                        i=1        i=1          i=1
                                                                                   P              P
                                       and so we have equality precisely when         n  x y = |     n  x y | and x
                                                                                      i=1 i i        i=1 i i
                                       and y are dependent. That is, when x and y are dependent and sgn(xi) =
                                       sgn(y ) for all i.  That is, when one is a non-negative multiple of the
                                             i
                                       other.
                                   1.3 Prove that |x−y| ≤ |x|+|y|. When does equality hold?
                                       Proof.
                                                      |x −y| = |x+(−y)| ≤ |x|+|−y| = |x|+|y|
                                       Conditions for equality are the same as in 1.2, for x and −y.
                                                                          1
                                     1.4 Prove that ||x| − |y|| ≤ |x − y|.
                                         Proof. Notice
                                                                 |x| = |x − y + y| ≤ |x − y| + |y|
                                         Thus
                                                                         |x| − |y| ≤ |x − y|
                                         Similarly,
                                                         |y| = |y − x + x| ≤ |y − x| + |x| = |x − y| + |x|
                                         Thus
                                                                        |y| − |x| ≤ |x − y|
                                                                        −|x−y|≤|x|−|y|
                                         So, combining these results yields
                                                                  −|x−y|≤|x|−|y|≤|x−y|
                                         which implies
                                                                        ||x| − |y|| ≤ |x − y|
                                         as desired.
                                     1.5 The quantity |y − x| is called the distance between x and y. Prove and
                                         interpret geometrically the “triangle inequality”:
                                                                    |z −x| ≤ |z −y|+|y −x|
                                         Proof.
                                                          |z −x| = |z −y +y −x| ≤ |z −y|+|y −x|
                                         Geometrically, we have
                                                                              y
                                                                                    |
                                                                                    y −x
                                                                           y|             |
                                                                         z −                  x
                                                                        |
                                                                                      |
                                                                                |z −x
                                                                     z
                                     1.6 Let f and g be integrable on [a,b].
                                                                             1          1
                                                             Rb            Rb     2    Rb     2
                                           (a) Prove that      f · g ≤      f2    ·     g2    .
                                                             a            a           a
                                                                               2
                                          (b) If equality holds, must f = λg for some λ ∈ R? What if f and g are
                                              continuous?
                                          (c) Show that Theorem 1-1(2) is a special case of (a).
                                         Proof.   (a) One way to prove this would be to observe that 1-1(2) implies
                                              that
                                                                                           !1                     !1
                                                 n                         n                 2    n                 2
                                                Xf(t )g(t )∆x  ≤         X(f(t ))2∆x             X(g(t ))2∆x
                                                        k    k     k             k       k              k       k
                                                                                                 k=1
                                                 k=1                       k=1
                                              Notice, by integrability, all the sums can be considered functions of
                                                                      ˙                                   Rb        Rb 2
                                              the tagged partition P, such that each will approach a f · g, a f
                                                    Rb                            
                                                        2                         ˙ 
                                              and      g , respectively, when      P → 0. Thus, by continuity of
                                                     a                           
                                              the square root function and the absolute value, taking the limit as
                                                
                                               ˙ 
                                                P →0willgive us the desired result.
                                                
                                              However, following Spivak’s hint, we observe that either there exists
                                              λ∈Rsuchthat                      Z b
                                                                                            2
                                                                           0 = a (f −λg)
                                              or, since (f − λg)2 is nonnegative, for all λ ∈ R
                                                                               Z b          2
                                                                           0 < a (f −λg)
                                              Notice from Lebesgue’s theorem of Riemann-integrability that in the
                                              first case we must have that f = λg almost everywhere on [a,b], and
                                              therefore
                                                                         Z bf · g = λZ bg2
                                                                           a             a
                                              and
                                                                           Z bf2 = Z bλ2g2
                                                                             a         a
                                              Which would give us that
                                                                     v             !      s                    s
                                               Z               Z         u Z             2      Z         Z           Z      Z
                                                 b            b      u         b               b        b           b      b
                                                                 2   t 2          2               2 2      2           2      2
                                                   f · g = λ       g   = λ           g     =        λ g       g =         f      g
                                                                  
                                               a             a                a               a        a           a      a
                                              which can be rewritten as
                                                                                     !1          !1
                                                                  Z             Z        2   Z        2
                                                                   b             b           b
                                                                     f · g =      f2           g2
                                                                         
                                                                  a             a           a
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