ALL 7 CHAPTERS COVERED
, 1
Chapter 1
1.1 (a) 0 x =0 x + [0 x + ( 0x)] = (0 + 0) x + ( 0 x) =0 x + ( 0:x) = 0:
(b) a0 = a0 + [a0+( a0)] = a(0 + 0) + ( a0) =a0 + ( a0) = 0:
(c) ( 1) x + x =( 1 + 1) x =0 x = 0:
1 1
(d) If a 6= 0 then a x = 0 implies x =a (a x) = a 0 = 0:
1.2 (a) Complex vector space, (b) real vector space, (c) not a vector space, (d)
real vector space.
1.3 If x1 ; :::; xn are linearly dependent then there are scalars b1 ; :::; bn ; not all
Pn
zeros, such that i=1 bi xi = 0: Assuming bk 6= 0 for some k 2 f1; :::; ng,
we can multiply by bk 1 to obtain xk = i6=k ai xi , where ai = bk 1 bi : The
P
converse is obvious. If the set of vectors x1 ; x2 ; x3 ; ::: is in nite , then it
is linearly dependent if, and only if, it has a nite subset which is linearly
dependent, and the desired conclusion follows.
1.4 Assume that fx1 ; ; xn g and fy1 ; ; ym g are bases of the same vector
space with n 6= m and show that this leads to a contradiction. If m > n;
express each yi , 0 i n, as a linear combination of x1 ; ; xn : The
resulting system of n linear equations can be solved uniquely for each xi ,
0 i n; as a linear combination of yi , 0 i n (why?). Since each
vector yn+1 ; :::; ym is also a linear combination of x1 ; ; xn (and hence
of y1 ; ; yn ), this contradicts the linear independence of fy1 ; ; ym g:
Similarly, If m < n then fx1 ; :::; xn g is linearly dependent. Hence m = n:
1.5 Assume an xn + ::: + a1 x + a0 = 0 for all x in the interval I. We can
di¤erentiate both sides of this identity n times to conclude that an = 0,
then n 1 times to obtain an 1 = 0, etc. Therefore all the coe¢cients ak
are zeros, and so f1; x; :::; xn : x 2 Ig is linearly independent for every n:
It follows that the in nite set f1; x; ::: : x 2 Ig is linearly independent.
1.6 It su¢ces to consider the case where both dim X and dim Y are nite.
If B is a basis of Y , then B lies in X: Since the vectors in B are linearly
independent, dim X cannot be less than the number of vectors in B, namely
dim Y:
1.7 Recall that a determinant is zero if, and only if, one of its rows (or columns)
is a linear combination of the other rows (or columns).
2 2 2
1.8 Since kx + yk = kxk + 2 Re hx; yi + kyk , the equality holds if, and only
if, Re hx; yi = 0: Consider x = (1; 1) and y = (i; i) in C2 .
, 2
1.9 (a) Let a(x + y) + b(x y) = (a + b)x + (a b)y =0: Because x and y
are linearly independent, it follows that a + b = 0 and a b = 0. But this
implies a = b = 0, hence the linear independence of x + y and x y:
2 2 2 2
(b) hx + y; x yi = kxk +hy; xi hx; yi kyk = kxk kyk . Therefore
x + y and x y are orthogonal if kxk = kyk.
p
1.10 (a) 0, (b) 2=3, (c) 8=3, (d) 14:
1.11 h'1 ; '3 i = h'1 ; '4 i = h'2 ; '4 i = h'3 ; '4 i = 0: Thus the largest orthogonal
subset is f'1 ; '3 ; '4 g:
p p
1.12 hf; f1 i = kf1 k = =2; hf; f2 i = kf2 k = 0; hf; f3 i = kf3 k = =2:
1.13 Let a + bx + cx2 = 0 for all x 2 [ 1; 1]: Setting x = 0; x = 1; and x = 1;
the resulting three equations yield the only solution a = b = c = 0: The
corresponding orthogonal functions are given by
f1 (x) = 1;
hx; 1i
f2 (x) = x 2 = x;
k1k
hx2 ; 1i hx2 ; xi 1
f3 (x) = x2 2 2 x = x2 :
k1k kxk 3
1.14 Let a + bx + c jxj = 0 for all x 2 [ 1; 1]: Setting x = 0; x = 1; and x = 1
yields a = b = c = 0: The corresponding orthogonal set is
f1 (x) = 1;
hx; 1i
f2 (x) = x 2 =x
k1k
hjxj ; 1i hjxj ; xi 1
f3 (x) = jxj 2 2 = jxj ;
k1k kxk 2
and the normalized set is
f1 (x) 1
=p ;
kf1 k 2
f2 (x) x
=p ;
kf2 k 2=3
f3 (x) 1
= p (jxj 1=2) :
kf3 k 6
The set f1; x; jxjg is not linearly independent on [0; 1] because jxj = x on
[0; 1]:
, 3
1.15 From the result of Exercise 1.3 we know that f1 ; :::; fn are linearly de-
pendent if, and only if, there is a number k 2 f1; :::; ng such that
P
fk = i6=k ai fi on I. By di¤erentiating this identity up to order n 1,
(j) P (j)
we arrive at the system of equations fk = i6=k ai fi ; 0 j n 1.
Writing this system in matrix form, and using the properties of determi-
nants, we conclude that the system is equivalent to the single equation
(j)
det(fi ) = 0 on I, where 1 i n and 0 j n 1:
1.16 Noting that both '1 and '2 are even whereas '3 is odd, we conclude
that h'1 ; '3 i = h'2 ; '3 i = 0: Moreover, h'1 ; '2 i = 0: The corresponding
orthonormal set is
'1 (x) 1
=p ;
k'1 k 2
r
'2 (x) 45 1
= x2 ;
k'2 k 8 3
'3 (x) 1
= p '3 (x):
k'3 k 2
1.17 Solving the pair of equations hx2 +ax+b; x+1i = 0 and hx2 +ax+b; x 1i =
0 gives a = 1; b = 1=6:
1.18 If f R: [a; b] ! C is a continuous function and kf k = 0, then it follows
b 2 2
that a jf (x)j dx = 0 and jf (x)j is continuous and nonnegative on [a; b]:
But this implies f (x) = 0 for all x 2 [a; b]: On the other hand, the non-
continuous function
0; x 2 [0; 1]nf1=2g
f (x) =
1; x = 1=2
clearly satis es kf k = 0; but f is not identically 0 on [0; 1]:
2 2 2 2
1.19 From the CBS inequality, kf + gk = kf k + 2 Rehf; gi + kgk kf k +
2
2 kf k kgk + kgk = (kf k + kgk)2 : The triangle inequality follows by taking
the square root of each side.
p
1.20 h1; xi = 1=2; k1k = 1; and kxk = 1= 3: Clearly h1; xi < k1k kxk :
1.21 Use the de nition of the Riemann integral, based on Riemann sums, to
show that f; and hence f g; is not integrable on [0; 1], whereas f 2 = 1 and
g 2 = 1 are both integrable.
p
1.22 (i) 1= 2, (ii) not in L2 (0; 1), (iii) 1, (iv) not in L2 (0; 1):
, 4
1.23 If kf k = 0 then f , being continuous, is identically 0 and the pair f , g is
linearly dependent. The same is true if kgk = 0. Hence we assume kf k = 6 0
and kgk =6 0: Now
2 Z b 2 Z b 2 Z b
f g f g fg
= 2 + 2 2
kf k kgk a kf k a kgk a kf k kgk
=1+1 2 = 0;
where we used hf; gi = kf k kgk in the second equality. The implies g = f
with = kgk = kf k :
2
Conversely, if g = f for some positive number , then hf; gi = kf k =
kf k kgk :
2 2 2
1.24 In general kf + gk = kf k + kgk + 2 Rehf; gi: If kf + gk = kf k + kgk
then we must have Rehf; gi = kf k kgk ; and if, furthermore, the functions
f and g are positive and continuous, then Rehf; gi = hf; gi = kf k kgk and
(by Exercise 1.23) f and g are linearly dependent.
Conversely, if the functions are linearly dependent then g = f for some
number , and kf + gk = k(1 + )f k = j1 + j kf k : For the equality
kf + gk = kf k + kgk to hold we must therefore have j1 + j = 1 + j j ;
which implies 0:
2 R1
1.25 The norm kx k = 0 x2 dx is nite if, and only if, 2 > 1; that is,
> 1=2:
1.26 < 1=2:
1.27 Suppose limx!1 f (x) = ` 6= 0, then limx!1 jf (x)j = j`j > 0 and there is
a positive integer n such that, for all x n;
jjf j j`jj < j`j =2
0 < j`j =2 < jf (x)j < 3 j`j =2:
R1 2
But this implies n jf (x)j dx = 1; which contradicts the integrability of
2
jf j on (0; 1):
Rb p
1.28 a jf (x)j dx = hjf (x)j ; 1i kf k k1k = b a kf k by the CBS inequality.
p
f (x) = 1= x is integrable on (0; 1) but f 2 (x) = 1=x is not.
1.29 Suppose jf (x)j M for all x 0: Then
Z 1 Z 1
f 2 (x)dx M jf (x)j dx < 1:
0 0
1
The function f (x) = (1 + x) is bounded on [0; 1), lies in L2 (0; 1), but
is not integrable on [0; 1):
, 5
1.30 Using familiar trigonometric identities,
sin3 x = sin x(1 cos2 x)
1
= sin x cos x sin 2x
2
1
= sin x (sin 3x + sin x)
4
3 1
= sin x sin 3x:
4 4
1.31 There are many answers to this exercise. Any odd function, such as f (x) =
ax; where a is a (non-zero) constant satis es hf; x2 + 1i = 0 since 2
px + 1 is
even. For this pchoice of f , the constant
p a has to satisfy kaxk = jaj 2=3 = 2,
that is, jaj = 6: Thus f (x) = 6x is one possible answer.
R 1 1=2
1.32 The L2 (0; 1) norm of a polynomial p is given by R0 p2 (x) e x dx : It
1 x
is therefore su¢cient to show that the integral I = 0 q(x)e dx is nite
for any polynomial q: This follows from integrating R 1 by parts, noting that
x 0 x
q(x)e
R1 0 ! 0 as x ! 1; to obtain
R 1 00 I = q(0) + 0
q (x)e dx. Similarly,
x 0 x
0
q (x)e dx = q (0) + 0
q (x)e dx: If q has degree n, we can use in-
duction to show
R1 x that, in the n-th step, the integral is reduced to a constant
multiple of 0 e dx = 1.
1.33 Using the monotonic property of the integral,
Z b Z b
2 2 2 2
kf k = jf (x)j (x)dx jf (x)j (x)dx = kf k :
a a
Therefore, if f 2 L2 (a; b) then f 2 L2 (a; b):
1.34 (a) The limit is the discontinuous function
8
> 1; jxj > 1
xn
>
0; jxj < 1
<
lim =
n!1 1 + xn > 1=2; x=1
>
:
unde ned, x = 1:
p
0; x=0
(b) limn!1 n
x=
1; x > 0:
(c) limn!1 sin nx does not exist, except when x is an integral multiple of
:
1.35 (a) Pointwise (not uniform), by Theorem 1.17(i), since the limit (Exercise
1.34(a)) is discontinuous.
, 6
(b) Uniform, since x1=n ! 1 for all x 2 [1=2; 1] and x1=n 1 1
(1=2)1=n ! 0:
(c) Pointwise (not uniform), since the limit is discontinuous at x = 0 (see
Exercise 1.34(b)).
0; x=0
1.36 fn is continuous for every n; whereas limn!1 fn (x) = is
1; 0 < x 1:
discontinuous, hence the convergence fn ! f is not uniform. Clearly,
Z 1 Z 1
lim fn (x)dx = 1 = lim fn (x)dx:
0 0
1.37 At x = 0, fn (0) = 0 for every n: When x > 0, fn (x) = n(1 x)=(n 1) ! 1:
Therefore the sequence fn converges pointwise to
0; x=0
f (x) =
1 x; 0 < x 1:
Since f is not continuous the convergence is not uniform.
1.38 fn (0) = fn (1) = 0 and, for all x 2 (0; 1); jfn (x)j = nx(1 x2 )n n(1
x2 )n ! 0 as n ! 1: Hence limn!1 fn (x) = 0 on [0; 1]: Since
Z 1
n 1
lim fn (x)dx = lim = ;
n!1 0 n!1 2n + 2 2
R1
whereas 0 limn!1 fn (x)dx = 0; the convergence fn ! 0 is not uniform
(by Theorem 1.17(ii)).
x a x u
1.39 0 ! 0, hence ! 0 on [0; a]:
n+x n n+x
x
For x 0, we also have limn!1 = 0 pointwise. In this case, assuming
n+x
x
0 < " < 1; the inequality < " cannot be satis ed when x n"=(1 "),
n+x
hence the convergence is not uniform.
Another approach: Since the statement jfn (x) f (x)j " for all x 2 I is
equivalent to the requirement that supx2I jfn (x) f (x)j ", we see that
u
fn ! f on I if, and only if, supx2I jfn (x) f (x)j ! 0 as n ! 1: When
x 2 [0; a]; sup fn (x) = a=n ! 0; but when x 2 [0; 1) we have fn (n) = 1=2,
hence sup fn (x) 1=2 9 0:
1.40 The sequence fn is de ned by
1=n; jxj n
fn (x) =
0; jxj > n:
, 7
Since 0 R fn (x) 1=n ! 0 for all x 2 R; the convergence fn ! 0 is
1
uniform. 1 fn (x)dx = 2; therefore
Z 1 Z 1
lim fn (x)dx 6= lim fn (x)dx;
1 1
the reason being that the domain of de nition of fn is not bounded.
u
1.41 Suppose fn ! f on [a; b]. Given any " > 0; it then follows that there is an
integer N1 such that
n N1 ) jfn (x) f (x)j < " for all x 2 [a; b];
u
which implies jfn (x) f (x)j ! 0: We can also nd an integer N2 such that
n N2 ) jfn (x) f (x)j < 1 for all x 2 [a; b]:
If N = maxfN1 ; N2 g, then
2
n N ) jfn (x) f (x)j < " for all x 2 [a; b];
2 u
which implies jfn (x) f (x)j ! 0:
1.42 (a) jfn (x)j 1=n2 for all x 2 R. Since 1=n2 converges,
P P
fn (x) con-
verges uniformly on R by the Weierstrasse M-test.
n
(b) If x 2 ( 1; 1) then there is an integer N such that jxj < 1=2 for all
n N; and hence
n
xn jxj n
n
= 2 jxj for all n N;
1+x 1 1=2
from which we conclude that the series converges by comparison with the
geometric series. It diverges on ( 1; 1] [ [1; 1) where fn (x) 9 0 (see
Exercise 1.34(a)).
P P
1.43 jan sin nxj jan j for all x 2 R: Since jan j converges, an sin nx con-
verges uniformly.
1.44
(n+1) Z (n+1)
jsin xj 1
Z
dx jsin xj dx
n x n n
Z
1
= sin xdx
n 0
2
= ! 0 as n ! 0:
n
, 8
R (n+1) 1
Let An = n x jsin xj dx: Because x 1 jsin xj > (x+) 1 jsin(x + )j
for every x > 0, we see that An An+1 for all n and An ! 0. Moreover,
(n+1) k=n
X Z (k+1)
sin x jsin xj
Z
dx = ( 1)k dx
0 x k x
k=0
k=n
X
= ( 1)k Ak :
k=0
R1 P1
Hence 0 x 1 sin xdx = k=0 ( 1)k Ak ; which converges by the alternat-
ing series test.
On the other hand,
Z (n+1) Z
jsin xj 1 2
dx sin xdx =
n x (n + 1) 0 (n + 1)
Z 1 1 (n+1) 1
jsin xj jsin xj 2
X Z X
) dx = dx = 1:
0 x n=0 n
x (n + 1)
k=0
1.45 Let r = R ": Then jan xn j jan j rn for all x 2 [ r; r]: Since 0 < r < R, the
numerical series jan j rn is convergent. By the M-test, with Mn = jan j rn ,
P
an xn is uniformly convergent in [ r; r]:
P
the series
1.46 Being uniformly convergent on [R "; R+"], the series an xn represents a
P
continuous function f on [R "; R+"]: Since this is true for every " > 0; the
series is continuous on ( R; R): Each term fn (x) = an xn is di¤erentiable,
P1
fn0 (x) = nan xn 1 ; and the series n=1 nan xn 1 has the same radius of
convergence R as the original series (by the root test). Now the power series
P1 n 1
n=1 nan x ; by the result of Exercise 1.45, is uniformly convergent on
[R "; R + "]. Therefore Theorem 1.17(iii) applies and we conclude that
P1 0 P1
f 0 (x) = ( n=0 an xn ) = n=1 nan xn 1 :
P1
1.47 With bn = (n + 1)an+1 we can write f 0 (x) = n=0 bn xn and repeat the
P1
argument in Exercise 1.46 to conclude that f 00 (x) = n=1 nbn x
n 1
=
P1 n 2
n=2 n(n 1)a n x ; and so on to any order of di¤erentiation. By induc-
tion we clearly have
1
X n! (k + 1)! (k + 2)!
f (k) (x) = an xn k
= k!ak + ak+1 x+ ak+2 x2 + :
(n k)! 1! 2!
n=k
At x = 0 this yields ak = f (k) (0)=k!; k 2 N:
, 9
1.48 Let the function f (x) be continuous and di¤erentiable up to any order on
R: According to Taylors theorem we can represent such a function at any
x 2 R by
f 0 (0) f 00 (0) 2 f (n) (0) n f (n+1) (c) n+1
f (x) = f (0) + x+ x + + x + x ;
1! 2! n! (n + 1)!
where c lies between 0 and x: If f (x) = ex ; then f (n) (x) = ex and f (n) (0) =
1 for all n. Thus we obtain
x x2 xn xn+1 c
ex = 1 + + + + + e :
1! 2! n! (n + 1)!
Since, for any x 2 R; xn+1 =(n + 1)! ! 0 as n ! 1, we arrive at the desired
power series representation of ex by taking the limit of the right-hand side
as n ! 1:
If f (x) = cos x, then f (n) (0) = 0 if n is odd and f (n) (0) = ( 1)n=2 if n is
even. The remainder term is bounded by xn+1 =(n + 1)! which tends to 0 as
P1
n ! 1; and we obtain cos x = n=0 ( 1)n x2n =(2n)!. Similarly we arrive
at the given representation for sin x:
1.49 Eulers formula is obtained by replacing x by ix in the power series which
represent ex ; cos x; and sin x; and using the equations i2n = ( 1)n and
i2n+1 = ( 1)n i.
1.50 (a) 1:
(b) 1.
(c) fn (0) = fn (1) = 0: For every x 2 (0; 1); fn (x) = nx(1 x)n ! 0 as
n ! 1: Therefore fn (x) ! 0 pointwise.
Z 1
2
kfn 0k = n2 x2 (1 x)2n dx
0
2 1
2n
Z
= x(1 x)2n+1 dx
2n + 1 0
Z 1
n2
= (1 x)2n+2 dx
(2n + 1)(n + 1) 0
n2
= ! 0 as n ! 1:
(2n + 1)(n + 1)(2n + 3)
L2
Hence fn ! 0:
P 2=3 2 2 P 4=3
P 4=3
1.51 (a) convergent, since k sin kx = ksin kxk k = k <
1.
, 1
Chapter 1
1.1 (a) 0 x =0 x + [0 x + ( 0x)] = (0 + 0) x + ( 0 x) =0 x + ( 0:x) = 0:
(b) a0 = a0 + [a0+( a0)] = a(0 + 0) + ( a0) =a0 + ( a0) = 0:
(c) ( 1) x + x =( 1 + 1) x =0 x = 0:
1 1
(d) If a 6= 0 then a x = 0 implies x =a (a x) = a 0 = 0:
1.2 (a) Complex vector space, (b) real vector space, (c) not a vector space, (d)
real vector space.
1.3 If x1 ; :::; xn are linearly dependent then there are scalars b1 ; :::; bn ; not all
Pn
zeros, such that i=1 bi xi = 0: Assuming bk 6= 0 for some k 2 f1; :::; ng,
we can multiply by bk 1 to obtain xk = i6=k ai xi , where ai = bk 1 bi : The
P
converse is obvious. If the set of vectors x1 ; x2 ; x3 ; ::: is in nite , then it
is linearly dependent if, and only if, it has a nite subset which is linearly
dependent, and the desired conclusion follows.
1.4 Assume that fx1 ; ; xn g and fy1 ; ; ym g are bases of the same vector
space with n 6= m and show that this leads to a contradiction. If m > n;
express each yi , 0 i n, as a linear combination of x1 ; ; xn : The
resulting system of n linear equations can be solved uniquely for each xi ,
0 i n; as a linear combination of yi , 0 i n (why?). Since each
vector yn+1 ; :::; ym is also a linear combination of x1 ; ; xn (and hence
of y1 ; ; yn ), this contradicts the linear independence of fy1 ; ; ym g:
Similarly, If m < n then fx1 ; :::; xn g is linearly dependent. Hence m = n:
1.5 Assume an xn + ::: + a1 x + a0 = 0 for all x in the interval I. We can
di¤erentiate both sides of this identity n times to conclude that an = 0,
then n 1 times to obtain an 1 = 0, etc. Therefore all the coe¢cients ak
are zeros, and so f1; x; :::; xn : x 2 Ig is linearly independent for every n:
It follows that the in nite set f1; x; ::: : x 2 Ig is linearly independent.
1.6 It su¢ces to consider the case where both dim X and dim Y are nite.
If B is a basis of Y , then B lies in X: Since the vectors in B are linearly
independent, dim X cannot be less than the number of vectors in B, namely
dim Y:
1.7 Recall that a determinant is zero if, and only if, one of its rows (or columns)
is a linear combination of the other rows (or columns).
2 2 2
1.8 Since kx + yk = kxk + 2 Re hx; yi + kyk , the equality holds if, and only
if, Re hx; yi = 0: Consider x = (1; 1) and y = (i; i) in C2 .
, 2
1.9 (a) Let a(x + y) + b(x y) = (a + b)x + (a b)y =0: Because x and y
are linearly independent, it follows that a + b = 0 and a b = 0. But this
implies a = b = 0, hence the linear independence of x + y and x y:
2 2 2 2
(b) hx + y; x yi = kxk +hy; xi hx; yi kyk = kxk kyk . Therefore
x + y and x y are orthogonal if kxk = kyk.
p
1.10 (a) 0, (b) 2=3, (c) 8=3, (d) 14:
1.11 h'1 ; '3 i = h'1 ; '4 i = h'2 ; '4 i = h'3 ; '4 i = 0: Thus the largest orthogonal
subset is f'1 ; '3 ; '4 g:
p p
1.12 hf; f1 i = kf1 k = =2; hf; f2 i = kf2 k = 0; hf; f3 i = kf3 k = =2:
1.13 Let a + bx + cx2 = 0 for all x 2 [ 1; 1]: Setting x = 0; x = 1; and x = 1;
the resulting three equations yield the only solution a = b = c = 0: The
corresponding orthogonal functions are given by
f1 (x) = 1;
hx; 1i
f2 (x) = x 2 = x;
k1k
hx2 ; 1i hx2 ; xi 1
f3 (x) = x2 2 2 x = x2 :
k1k kxk 3
1.14 Let a + bx + c jxj = 0 for all x 2 [ 1; 1]: Setting x = 0; x = 1; and x = 1
yields a = b = c = 0: The corresponding orthogonal set is
f1 (x) = 1;
hx; 1i
f2 (x) = x 2 =x
k1k
hjxj ; 1i hjxj ; xi 1
f3 (x) = jxj 2 2 = jxj ;
k1k kxk 2
and the normalized set is
f1 (x) 1
=p ;
kf1 k 2
f2 (x) x
=p ;
kf2 k 2=3
f3 (x) 1
= p (jxj 1=2) :
kf3 k 6
The set f1; x; jxjg is not linearly independent on [0; 1] because jxj = x on
[0; 1]:
, 3
1.15 From the result of Exercise 1.3 we know that f1 ; :::; fn are linearly de-
pendent if, and only if, there is a number k 2 f1; :::; ng such that
P
fk = i6=k ai fi on I. By di¤erentiating this identity up to order n 1,
(j) P (j)
we arrive at the system of equations fk = i6=k ai fi ; 0 j n 1.
Writing this system in matrix form, and using the properties of determi-
nants, we conclude that the system is equivalent to the single equation
(j)
det(fi ) = 0 on I, where 1 i n and 0 j n 1:
1.16 Noting that both '1 and '2 are even whereas '3 is odd, we conclude
that h'1 ; '3 i = h'2 ; '3 i = 0: Moreover, h'1 ; '2 i = 0: The corresponding
orthonormal set is
'1 (x) 1
=p ;
k'1 k 2
r
'2 (x) 45 1
= x2 ;
k'2 k 8 3
'3 (x) 1
= p '3 (x):
k'3 k 2
1.17 Solving the pair of equations hx2 +ax+b; x+1i = 0 and hx2 +ax+b; x 1i =
0 gives a = 1; b = 1=6:
1.18 If f R: [a; b] ! C is a continuous function and kf k = 0, then it follows
b 2 2
that a jf (x)j dx = 0 and jf (x)j is continuous and nonnegative on [a; b]:
But this implies f (x) = 0 for all x 2 [a; b]: On the other hand, the non-
continuous function
0; x 2 [0; 1]nf1=2g
f (x) =
1; x = 1=2
clearly satis es kf k = 0; but f is not identically 0 on [0; 1]:
2 2 2 2
1.19 From the CBS inequality, kf + gk = kf k + 2 Rehf; gi + kgk kf k +
2
2 kf k kgk + kgk = (kf k + kgk)2 : The triangle inequality follows by taking
the square root of each side.
p
1.20 h1; xi = 1=2; k1k = 1; and kxk = 1= 3: Clearly h1; xi < k1k kxk :
1.21 Use the de nition of the Riemann integral, based on Riemann sums, to
show that f; and hence f g; is not integrable on [0; 1], whereas f 2 = 1 and
g 2 = 1 are both integrable.
p
1.22 (i) 1= 2, (ii) not in L2 (0; 1), (iii) 1, (iv) not in L2 (0; 1):
, 4
1.23 If kf k = 0 then f , being continuous, is identically 0 and the pair f , g is
linearly dependent. The same is true if kgk = 0. Hence we assume kf k = 6 0
and kgk =6 0: Now
2 Z b 2 Z b 2 Z b
f g f g fg
= 2 + 2 2
kf k kgk a kf k a kgk a kf k kgk
=1+1 2 = 0;
where we used hf; gi = kf k kgk in the second equality. The implies g = f
with = kgk = kf k :
2
Conversely, if g = f for some positive number , then hf; gi = kf k =
kf k kgk :
2 2 2
1.24 In general kf + gk = kf k + kgk + 2 Rehf; gi: If kf + gk = kf k + kgk
then we must have Rehf; gi = kf k kgk ; and if, furthermore, the functions
f and g are positive and continuous, then Rehf; gi = hf; gi = kf k kgk and
(by Exercise 1.23) f and g are linearly dependent.
Conversely, if the functions are linearly dependent then g = f for some
number , and kf + gk = k(1 + )f k = j1 + j kf k : For the equality
kf + gk = kf k + kgk to hold we must therefore have j1 + j = 1 + j j ;
which implies 0:
2 R1
1.25 The norm kx k = 0 x2 dx is nite if, and only if, 2 > 1; that is,
> 1=2:
1.26 < 1=2:
1.27 Suppose limx!1 f (x) = ` 6= 0, then limx!1 jf (x)j = j`j > 0 and there is
a positive integer n such that, for all x n;
jjf j j`jj < j`j =2
0 < j`j =2 < jf (x)j < 3 j`j =2:
R1 2
But this implies n jf (x)j dx = 1; which contradicts the integrability of
2
jf j on (0; 1):
Rb p
1.28 a jf (x)j dx = hjf (x)j ; 1i kf k k1k = b a kf k by the CBS inequality.
p
f (x) = 1= x is integrable on (0; 1) but f 2 (x) = 1=x is not.
1.29 Suppose jf (x)j M for all x 0: Then
Z 1 Z 1
f 2 (x)dx M jf (x)j dx < 1:
0 0
1
The function f (x) = (1 + x) is bounded on [0; 1), lies in L2 (0; 1), but
is not integrable on [0; 1):
, 5
1.30 Using familiar trigonometric identities,
sin3 x = sin x(1 cos2 x)
1
= sin x cos x sin 2x
2
1
= sin x (sin 3x + sin x)
4
3 1
= sin x sin 3x:
4 4
1.31 There are many answers to this exercise. Any odd function, such as f (x) =
ax; where a is a (non-zero) constant satis es hf; x2 + 1i = 0 since 2
px + 1 is
even. For this pchoice of f , the constant
p a has to satisfy kaxk = jaj 2=3 = 2,
that is, jaj = 6: Thus f (x) = 6x is one possible answer.
R 1 1=2
1.32 The L2 (0; 1) norm of a polynomial p is given by R0 p2 (x) e x dx : It
1 x
is therefore su¢cient to show that the integral I = 0 q(x)e dx is nite
for any polynomial q: This follows from integrating R 1 by parts, noting that
x 0 x
q(x)e
R1 0 ! 0 as x ! 1; to obtain
R 1 00 I = q(0) + 0
q (x)e dx. Similarly,
x 0 x
0
q (x)e dx = q (0) + 0
q (x)e dx: If q has degree n, we can use in-
duction to show
R1 x that, in the n-th step, the integral is reduced to a constant
multiple of 0 e dx = 1.
1.33 Using the monotonic property of the integral,
Z b Z b
2 2 2 2
kf k = jf (x)j (x)dx jf (x)j (x)dx = kf k :
a a
Therefore, if f 2 L2 (a; b) then f 2 L2 (a; b):
1.34 (a) The limit is the discontinuous function
8
> 1; jxj > 1
xn
>
0; jxj < 1
<
lim =
n!1 1 + xn > 1=2; x=1
>
:
unde ned, x = 1:
p
0; x=0
(b) limn!1 n
x=
1; x > 0:
(c) limn!1 sin nx does not exist, except when x is an integral multiple of
:
1.35 (a) Pointwise (not uniform), by Theorem 1.17(i), since the limit (Exercise
1.34(a)) is discontinuous.
, 6
(b) Uniform, since x1=n ! 1 for all x 2 [1=2; 1] and x1=n 1 1
(1=2)1=n ! 0:
(c) Pointwise (not uniform), since the limit is discontinuous at x = 0 (see
Exercise 1.34(b)).
0; x=0
1.36 fn is continuous for every n; whereas limn!1 fn (x) = is
1; 0 < x 1:
discontinuous, hence the convergence fn ! f is not uniform. Clearly,
Z 1 Z 1
lim fn (x)dx = 1 = lim fn (x)dx:
0 0
1.37 At x = 0, fn (0) = 0 for every n: When x > 0, fn (x) = n(1 x)=(n 1) ! 1:
Therefore the sequence fn converges pointwise to
0; x=0
f (x) =
1 x; 0 < x 1:
Since f is not continuous the convergence is not uniform.
1.38 fn (0) = fn (1) = 0 and, for all x 2 (0; 1); jfn (x)j = nx(1 x2 )n n(1
x2 )n ! 0 as n ! 1: Hence limn!1 fn (x) = 0 on [0; 1]: Since
Z 1
n 1
lim fn (x)dx = lim = ;
n!1 0 n!1 2n + 2 2
R1
whereas 0 limn!1 fn (x)dx = 0; the convergence fn ! 0 is not uniform
(by Theorem 1.17(ii)).
x a x u
1.39 0 ! 0, hence ! 0 on [0; a]:
n+x n n+x
x
For x 0, we also have limn!1 = 0 pointwise. In this case, assuming
n+x
x
0 < " < 1; the inequality < " cannot be satis ed when x n"=(1 "),
n+x
hence the convergence is not uniform.
Another approach: Since the statement jfn (x) f (x)j " for all x 2 I is
equivalent to the requirement that supx2I jfn (x) f (x)j ", we see that
u
fn ! f on I if, and only if, supx2I jfn (x) f (x)j ! 0 as n ! 1: When
x 2 [0; a]; sup fn (x) = a=n ! 0; but when x 2 [0; 1) we have fn (n) = 1=2,
hence sup fn (x) 1=2 9 0:
1.40 The sequence fn is de ned by
1=n; jxj n
fn (x) =
0; jxj > n:
, 7
Since 0 R fn (x) 1=n ! 0 for all x 2 R; the convergence fn ! 0 is
1
uniform. 1 fn (x)dx = 2; therefore
Z 1 Z 1
lim fn (x)dx 6= lim fn (x)dx;
1 1
the reason being that the domain of de nition of fn is not bounded.
u
1.41 Suppose fn ! f on [a; b]. Given any " > 0; it then follows that there is an
integer N1 such that
n N1 ) jfn (x) f (x)j < " for all x 2 [a; b];
u
which implies jfn (x) f (x)j ! 0: We can also nd an integer N2 such that
n N2 ) jfn (x) f (x)j < 1 for all x 2 [a; b]:
If N = maxfN1 ; N2 g, then
2
n N ) jfn (x) f (x)j < " for all x 2 [a; b];
2 u
which implies jfn (x) f (x)j ! 0:
1.42 (a) jfn (x)j 1=n2 for all x 2 R. Since 1=n2 converges,
P P
fn (x) con-
verges uniformly on R by the Weierstrasse M-test.
n
(b) If x 2 ( 1; 1) then there is an integer N such that jxj < 1=2 for all
n N; and hence
n
xn jxj n
n
= 2 jxj for all n N;
1+x 1 1=2
from which we conclude that the series converges by comparison with the
geometric series. It diverges on ( 1; 1] [ [1; 1) where fn (x) 9 0 (see
Exercise 1.34(a)).
P P
1.43 jan sin nxj jan j for all x 2 R: Since jan j converges, an sin nx con-
verges uniformly.
1.44
(n+1) Z (n+1)
jsin xj 1
Z
dx jsin xj dx
n x n n
Z
1
= sin xdx
n 0
2
= ! 0 as n ! 0:
n
, 8
R (n+1) 1
Let An = n x jsin xj dx: Because x 1 jsin xj > (x+) 1 jsin(x + )j
for every x > 0, we see that An An+1 for all n and An ! 0. Moreover,
(n+1) k=n
X Z (k+1)
sin x jsin xj
Z
dx = ( 1)k dx
0 x k x
k=0
k=n
X
= ( 1)k Ak :
k=0
R1 P1
Hence 0 x 1 sin xdx = k=0 ( 1)k Ak ; which converges by the alternat-
ing series test.
On the other hand,
Z (n+1) Z
jsin xj 1 2
dx sin xdx =
n x (n + 1) 0 (n + 1)
Z 1 1 (n+1) 1
jsin xj jsin xj 2
X Z X
) dx = dx = 1:
0 x n=0 n
x (n + 1)
k=0
1.45 Let r = R ": Then jan xn j jan j rn for all x 2 [ r; r]: Since 0 < r < R, the
numerical series jan j rn is convergent. By the M-test, with Mn = jan j rn ,
P
an xn is uniformly convergent in [ r; r]:
P
the series
1.46 Being uniformly convergent on [R "; R+"], the series an xn represents a
P
continuous function f on [R "; R+"]: Since this is true for every " > 0; the
series is continuous on ( R; R): Each term fn (x) = an xn is di¤erentiable,
P1
fn0 (x) = nan xn 1 ; and the series n=1 nan xn 1 has the same radius of
convergence R as the original series (by the root test). Now the power series
P1 n 1
n=1 nan x ; by the result of Exercise 1.45, is uniformly convergent on
[R "; R + "]. Therefore Theorem 1.17(iii) applies and we conclude that
P1 0 P1
f 0 (x) = ( n=0 an xn ) = n=1 nan xn 1 :
P1
1.47 With bn = (n + 1)an+1 we can write f 0 (x) = n=0 bn xn and repeat the
P1
argument in Exercise 1.46 to conclude that f 00 (x) = n=1 nbn x
n 1
=
P1 n 2
n=2 n(n 1)a n x ; and so on to any order of di¤erentiation. By induc-
tion we clearly have
1
X n! (k + 1)! (k + 2)!
f (k) (x) = an xn k
= k!ak + ak+1 x+ ak+2 x2 + :
(n k)! 1! 2!
n=k
At x = 0 this yields ak = f (k) (0)=k!; k 2 N:
, 9
1.48 Let the function f (x) be continuous and di¤erentiable up to any order on
R: According to Taylors theorem we can represent such a function at any
x 2 R by
f 0 (0) f 00 (0) 2 f (n) (0) n f (n+1) (c) n+1
f (x) = f (0) + x+ x + + x + x ;
1! 2! n! (n + 1)!
where c lies between 0 and x: If f (x) = ex ; then f (n) (x) = ex and f (n) (0) =
1 for all n. Thus we obtain
x x2 xn xn+1 c
ex = 1 + + + + + e :
1! 2! n! (n + 1)!
Since, for any x 2 R; xn+1 =(n + 1)! ! 0 as n ! 1, we arrive at the desired
power series representation of ex by taking the limit of the right-hand side
as n ! 1:
If f (x) = cos x, then f (n) (0) = 0 if n is odd and f (n) (0) = ( 1)n=2 if n is
even. The remainder term is bounded by xn+1 =(n + 1)! which tends to 0 as
P1
n ! 1; and we obtain cos x = n=0 ( 1)n x2n =(2n)!. Similarly we arrive
at the given representation for sin x:
1.49 Eulers formula is obtained by replacing x by ix in the power series which
represent ex ; cos x; and sin x; and using the equations i2n = ( 1)n and
i2n+1 = ( 1)n i.
1.50 (a) 1:
(b) 1.
(c) fn (0) = fn (1) = 0: For every x 2 (0; 1); fn (x) = nx(1 x)n ! 0 as
n ! 1: Therefore fn (x) ! 0 pointwise.
Z 1
2
kfn 0k = n2 x2 (1 x)2n dx
0
2 1
2n
Z
= x(1 x)2n+1 dx
2n + 1 0
Z 1
n2
= (1 x)2n+2 dx
(2n + 1)(n + 1) 0
n2
= ! 0 as n ! 1:
(2n + 1)(n + 1)(2n + 3)
L2
Hence fn ! 0:
P 2=3 2 2 P 4=3
P 4=3
1.51 (a) convergent, since k sin kx = ksin kxk k = k <
1.
