1. LET F(X)= 2X+2 / 2X^2 - X - 3. Find the x values at which f is not continous and classify those ( removeable or nonremoveable ) 2. Find the limit if it exists: lim x-->0 sin^8x(2-2cosx) / x^5 Solution F(X)= 2X+ 2 / 2X^2 - X - 3 it is clear that at x=0 the function is not defined so it is not continuoous at x=0 2. lim x-->0 sin^8x(2-2cosx) / x^5 lim x-->0 sin8 x(2-2cosx) * (x(2-2cosx))8 /(x(2-2cosx))8 x^5 =lim x-->0 (x(2-2cosx))8 / x^5=0.

1. 1. Let H be a real Hilbert space with inner product. and S,T be in H. Prove that ST = TS iff ST is
in H. 2. Let H = R^2, S be the 2 x 2 matrix with entries all one and T be the 2 x 2 matrix with
entries all one expect the entry in the first row and first column, which 2. Prove that S, T are in
H+ and S <= T, but not S^2 <= T^2. 3. The closure of a subalgebra of a real Hilbert space H is a
subalgebra of H. 4. Let T be in H, where H is a real Hilbert space. Let A be the closure of
{p(T):p is a real polynomial}. Prove that A is a subalgebra of H that contains I, the identity
operator.
Solution
Inner Product Spaces Inner product spaces (IPS) are generalizations of the three
dimensional Euclidean space, equipped with the notion of distance between points represented
by vectors and angles between vectors, made possible through the concept of an inner product.
These are broadly categorized as - real inner product spaces and complex inner product spaces
and may be defined as follows. Let K commonly denote any of the fields R, or, C. Let V be a
vector space over K. A map (. , .) : V ? V ® K, satisfying: (a) (x, x) ? 0, and (x, x) = 0, iff x = 0;
(b) (x, y) = (y,x)* (star denoting complex conjugate); and, (c) (cx+y, z) = c(x, z) + (y, z), for all
x, y, z Î V and c Î K, is called an inner-product in V. Note that (a) asserts that (x, x) is a real
number and is non-negative, even when K = C. A vector space V with an inner product on it is
called an inner product space. If K = R, V is called a real inner-product space and if K = C, V is
called a complex inner-product space. The scalar (x, y) is called the inner product of x and y.
Note that in (b) the bar denotes complex conjugation, and so when K = R, (b) simply reads as
(x,y) = (y,x). In the sequel, unless otherwise specified, an inner product refers to the field K
which could be either of the fields R or C. Taking c = 1 in (c), (x+y, z) = (x,z) + (y,z). Putting x
= 0 in it (0, z) = 0 and hence with y = 0 in (c), (cx,z) = c(x,z). The distributivity and conjugate
linearity with respect to the second argument, namely (x, y+z) = (x, y) + (x, z), and (x, cz) = c(x,
z), follow from the axiom (b) for an inner product. Examples: In the following the verification
of the inner product axioms is elementary and is left as an exercise. (1) The Euclidean space Rn
and the Unitary space Cn may be commonly defined as the inner product space Kn over the field
K, with the inner product given by (x, y) = y* x = S 1? i? n xiyi* , (x, y Î K ). This inner product
is known as the standard inner product in K . In the sequel, unless otherwise specified, K would
stand for the above standard inner product space. Note that when K = R, the complex
conjugation is redundant and (x, y) = y?x = x?y. (2) The sequence space l2(K) consisting of
infinite square summable sequences x = (x1, x2, ? , xn, ? ), i.e., S i? 1 |xn|2 < ? , with the inner
product (x, y) = S i? 1 xiyi* , (x, y Î l2(K)), is an infinite dimensional generalization of the inner
product space K . (3) The space L2(X, m ), over K, of functions f that are square integrable on a

2. domain X with respect to the measure m , i.e., ò X |f|2 dm < ? , with the inner product (f, g) = ò
X fg* dm , is a further generalization of l2(K). The following is the most improtant result about
an inner product: Cauchy's inequality. If x, y Î V, an inner product space, |(x, y)|2 ? (x, x)(y, y),
with equality iff x and y are linearly dependent. Proof: If x and y are linearly dependent, without
loss of generality we may take y = l x, in which case both sides of the inequality equal |l |2(x, x) .
If x and y are linearly independent, (by the axiom (a)) for any scalar l , 0 < (y-l x, y-l x) = (y, y) -
2Re [l (x,y)] + |l |2(x,x). Taking l = (y, x)/(x, x), this inequality reduces to 0 < (y,y) -
|(x,y)|2/(x,x), which, since (x,x) > 0 (due to (a)), is equivalent to the Cauchy's inequality, with
strict inequality, completing the proof. # The Cauchy's inequality is also termed as Cauchy-
Schwarz, or, Cauchy-Schwarz-Bunyakovsky inequality. PROBLEMS 1. Verify the following
for an inner product: (1) (0, y) = 0 = (x, 0); (2) (cx, y) = c(x, y), and (x, cy) = c* (x,y); (3) (cx,
cx) = |c|2 (x, x); (4) (S cjvj , v) = S cj(vj, v), and, (v, S cjvj) = S cj* (v, vj ); (5) (S aiui, S bjvj) =
S ai S bj* (ui, vj); (6) (x+y, x+y) = (x, x) + (y, y) + 2 Re(x,y); (7) if (x, y) = 0 for all x Î V, then y
= 0; (8) if Re (x, y) ? 0 for all x Î V, then y = 0; and, (9) if (x, y) = (x, z) for all x Î V, then y = z.
2. Show that the sum of two inner products on the same vector space V and a positive scalar
multiple of an inner product on V is an inner product on V. Is the difference of two inner
products an inner product? 3. Let (. , .) and (. , .) be two inner products on a vector space V.
Show that there exist a scalar d such that (x, y) = (x, y) - c(x, y) is an inner product on V for all c
< d, but for no c ? d. How many such scalars d are there? 4. Describe explicitly all real inner
products on R1 and on C1, and all complex inner products on C1. 5. Let (. , .) be an inner
product on a vector space V over C. Show that (. , .) continues to remain an inner product if V is
regarded as a vector space over R. 6. Verify that the standard inner product on Kn is an inner
product and that it is characterized by (ei, ej) = d ij, where ei 's are the standard unit (basis)
vectors. 7. Let {v1, v2, ? , vn} be an ordered basis of an inner product space V. Show than given
any n-scalars c1, ? , cn, there exists a unique v = S 1? i? n xjvj Î V such that (v, vj) = cj, 1 ? j ? n.
Hence derive a system of linear equations for determining xj 's. 8. Determine all linear
operators T on Kn for which (Tv, v) = 0 for all v. 9. If A Î Kn? n, prove that fA(x, y) = y* Ax is
an inner product on the vector space Kn over K iff A is positive definite, i.e., x* Ax is positive
for all non-zero x Î C n. 10. Verify for an inner product that (u+v, u+v) + (u-v, u-v) = 2(u, u) +
2(v, v). Interpret this as saying that the sum of squares of lengths of diagonals equals the sum of
squares of lengths of all the four sides of a parallelogram. Deduce Pythagoras theorem from this.
11. The matrix M of an inner product with respect to an ordered basis b is defined by: (u, v) =
[v]b * M[u]b . If K = C (R), prove that M is hermitian (real symmetric) and is unique. 12. Show
that (x, y) = x1y1 -3x2y1 -3x1y2 +10x2y2, defines an inner product on R2. What is the matrix of
this inner product relative to the standard basis in R2? What are all bases with respect to which
the matrix of this inner product becomes the identity matrix? 13. For the real inner product (f, g)

3. = ò [0,1] f(t)g(t)dt, on L2[0, 1], verify (S 0? j? n-1 ajxj, S 0? k? n-1 bkxk) = S 0? j,k? n
(ajbk)/(j+k+1). 14. Prove that the Hilbert matrix (1/(i+j-1))1? i,j? n is positive definite. 15.
Determine the inner product on R2, the quadratic form of which is given by P xP 2 = (2x1-3x2)2
+ (3x1-2x2)2. 16. If (u, v) is an inner product in Kn such that |(u, v)| ? C|v*u|, for all u, v Î Kn,
where C is a constant, prove that (u, v) = cv*u, for all u, v Î Kn, where c is a constant. 17. Which
of (a) (x, y) = x* y, (b) (x, y) = y* x, (c) (x, y) = y?x, and, (d) (x, y) = x?y, is an inner product in
Cn? Which of these are inner products in R n. 18. Let {ui}1? i? n be a basis of an inner product
space V. Show that = S 1? i? n (Sui, Tui) is an inner product in L(V). 19. Show that the
standard basis {Eij} of Kn? n= L(Kn), is orthonormal with respect to the inner product: = S 1? i?
n (Sei , Tei). Norm Induced by an Inner Product An inner product on a vector space V induces
a natural norm in V by defining P xP 2 = (x, x) , x Î V. Of the norm axioms: (a) P xP ? 0, and P
xP = 0 iff x = 0; (b) P cxP = |c|P xP ; and, (c) P x+yP ? P xP + P yP , (a)-(b) trivially follow from
the inner product axioms. For the triangle inequality (c), we have P x+yP 2 = P xP 2 + P yP 2 + 2
Re(x, y) ? P xP 2 + P yP 2 + 2|(x,y)| ? (P xP + P yP )2, as follows from the Cauchy-Schwarz
inequality. In the sequel, unless otherwise specified, the norm in an inner-product space context
would mean the induced norm defined as above. A sequence of vectors {x(n)} in a normed
(vector) space V (over K) is called Cauchy if given any e > 0, there exists an N such that P x(m)
-x(n)P < e , for all m, n > N. A sequence {x(n)} is said to converge if there exists an x Î V such
that P x(n) -xP ® 0, as n ® ? . A normed space V is said to be complete if every Cauchy
sequence in it converges. A complete inner product space is called a Hilbert space. (In the
literature, however, several authors exclude finite dimensional inner product spaces from being
called Hilbert spaces). Bolzano-Weierstrass theorem and the equivalence of any two norms in a
finite dimensional space imply that all finite dimensional vector spaces over K are complete. A
subspace W of a normed space V is called closed if W ' w(n) ® w, implies that w Î W (i.e., a
sequence in W can converge only to an element in W itself). PROBLEMS 1. Let V be an inner
product space. For what real values of q does the function d(u, v) = (u-v, u-v)q , (u, v Î V),
defines a distance in V, i.e., there hold: (1) d(u, v) ? 0; (2) d(u, v) = 0 iff u = v; (3) d(u, v) = d(v,
u) (symmetry); and, (4) d(u, w) ? d(u, v) + d(v, w), u, v, w Î V (triangle inequality). 2. Let V be
a complex inner product space. If u, v Î V are such that (u, v) is real, i.e., u has a zero axial-twist
with respect to y, show that P u+ivP 2 = P uP 2 + P vP 2 and that u + iv is orthogonal to u - iv iff
u is orthogonal to v and P uP = P vP . 3. Let V be an inner product space and G a finite
dimensional subspace of V. Let v Î VG and g Î G . Prove that P v-g0P ? P v-gP , for all g Î G ,
i.e., g0 is a best approximation of v from G, iff there exists an element u Î V such that (i) P uP =
1, (ii) (g, u) = 0, for all g Î G , and, (iii) (v, u) = P v-g0P . 4. Let V be an inner product space and
v1, v2, ? , vn Î V be linearly independent. Let G = < v1, v2, ? , vn >, the span of the vectors v1,
v2, v3, ? , vn. Show thatinf P v - gP = det ([(vi, vj)]0? i,j? n)/det ([(vi, vj)]1? i,j? n] ). 5.

4. Consider L2(0,1), the Hilbert space of square integrable real valued functions on the interval
(0,1) with the inner product (f, g) = ò [0,1] f(x)g(x)dx. Let a i > -1/2, 0 ? i ? n, and en = inf {P xa
0 - S 1? i? n cixa iP : ci Î R}. Prove that en2 = D 0/D 1 = [1/(2a 0+1)]P 1? i? n {(a 0 -a i)2/(a 0+a
i+1)}, where D k = P k? i = {cv: c Î K } spanned by a vector v of V . How does it look like?
Unlike when the field K is R in which case could be imagined as a line passing through origin,
in the case K = C , it requires something like an imaginary surface of a right circular cylinder
whose axis lies along the direction of v passing through the origin, so that assuming its base to be
in the plane perpendicular to the axis through the origin and taking v to be represented by an
arbitrary point on the surface at a positive height ||v|| marking the angle zero on the circle, the
vector cv = reiq v, (q Î [-p , p ]) could be represented as a point on the cylinder at signed height r
and making angle q in the anticlockwise sense. The angle q could be imagined as an axial twist
of cv over v. Then, interpreting the polar representation (x,y)/( P xP P yP ) = reif , (-1 ? r ? 1, -p
/2 ? f ? p /2), we find that the vector x has an additional minimal magnitude axial-twist of f
radians over y. Thus if we wish to do meaningful real geometric operations compatible with the
inner product we should work with the pair of vectors e-if x, y, or, the vectors x, eif y, the cosine
of the real angle q between which is given by cos q = r Î [-1,1]. The principal value of q Î [0, p ]
then may be defined as the required angle between x and y. In the case of real inner-product
spaces this notion is compatible with the standard one.