FUNCTIONAL ANALYSIS DR. RITU AGARWAL MALAVIYA NATIONAL INSTITUTE OF TECHNOLOGY JAIPUR Contents 1. About 1 2. Syllabus 1 References 1 2.1. Origin 2 3. Abstract space 2 4. Metric Space 2 5. Semi-Metric 6 6. Separable spaces 6 7. Completeness 8 1. About Functional analysis is the abstract branch of mathematics that has originated from classical analysis. It is study of certain topological algebraic structures and of methods by which knowledge of these structures can be applied to analytic problems. 2. Syllabus Normed Spaces, continuity of a linear mapping. Banach spaces, Linear Transformations and functionals and Normed bounded linear transformation, dual spaces, Hahn Banach theorem. Hilbert Spaces. Orthonormal sets, Bessels Inequality, Parsevals relation, Riesz Representation theorem, Relationship between Banach Spaces, Hilbert Spaces. Adjoint operators in Hilbert Spaces, Self adjoint operators, positive operators, Projection Operators and orthogonal projections in Banach & Hilbert spaces, Fixed point theorems and their applications, Best approximations in Hilbert Spaces. Gatebux and Frechat Derivatives. Solution of boundary value problems. Optimization problems. Applications to Integral and differential equations. References [1] Kreyszig, Introductory Functional Analysis With Applications, Wiley India Edition, 1989. [2] Walter Rudin, Functional Analysis, Tata Mc-Graw Hill, New Delhi. [3] Balmohan Vishnu Limaye, Functional analysis, New Age International, 1996. [4] Aldric Loughman Brown andA. Page, Elements of functional analysis, Van Nostrand-Reinhold, 1970 [5] J.B. Conway, A Course in Functional Analysis, 2nd Edition, Springer, Berlin, 1990. Date : January 27, 2017. Email:[email protected]Web: drrituagarwal.wordpress.com. 1
Transcript
FUNCTIONAL ANALYSIS
DR. RITU AGARWALMALAVIYA NATIONAL INSTITUTE OF TECHNOLOGY JAIPUR
Contents
1. About 12. Syllabus 1References 12.1. Origin 23. Abstract space 24. Metric Space 25. Semi-Metric 66. Separable spaces 67. Completeness 8
1. About
Functional analysis is the abstract branch of mathematics that has originated from classical analysis. Itis study of certain topological algebraic structures and of methods by which knowledge of these structurescan be applied to analytic problems.
2. Syllabus
Normed Spaces, continuity of a linear mapping. Banach spaces, Linear Transformations and functionalsand Normed bounded linear transformation, dual spaces, Hahn Banach theorem. Hilbert Spaces.Orthonormal sets, Bessels Inequality, Parsevals relation, Riesz Representation theorem, Relationshipbetween Banach Spaces, Hilbert Spaces. Adjoint operators in Hilbert Spaces, Self adjoint operators,positive operators, Projection Operators and orthogonal projections in Banach & Hilbert spaces, Fixedpoint theorems and their applications, Best approximations in Hilbert Spaces. Gatebux and FrechatDerivatives.Solution of boundary value problems. Optimization problems. Applications to Integral and differentialequations.
References
[1] Kreyszig, Introductory Functional Analysis With Applications, Wiley India Edition, 1989.[2] Walter Rudin, Functional Analysis, Tata Mc-Graw Hill, New Delhi.[3] Balmohan Vishnu Limaye, Functional analysis, New Age International, 1996.[4] Aldric Loughman Brown andA. Page, Elements of functional analysis, Van Nostrand-Reinhold, 1970[5] J.B. Conway, A Course in Functional Analysis, 2nd Edition, Springer, Berlin, 1990.
Date: January 27, 2017.Email:[email protected] Web: drrituagarwal.wordpress.com.
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2 DR. RITU AGARWAL
2.1. Origin. It originates from the calculus of variations in the study of operators on function spacesdefined by differentiation and integration. The name functional analysis was coined by the French math-ematician Paul Levy (1886-1971). Early pioneers were:Italian Vito Volterra (1860-1940),Swedish Erik Ivar Fredholm (1866-1927),German David Hilbert (1862-1943) andHungarian Frigyes Riesz (1880-1956).
2.1.1. Influential contributions. The Polish mathematician Stefan Banach (1892-1945) was influential inbringing the notions of topology into functional analysis, and he is known for the seminal book Theoriedes operations lineaires of 1932.Fundamental contributions to the study of operators on Hilbert spaces were made by the Hungarianmathematician John von Neumann (1903-57).
2.1.2. Inspiration. Inspiration to this work came from many sides, not least from the development ofquantum mechanics in physics in the 1920s by physicists such asNiels Bohr (1885-1962),Paul Dirac (1902-84),Werner K. Heisenberg (1901-76)Erwin Schrodinger (1887-1961)
3. Abstract space
Definition 3.1 (Mathematical Analysis). Mathematical analysis leads to exact results by approximatecomputations. It is based on the notions of approximation and limit process. For instance, the derivativeis the limit of differential quotients, and the integral is the limit of Riemann sums.
Definition 3.2. Set of unspecified elements satisfying certain axioms
Example:
• Groups
• Rings
• Fields
• Vector space
• Metric space
• and many more.
4. Metric Space
Definition 4.1. Metric Space A function d : X ×X −→ R is said to be a metric on non-empty set X ifit satisfies the following axioms:
• Non-negative: d(x, y) ≥ 0.
• d(x, y) = 0 iff x = y
FUNCTIONAL ANALYSIS 3
• Symmetry: d(x, y) = d(y, x) for all x, y
• Triangle inequality: d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X.
A set X with metric d defined on it is called Metric Space and is denoted by (X, d).
Example 1. Examples of metric spaces:
a. Real line R is a metric space with metric d(x, y) = |x− y|, x, y ∈ R.
b. Euclidean space R2 is a metric space with metric• d1(x, y) =
√(ξ1 − η1)2 + (ξ2 − η2)2
• and d2(x, y) = |ξ1 − η1|+ |ξ2 − η2|x = (ξ1, ξ2), y = (η1, η2) ∈ R2.
c. Euclidean space Rn is a metric space with metric d(x, y) =√
(ξ1 − η1)2 + ...+ (ξn − ηn)2, x =(ξ1, ..., ξn), y = (η1, ..., ηn) ∈ Rn.
d. Unitary Space Cn is a metric space with metric d(x, y) =√|ξ1 − η1|2 + ...+ |ξn − ηn|2, x = (ξ1, ..., ξn), y =
(η1, ..., ηn) ∈ Cn.
e. Function space C[a, b]: This is the set of all real valued functions x, y, ... which are functions ofindependent real variable t and are defined and continuous on a given closed interval [a, b]. Themetric on this set can be defined as:• d1(x, y) = maxt∈[a,b] |x(t)− y(t)|
• d2(x, y) =∫ ba|x(t) − y(t)|dt represents the area between two curves bounded by the ordinates
x = a and x = b.f. Sequence space S of all bounded and unbounded sequences of complex numbers with metric d defined
as
d(x, y) =∞∑i=1
1
2i|ξi − ηi|
1 + |ξi − ηi|
where x = (ξi), y = (ηi) ∈ S, is a metric space.
[Hint: To prove triangle inequality in this case, use the result |a+ b| ≤ |a|+ |b| ⇒ |a+b|1+|a+b| ≤
|a|+|b|1+|a|+|b| ≤
|a|1+|a| + |b|
1+|b| as f(t) = t1+t
is monotonically increasing function of t.]
g. B[a, b] set of all bounded functions defined on [a, b] with metric d defined as d(x, y) = supt∈[a,b] |x(t)−y(t)|, x, y ∈ B[a, b] is a metric space.
h. Sequence space l∞
l∞ =
{x = (ξi)
∞i=1 s.t. sup
i|ξi| <∞
}with metric d defined as d(x, y) = supi |ξi − ηi|, x, y ∈ l∞.
i. Sequence space lp
lp =
{x = (ξi)
∞i=1 s.t.
∞∑i=1
|ξi|p <∞
}
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with metric d defined as d(x, y) =
(∞∑i=1
|ξi − ηi|p)1/p
, x, y ∈ lp, p ≥ 1.
Remark 4.2. Different metric can be defined on a given set X.
In order to prove triangle inequality for Example 1(i), first we are required to prove following results.
Definition 4.3 (Conjugate Exponents). Let p > 1 and q ∈ R be such that
1
p+
1
q= 1. (4.1)
Then p and q are called conjugate exponents. Also, (4.1) can be written as
(p− 1)(q − 1) = 1
Theorem 4.4. Let α and β be any non-negative real numbers and if p and q are conjugate exponentsthen
αβ ≤ αp
p+βq
q(4.2)
Proof: If α = 0, β = 0, the result is obvious.
Let α 6= 0, β 6= 0, consider the function u = tp−1 i.e. t = uq−1. Then,αβ = Total area of the region 1 and 2
≤∫ α0tp−1dt+
∫ β0uq−1du = αp
p+ βq
q.
Theorem 4.5 (Holder’s inequality). Let x = (ξj) ∈ lp, p > 1 and y = (ηj) ∈ lq, p and q are conjugateexponents. Then (ξjηj) ∈ l1 and
∞∑j=1
|ξjηj| ≤
(∞∑k=1
|ξk|p)1/p( ∞∑
m=1
|ηm|q)1/q
(4.3)
Proof: Let x = (ξj) ∈ lp, and y = (ηj) ∈ lq be two sequences such that∞∑k=1
|ξk|p = 1 and∞∑m=1
|ηm|q = 1
Setting α = (ξj), and β = (ηj) in (4.2), we get
|ξj ηj| ≤|ξj|p
p+|ηj|q
q⇒
∞∑j=1
|ξj ηj| ≤ 1. (4.4)
Now, let x = (ξj) ∈ lp, p > 1 and y = (ηj) ∈ lq. Then substituting
ξj =ξj
(∑∞
k=1 |ξk|p)1/p
andηj
(∑∞
m=1 |ηm|q)1/q
in (4.4), we get the desired result (4.3).
Theorem 4.6 (Minkowski Inequality). Let x = (ξj), y = (ηj) ∈ lp p ≥ 1. Then(∞∑j=1
|ξj + ηj|p)1/p
≤
(∞∑k=1
|ξk|p)1/p
+
(∞∑m=1
|ηm|p)1/p
(4.5)
FUNCTIONAL ANALYSIS 5
Proof: Result is obvious for p = 1. For p > 1, let ξj + ηj = ωj.
Applying Holder’s inequality to x = (ξj) ∈ lp and |ωj|p−1 ∈ lq, we get
∞∑j=1
|ξj||ωj|p−1 ≤
(∞∑k=1
|ξk|p)1/p( ∞∑
m=1
(|ωm|p−1)q)1/q
=
(∞∑k=1
|ξk|p)1/p( ∞∑
m=1
|ωm|p)1/q
(4.6)
Similarly,∞∑j=1
|ηj||ωj|p−1 ≤
(∞∑k=1
|ηk|p)1/p( ∞∑
m=1
|ωm|p)1/q
(4.7)
Adding (4.6) and (4.7) and simplifying, we get the desired result.
Remark 4.7. In order to prove the triangle inequality for lp space, replace ξj by ξj − ζj and ηj by ζj − ηjin (4.5).
Definition 4.8 (Metric Subspace). Let (X, d) be a metric space and Y ⊂ X is non-empty proper subsetof X. Suppose d is the restriction of d on Y × Y such that
d(y1, y2) = d(y1, y2) ∀y1, y2 ∈ Y
Example 2. If A is a subspace of l∞ consisting of all sequences of 0 and 1. The induced metric on A is adiscrete metric i.e.
d(x, y) =
{0, x = y
1, x 6= y
Definition 4.9 (Continuous mapping). Let (X, d) and (Y, d) be two metric spaces. A mapping T : X →Y is said to be continuous at a point x0 ∈ X, if for every ε > 0, ∃ δ > 0 such that
d(Tx, Tx0) < ε whenever d(x, x0) < δ
Theorem 4.10. A mapping T of a metric (x, d) into a metric space (Y, d) is continuous if and only ifinverse iamge of an open subset of y is an open subset of X.
Proof: Let f is continuous Y is open set in T . Also, let p ∈ f−1(Y ). Then y = f(p) for some y ∈ Y .Since Y is open and y ∈ Y , for some ε > 0, BT (y; ε) ⊆ Y .Since f is continuous, ∃δ > 0 such that f(BS(p; δ)) ⊆ BT (y; ε).⇒ BS(p; δ) ⊆ f−1(f(BS(p; δ))) ⊆ f−1(BT (y; ε)) ⊆ f−1(Y )⇒ p is an interior point of f−1(Y ).Conversely let f−1(Y ) is open in S for every open subset Y in T . Choose p ∈ S and y = f(p).For every ε > 0, BT (y; ε) is open in T . ⇒ f−1(BT (y; ε)) is open in S.Now p ∈ f−1(BT (y; ε))⇒ ∃ δ > 0 such that BS(p; δ) ⊆ f−1(BT (y; ε))⇒ f(BS(p; δ)) ⊆ BT (y; ε)f is continuous at p.
Remark 4.11. The image of an open set under continuous mapping is not necessarily open.E.g. Constant function
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5. Semi-Metric
Definition 5.1 (Semi-Metric). A function d : X ×X → R is said to be semi-metric if d(x, y), x, y ∈ Xis a real valued, non-negative finite number such that it is symmetric and satisfies triangle inequality, i.e.d(x, x) = 0, d(x, y) = d(y, x) and d(y, z) ≤ d(x, y) + d(y, z).
A space X equipped with semi-metric d is said to be semi-metric space (X, d).
Example 3. X = L[0, 1] is the set of all Lebesgue integrable functions f on [0, 1] i.e. f ∈ L[0, 1] =⇒∫ 1
0|f(x)dx is finite.
The metric on this space can be defined as :
d(f, g) =
∫ 1
0
|(f − g)(x)|dx (5.1)
d(f, g) = 0 =⇒ f = g almost everywhere.(L[0, 1], d) is semi-metric.
Remark 5.2. For the same definition given by (5.1), the space (C[0,1],d) is a metric space.
Theorem 5.3 (Equivalence relation using semi-metric). Let (X, d) be a semi-metric space. Define arelation R as xRy or x ∼ y if d(x, y) = 0.Then R is an equivalence relation and decomposes the set X into disjoint equivalent classes Ex = {x′ :d(x, x′) = 0}.Take E = {Ex, Ey, ...}. Define distance function ρ on E as ρ(Ex, Ey) = d(x, y). The ρ is well-definedand (E, ρ) is a metric space.
Proof: First prove that R is an equivalence relation where xRy if d(x, y) = 0.Next, to show that ρ is well-defined, consider x′Rx and y′Ry. Thend(x′, y′) = d(x, y).Further, (E, ρ) is a metric space. This can be proved using the concept that ‘two equivalence classes areeither identical or disjoint’.
6. Separable spaces
Definition 6.1 (Dense Set). A subset M of a metric space X is said to be dense in X if M = X
Example: Q ⊂ R is dense in R.
Definition 6.2 (Separable Space). A metric space X is said to be separable if it has a countable subsetM which is dense in X.
Example 4. The real line R is separable.The set Q of rational numbers is countable and dense in R.
Example 5. The complex plane C is separable.A countable dense subset of C is the set of all complex numbers whose real and imaginary parts are bothrational.
Example 6. A discrete metric space X is separable if and only if X is countable.The type of metric implies that no proper subset of X can be dense in X. Hence the only dense set in Xis X itself.
Example 7. The space l∞ is not separable.
Proof: Let A ⊂ l∞ be the set containing sequences of zeros and ones as y = (η1, η2, ...). Then y ∈ l∞.With each y, we associate a point ∩y ∈ [0, 1] whose binary representation is
η121
+η222
+η323
+ ...
FUNCTIONAL ANALYSIS 7
Since the interval [0, 1] is uncountable, there are uncountably many sequences of zeros and ones in A andmetric will be discrete metric (see example 2).Any two unequal elements in A are unit distance apart. Let each of these sequences be the centre of asmall ball of radius 1/3, then these balls do not intersect and we have uncountably many of them. If Mis dense in l∞, each of these non-intersecting balls must contain an element of M. Hence M cannot becountable. Since M was arbitrary dense set, l∞ cannot have dense subsets which are countable.Consequently, l∞ is not separable.
Example 8. The space lp, p ≥ 1 is separable.
Proof: Let M be the set of all sequences y of the form
y = (η1, η2, ..., ηn, 0, 0, ...)
where n is any positive integer and ηi’s are rational. Then M is countable. We show that M is dense inlp.Let x = (ξi) ∈ lp be arbitrary.. Then for every ε > 0 there is an n (depending on ε) such that
∞∑j=n+1
|ξj|p <εp
2
where the L.H.S. is the remainder of a converging series.Since the rationals are dense in R, for each ξj, there is a rational ηj close to it. Hence, we can find y ∈Msatisfying
n∑j=1
|ξj − ηj|p <εp
2
It follows that
{d(x, y)}p =n∑j=1
|ξj − ηj|p <εp
2+
∞∑j=n+1
|ξj|p < εp
We thus have d(x, y) < ε and see that M is dense in lp.
Exercise 6.3. Show that the space (B[a, b], d) of bounded functions defined over closed interval [a, b]and metric d(f, g) = supt∈[a,b] |f(t)− g(t)| is not separable.
Theorem 6.4. If (X, d) is a separable metric space and Y ⊆ X, then (Y, d) is separable in the inducedmetric d.
Proof: Given (X, d) is a separable metric space. So it has a countable subset E = {x1, x2, ...} which isdense in X.Case I: If E is contained in Y , then there is nothing to prove.Case II: If E is not contained in Y . We construct a countable subset yn,m of Y whose points are arbitraryclose to the set E.To show: There exists a countable subset of Y which is dense in Y . For positive integers n and m, letBn,m = B(xn, 1/m) be open ball centered at xn ∈ E with radius 1/m. Choose yn,m ∈ Bn,m∩Y wheneverit is non-empty. A sequence will be obtained as y1,m, y2,m, y3,m, ... ∈ Bn,m ∩ Y . The balls B(xn, 1/m) arecountable in number as x1, x2, ... are countable and some of these have empty intersection with Y , leavethose balls.{yn,m} is a countable subset of Y . It remains to show that yn,m is dense in Y .Let y ∈ Y and r > 0. Let m be so large that 1
m< r
2and find xn ∈ B(y, 1/m) then y ∈ B(xn, 1/m) = Bn,m
and hence y ∈ Y ∩Bn,m. Now
d(y, yn,m) ≤ d(y, xn) + d(xn, yn,m) <1
m+
1
m< r
Therefore, yn,m ∈ B(y, r). Since y ∈ Y and r > 0 are arbitrary, we see that {yn,m} is dense in Y .
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7. Completeness
Definition 7.1 (Convergent sequence). A sequence (xn) in a metric space (X, d) is said to be convergentif ∃x ∈ X such that
d(xn, x)→ 0 as n→∞Such an x is called limit of (xn) and we write limn→∞ xn = x.
Definition 7.2 (Cauchy sequence). A sequence (xn) in a metric space (X, d) is said to be a Cauchysequence if
d(xn, xm) < ε for n,m ≥ n0 ( depending on ε)
Remark 7.3. Every convergent sequence is Cauchy sequence but every Cauchy sequence is not necessarilyconvergent. (Why??)
A non-empty subset M ⊂ X is said to be bounded if its diameter δM = supx,y∈M d(x, y) is finite.
Definition 7.4 (Bounded sequence). A sequence (xn) in a metric space (X, d) is said to be bounded ifcorresponding point set {x1, x2, ...} is a bounded subset of X.
Remark 7.5. If M is bounded, M ⊂ B(x0; r) where x0 is arbitrary and r is sufficiently large positive realnumber.
Theorem 7.6. Let (X,d) be a metric space. Then
(i.) A convergent sequence in X is bounded and its limit is unique.
(ii.) If xn → x and yn → y in X then d(xn, yn)→ d(x, y).
Proof: (i) Suppose that xn → x. Then taking ε = 1, we can find an n0 such that d(xn, x) < 1 for alln ≥ n0. Hence by triangle inequality d(xn, x) < 1 + a where
a = max{d(x1, x), d(x2, x), ..., d(xn0 , x)}
This shows that (xn) is bounded. Further, assume that x and z be two limits of (xn). Then
0 ≤ d(x, z) ≤ d(x, xn) + d(xn, z)→ 0 + 0
Hence, follows the uniqueness of the limit of (xn).(ii) Since, xn → x and yn → y, by triangle inequality, we get
Definition 7.7 (Completeness). A metric space (X,d) is said to be complete if every Cauchy sequencein X is convergent (i.e. the limit is an element of X).
Theorem 7.8. Let M be a nonempty subset of a metric space (X,d) and M is closure of M. Then
(a) x ∈ M iff there is a sequence {xn} in M such that xn → x.
(b) M is closed iff the situation xn ∈M , xn → x implies that x ∈M .
FUNCTIONAL ANALYSIS 9
Proof: (a) Let x ∈ M . If x ∈M , a sequence of type is (x, x, ...). If x /∈M , it is an accumulation pointof M. Hence for each n = 1, 2, ... the ball B(x; 1/n) contains xn ∈ M and xn → x because 1/n → 0 asn→∞.
Conversely, if (xn) is in M and xn → x then x ∈M or every neighbourhood of x contains points xn 6= xso that x is a point of accumulation of M. Hence, x ∈ M , by the definition of closure.
(b) M is closed if and only if M = M , so that (b) follows readily from (a).
Theorem 7.9. A subspace M of a complete metric space (X,d) is complete iff M is closed in X.
Proof: First let M be complete. Let x ∈ M then there is a sequence (xn) converging to x. Since (xn) isa Cauchy sequence and M is complete, it converges in M. This implies, x ∈M i.e. M ⊂M ⇒M is closed.
Conversely, let M be closed and (xn) is a Cauchy sequence in M. Since X is complete, (xn)→ x ∈ Xwhich implies x ∈ M = M as M is closed. Hence the arbitrary Cauchy sequence (xn) converges in Mwhich proves completeness of M.
Example 9. Completeness for R is already proved. Set R is complete ordered field.
• Euclidean space (Rn, d) is complete for the usual metric d(x, y) =
(2)i , ... behaves as a Cauchy sequence of real numbers which is a complete
space. Hence the sequence (ξ(m)i ) is convergent in R (or C) and converges to ξi ∈ R(say) i.e.
10 DR. RITU AGARWAL
|ξ(m)i − ξi| < ε for all m > N
Let x = (ξ1, ξ2, . . .). To show xm → x as m→∞ and x ∈ l∞. From above
supi|ξ(m)i − ξi| < ε for m > N
⇒ d(xm, x) < ε ⇒ xm → x.
Further, since ξi = ξi − ξ(m)i + ξ
(m)i ,
|ξi| = |ξi − ξ(m)i |+ |ξ
(m)i | < ε+Km
where Km = supi ξ(m)i <∞.
⇒ supi ξi <∞ ⇒ x ∈ l∞.• Space of bounded sequences lp, 1 ≤ p <∞ is complete
• Space of convergent sequences C, with the metric d∞ induced from l∞, is complete.
Some spaces may be complete with respect to one metric but not with another metric defined on it.
Example 10. Function space C[a, b]: This is the set of all real valued functions x, y, ... which are functionsof independent real variable t and are defined and continuous on a given closed interval [a, b]. The metricon this set can be defined as:
• d1(x, y) = maxt∈[a,b] |x(t)− y(t)|
• d2(x, y) =∫ ba|x(t) − y(t)|dt represents the area between two curves bounded by the ordinates
x = a and x = b.
(C[a, b], d1) is complete metric space but (C[a, b], d2) is not.
Proof: (a) Let (xn) be a Cauchy sequence in C[a, b]. So given ε > 0, ∃n0 ∈ N such that d1(xm, xn) < ε,m,n > n0. That is
maxa≤t≤b
|xm(t)− xn(t)| < ε
This implies
|xm(t)− xn(t)| < ε ∀ t ∈ [a, b] . . . (i)
Fix t = t0, xm(t0) ∈ R, therefore (xm(t0)) is a Cauchy sequence of real numbers and hence it convergesto some x(t0) ∈ R (say).(x is well defined): In this way, we can associate with each t ∈ [a, b], an unique real number x(t). Thisdefines a pointwise function x on [a, b].Letting n→∞ in (i), it follows that,
|xm(t)− x(t)| < ε ∀ t ∈ [a, b] . . . (ii)
(xm) converges to x uniformly on [a, b] and (xm) is a sequence of continuous functions. Hence the limitfunction x is also continuous on [a, b]. That is x ∈ C[a, b].Further,
(ii) =⇒ maxa≤t≤b
|xm(t)− xn(t)| < ε
=⇒ xmd1−→ x.
Remark 7.10. This metric on C[a, b] is sometimes called uniform metric also since convergence of xm → xis uniform.
FUNCTIONAL ANALYSIS 11
(b) To show that (C[a, b], d2) is not complete, we here give a counter example.Let (xm) be a sequence in C[0, 1] defined as follows:
xm(t) =
0, t ∈ [0, 1/2]
(t− 1/2)/m, t ∈ (1/2, am]
1, t ∈ (am, 1]
(7.1)
where am = 12
+ 1m
. (xm) is a Cauchy sequence under the metric d2 in C[0, 1]. Let m > n then 1m< 1
n.
d2(xm, xn) =
∫ 1
0
|xm(t)− xn(t)|dt =
∫ 1/m
1/2
|xm(t)− xn(t)|dt+
∫ 1/n
1/m
|xm(t)− xn(t)|dt→ 0 as m,n→∞
Let m→∞ in (7.1) then
x(t) = limm→∞
xm(t) =
{0, t ∈ [0, 1/2]
1, t ∈ (1/2, 1](7.2)
Clearly x(t) is not a continuous function while xm → x not in C[0, 1]. Hence, (C[a, b], d2) is not complete.Other examples of incomplete metric spaces:
1. X = Q, the set of rational numbers is incomplete. But Q can be enlarged to R a complete metricspace. So, this gives a way to convert it to complete metric space.2. Set of polynomials is incomplete. This can be proved by giving counter example as follows:Consider the sequence of polynomials defined by (Pn(z)), Pn(z) = 1+z+ z2
2!+ z3
3!+ . . .. limn→∞ Pn(z) = ez
which is clearly not a polynomial but a transcendental function.
Definition 7.11 (Isometric mapping). Let (X, d) and (Y, d) be two metric spaces and T be a mappingfrom X to Y. T is said to be isometric or an isometry if it preserves the distance, i.e.
d(Tx, Ty) = d(x, y)
Isomtery is one-one for if Tx = Ty =⇒ d(Tx, Ty) = 0 = d(x, y) =⇒ x = y. If the isometry T is onto,the spaces x and y are said to be isomteric spaces.
Remark 7.12. Two isometric spaces may be regarded as two copies of same abstract space in the studywhere nature of points does not matter.
Example 11. Show that C[0, 1] and C[a, b] are isometric spaces.
Proof: Define T (t) = t−ab−a , t ∈ [a, b] and T (t) ∈ [0, 1]. So that whenever x(t) ∈ C[a, b], Tx = x(Tt) ∈
C[0, 1]. Then d(Tx, Ty) = d(x( t−ab−a), y( t−a
b−a)) = d(x, y).
Definition 7.13 (Homeomorphism). A homeomorphism is a continuous bijective mapping T : X → Ywhose inverse is also continuous.
Remark 7.14. If two spaces are isometric then they have same nature regarding completeness but this isnot necessary in case of homeomorphism.
Example 12. Illustrate with one example that a complete and incomplete metric spaces may be homeo-morphic.
Solution: Consider the metric spaces X = R and Y = (−1, 1). Clearly, X is complete while Y not.Define a mapping T : X → Y as T (x) = y = 2
πtan−1x. Then T is a continuous mapping and bijection
also. Further T−1y = x = tan(πy/2) is also continuous. T is a homeomorphism.
Definition 7.15 (Equivalent metric). A metric d is stronger than other metric d′ defined on a non-emptyset X if for every ε > 0, ∃δ > 0 such that Bd(x, δ) ⊂ Bd′(x, ε) that is every open subset of X w.r.to d′ isalso open w.r.to d.Two metrics defined on X are said to be equivalent if each is stronger than the other.
12 DR. RITU AGARWAL
Mathematically, two metrics d and d′ are equivalent to each other if there exists constants α and β suchthat
αd(x, y) ≤ d′(x, y) ≤ βd(x, y) (7.3)
Remark 7.16. If a sequence is convergent in the strong metric, it has to converge in a weak metric. i.e.
whenever xnd−→ x =⇒ xn
d′−→ x.
Example 13. Show that the completeness of a metric may not be shared by an equivalent metric.
Solution: Consider X = (0, 1]. Let d′(x, y) = |x− y| and d(x, y) =∣∣∣ 1x − 1
y
∣∣∣ for x, y ∈ (0, 1].
Clearly, xnd−→ x⇔ xn
d′−→ x. Hence d and d′ are equivalent metrics. (X, d′) is complete but (X, d) not.
Exercise 7.17. Show that the Cauchy sequences in (X, d) and (X, d′) are same where d and d′ areequivalent metrics.
Exercise 7.18. Show that Cauchy Schwarz inequality implies
(|ξ1|+ |ξ2|+ ...+ |ξn|)2 ≤ n(|ξ1|2 + |ξ2|2 + ...+ |ξn|2).Show that the metric d(x, y) =
