Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Additional Topics for IIT JAM Mathematics

Mathematics : Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

The document Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is a part of the Mathematics Course Additional Topics for IIT JAM Mathematics.
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Problems based on Module –I (Metric Spaces) 

Ex.1. Let d be a metric on X. Determine all constants K such that
(i) kd , (ii) d + k is a metric on X
Hint.1:
  Use definition . Ans (i) k>0 (ii) k=0 

Ex.2. Show that
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.2: 
Use Triangle inequality d Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Similarity, Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.3. Find a sequence x which is in lp with p >1  but Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.3:
Choose x = (xk) where xk = 1/k

Ex.4. Let (X,d) be a metric space and A,B are any two non empty subsets of X. Is D(A,B) = inf d (a,b)Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduReva metric on the power set of X?
Hint.4:
No. Because D(A,B)=0 ≠ > A=B e.g. Choose Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev where ai ≠ bclearly Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.5. Let ( X, d) be any metric space. Is Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduReva Metric space where Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.5:
Yes. ;

Ex.6. Let (X1, d1) and (X2, d2) be metric Spaces and X = X1 X2. Are Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev as defined below A metric on X?
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.6:
Yes. 

Ex.7. Show that in a discrete metric space X, every subset is open and closed.
Hint.7:
Any subset is open since for any a ∈ A,  the open ball Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev Similarly Ais open, so that (Ac)= A  is closed. 

Ex.8. Describe the closure of each of the following Subsets:
(a) The integers on R.
(b) The rational numbers on R.
(c) The complex number with real and imaginary parts as rational inQuestion For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev.
(d) The disk Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.8:
use Definition
Ans (a) The integer, (b) R, (c) Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.9. Show that a metric space X is separable if and only if X has a countable subset Y with the property: For every ∈  > 0  and every x ∈ X  there is a y ∈ Y such that d (x ,y) < ∈.
Hint.9: 
Let X be separable .So it has a countable dense subset Y i.e. Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  be given.Since Y is dense in X and Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev so that the ∈ neibourhood B(x;∈)  of x contains a y ∈ Y, and d (x, y) < ∈. Conversely, if X has a countable subset Y with the property given in the problem, every x ∈ X  is a point of Y or an accumulation point of Y. Hence  x ∈ Y, s result follows.

Ex.10. If (xn) and (yn)  are Cauchy sequences in a metric space (X,d), show that (an), where a= d(xn.yn) converges.
Hint.10:
Since Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev which shows that (an) is a Cauchy sequence of real  numbers . Hence convergent. 

Ex.11. Let  ab ∈ R and  a < b.Show that the open interval (a, b) is an incomplete subspace of R.
Hint.11: 
Choose Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev which is a Cauchy sequence in (a,b) but does not converge.

Ex.12. Let  Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev be the subspace consisting of all sequence Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev with at most finitely many nonzero terms .Find a Cauchy sequence in M which does not converge in M, so that M is not complete.
Hint.12: 
Choose (xn), where Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev which is Cauchy in M but does not converge. 

Ex.13. Show that the set X of all integers with metric d defined by d (m,n) = |m - n| is a complete metric space.
Hint.
13: Consider a sequence  Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Where xk = α  for k ≥ n, α  is an integer.This is a Cauchy and converges to α ∈ X.

Ex.14.Let X be the set of all positive integers and d (m, n) = |m-1 - n-1|. Show that (X , d) is not complete.
Hint.14:
Choose (xn) when x= n which is Cauchy but does not converge.

Ex.15. Show that a discrete metric space is complete.
Hint.15: 
Constant sequence are Cauchy and convergent. 

Ex.16. Let X be metric space of all real sequences Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev each of which has only finitely Nonzero terms, and Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev when y = (ηj). Show that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev 
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  
for j > n is Cauchy but does not converge.
Hint.16: 
For every ∈ > 0, ,there is an N s. t. for n > m >N, Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev 
But (xn) does not converge to any Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev Because Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev so that for n > Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
 Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  And d(xn, x) → 0 as n → ∞ is imposible.

Ex.17. Show that,by given a example ,that a complete and an incomplete metric spaces may be Homeomorphic.
Hint.17:
(Def) A homeomorphism is a continuous bijective mapping.
T: X → Y : whose inverse is continuous; the metric space X and Y are then said to be  homeomorphic. e g. . A mapping T: R: → (-1 ,1)  defined as Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev with metric d(x,y) = |x - y| on R. .Clearly T is 1-1,into & bi continuous so Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  But R is complete while (-1,1) is an incomplete metric space. 

Ex.18. If (X , d) is complete, show that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev where Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is complete.
Hint.18:
 
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hence if (xn) is Cauchy is Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev it is Cauchy in (X, d), and its limit in Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Problems on Module-II (Normed and Banach Spaces) 

Ex.1. Let (X, Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  i= 1, 2, ∞ be normed spaces of all ordered pairs
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev of real numbers where
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  are defined as 

Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
How does unit sphere in these norms look like?
Hint.1: 
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.2. Show that the discrete metric on a vector space X ≠{0} can not be obtained from a norm.
Hint.2:  Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.3. In l ,let γ be the subset of all sequences with only finitely many non zero terms. Show that γ is a subspace of  l but not a closed subspace.
Hint.3: Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  where Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  has 0 value after j>n. Clearly 

Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev as well as x(n)∈ γ  but Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.4. Give examples of subspaces of l and l2 which are not closed.
Hint.4:
Let γ be the subset of all sequences with only finitely many non zero terms.
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev 
but not closed. 

Ex.5. Show that a discrete metric space X consisting of infinitely many points is not compact.
Hint.5:
By def. of Discrete metric, any sequences (nx) cannot have convergent subsequence as Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.6. Give examples of compact and non compact curves in the plane R2.
Hint.6: 
As Ris of finite dimension, So every closed & bounded set is compact.
Choose Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  which is compact But  Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  is not compact. 

Ex.7. Show that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev are locally compact.
Hint.7:
(def.) A metric space X is said to be locally compact if every point of X has a compact neighbourhood. Result follows (obviously). 

Ex.8. Let X and Y be metric spaces. X is compact and T: X→ Y bijective and continuous. Show that T is homeomorphism.
Hint.8:
Only to show T-1 is continuous i.e. Inverse image of open set under T-1 is open.
OR.
If Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev It will follow from the fact that X is compact.

Ex.9. What are the domain, range and null space of T1,T2,T3 in exercise 9.
Hint.9:
The domain is R. The ranges are the Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRevThe null spaces are the Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRevthe origin.

Ex.10. Let T : X → Y be a linear operator. Show that the image of a subspace V of X is a vector space, and so is the inverse image of a subspace W of X.
Hint.10.
 
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
αx+ βx2 is an element of that inverse image. 

Ex.11. Let X be the vector space of all complex 2 x 2 matrices and define T: X → X by Tx=bx, where b ∈ X  is fixed and bx denotes the usual product of matrices. Show that T is linear. Under what condition does T-1 exist?
Hint.11.
|b| ≠ 0

Ex.12. Let T : D(T) → Y be a linear operator whose inverse exists. If {x1,x2,....,xnis a Linearly Independant set in D(T), show that the set {Tx1,Tx2,....Txn} is L.I.
Hint.12.

 Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.13. Consider the vector space X of all real-valued functions which are defined on R and have derivatives of all orders everywhere on R. Define T : X →  X by y(t) = Tx(t) = x'(t), show that R(T) is all of X but T-1  does not exist.
Hint.13:
R (T) = X  since for every y ∈ X we have y = Tx, where x(t) = Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRevBut n T-does not exist since Tx=0 for every constant function. 

Ex.14. Let X and Y be normed spaces. Show that a linear operator T: X → Y is bounded if and only if T maps bounded sets in X into bounded set in Y.
Hint.14:
Apply definition of bounded operator.

Ex.15. If T ≠ 0 is a bounded linear operator, show that for any Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRevwe have the strict inequality Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.15:
Since Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.16. Find the norm of the linear functional f defined on C[-1, 1] by
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.16: Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev 
For converse, choose x(t) = -1 on [-1,1]. So Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev 
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Problems on Module III (IPS/Hilbert space)

Ex.1. If x ⊥ y in an IPS X,Show that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.1: Use Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev and the fact that < x y > = 0,  if x ⊥ y. 

Ex.2. If X in exercise 1 is a real vector space, show that ,conversely, the given relation implies that x ⊥ y . Show that this may not hold if X is complex. Give examples.
Hint. 2:
By Assumption,
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.3. If an IPS X is real vector space, show that the condition Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  implies <x +y,x-y>= 0. What does this mean geometrically if X = R2?
Hint.3:
Start <x +y,x-y> = <x,x> + <y,-y> = Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  as X is real. Geometrically: If x & y are the vectors representing the sides of a parallelogram, then  x+y and x-y will represent the diagonal which are⊥.
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.4. (Apollonius identity): For any elements x, y, z in an IPS X, show that
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint 4:
Use Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev OR use parallelogram equality. 

Ex.5. Let x ≠ 0 and y ≠ 0. If x ⊥ y, show that {x,y} is a Linearly Independent set.
Hint.5:
Suppose Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev where α12 are scalars. Consider
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev .Similarly, one can show that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  is L.I.set.

Ex.6. If in an IPS X, <x,u> = <x,v> for all x, show that u = v.
Hint.6 :
Given <x,u-v> = 0. Choose x = u - v.
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.7. Let X be the vector space of all ordered pairs of complex numbers. Can we obtain  the norm defined on X by Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  from an Inner product?
Hint. 7:
No. because the vectors x= (1,1), y = (1,1) do not satisfy parallelogram equality. 

Ex.8. If X is a finite dimensional vector space and (ej ) is a basis for X, show that an inner product on X is completely determined by its values Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev. Can we choose scalars γjk in a completely arbitrary fashion?
Hint.8:
Use Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Open it so we get that it depends on Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
II Part: Answer:- NO. Because Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.9. Show that for a sequence (xn ) in an IPS X , the conditions
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.9 : 
We have
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.10. Show that in an IPS X,
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  for all scalars α.
Hint.10 :
From
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  condition follows as x ⊥ y.
Conversely, Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Choose α = 1  if the space is real which implies x ⊥ y.
Choose α = 1, α = i  if the space is complex then we get <x,y> = 0 ⇒x ⊥ y.

Ex.11. Show that in an IPS X, Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev for all scalars.
Hint.11 :
Follows from the hint given in Ex.-10. 

Ex.12. Let V be the vector space of all continuous complex valued functions on J = [a,b]. Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev 
where Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev Show that the identity mapping Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev of Xonto X is continuous. Is it Homeomorphism?
Hint.12 : 
Since  
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hence I is continuous.
Part-II: Answer No. because X2 is not complete. 

Ex.13. Let H be a Hilbert space, Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  a convex subset, and (xn) is a sequence in M  such that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev,  where Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  Show that (xn) converges in H.
Hint.13 :
(xn) is Cauchy, since from the assumption and the parallelogram equality, we have,
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Ex.14. If (ek) is an orthonormal sequence in an IPS X, and x ∈ X, show that x-y with y given by Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is orthogonal to the subspace Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.14 :

 Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.15. Let (ek) be any orthonormal sequence in an IPS X. Show that for any , x y ∈ X, Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.15: Use Cauchy Schwaz’s Inequality & Bessel’s Inequality, we get
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev 

Ex.-16. Show that in a Hilbert Space H,convergence of Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  implies convergence ofQuestion For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.16 : 
Let Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  is a Cauchy. Since H is complete, henceQuestion For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev will converge.
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  converge in H.

Problems On Module IV (On Fundamental theorems)

Ex.1. Let Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev be a sequence of bounded linear functionals defined as Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev where Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev. show that (fn)converge strongly to 0 but not uniformly.
Hint.1 : Since Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
ie Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Ex.2. Let Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev where X is a Banach space and Y a normed space. If (Tn) is strongly  convergent with limit T, then Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev 
Hint.2
T linear follows
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
T is bounded :-  Since Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
So (Tnx)  is bounded for every x.Since X is complete, so Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is bounded by uniform bounded ness theorem. Hence Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev Taking limit ⇒ T is bounded. 

Ex.3. If Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev. Show that (xn)  is point wise convergent on [a,b].
Hint .3 :
A bounded linear functional on Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.4. If Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev in a normed space X. Show thatQuestion For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.4 :
Use Lemma:- ‘’Let Y be a proper closed sub-space of a normed space X and let Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev be arbitrary point and Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  then there exists an Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev, dual of X such that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev for all y ∈ Y  and Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev 
suppose Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev which is a closed sub space of X. so by the above result , 

Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev hence there exists Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev 
which is a contradiction that

Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.5. Let Tn = Sn, where the operator Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is defined by S Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev Find a bound for Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint.5 : Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  For converse, choose x =Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
(iv) Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.6. Let X be a Banach space, Y a normed space and Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  such that (Tnx) is Cauchy in Y for every x ∈ X. .show that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is bounded.
Hint. 6 : 
Since (Tnx)is Cauchy in Y for every x, so it is bounded for each x ∈ X. Hence by uniform bounded ness theoremQuestion For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is bounded.

Ex.7. If (xn) in a Banach space X is such that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  is bounded for allQuestion For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev.Show that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  is bounded.
Hint.7 :
Suppose Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev ThenQuestion For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRevis bounded for every Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev. So by uniform bounded ness theoremQuestion For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is bounded and Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev.

Ex.8. If a normed space X is reflexive, Show that X' is reflexive.
Hint. 8 :
Let Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev there is an x ∈ X such that g = Cx  since X is reflexive. Hence h(g) = h (Cx) = f(x)  defines a bounded linear functional f on X and Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev whereQuestion For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is the canonical mapping. Hence C1 is surjective, so that X' is reflexive.

Ex.9. If x0 in a normed space X is such that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev of norm1. show that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev
Hint. 9 :
suppose Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev Then by Lemma: Let X be a normed space and letQuestion For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRevbe any elementof X. Then there exist a bounded linear functional Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev 
Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev would imply the existence of an Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.10. Let Y be a closed sub space of a normed space X such that every f ∈ X which is zero every where on Y is zero every where on the whole space X. Show that Y = X
Hint. 10 : 
If , Y ≠ X there is an Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev since Y is closed.
Use the Lemma as given in Ex 4 ( Hint ).
By this Lemma, there is on Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev which is zero on Y but not zero at x0, which contradicts our assumption. 

Ex.11. Prove that Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRevWhere Tis the adjoint operator of T.
Hint. 11 : Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev 

Similarly others.

Ex.12. Prove (ST)= TxSx
Hint. 12 : Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.13. Show that (Tn)x = (Tx)n.
Hint. 13 : Follows from Induction.

Ex.14. Of what category is the set of all rational number (a) in Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev, ( b ) in itself, (Taken usual metric)
Hint 14 :
(a) first (b) first. 

Ex.15. Find all rare sets in a discrete metric space X. 
Hint.15 :  Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev because every subset of X is open

Ex.16. Show that a subset M of a metric space X is rare in X if and only if is Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is dense in X.
Hint. 16 :
The closure of Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is all of X if and if Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev has no interior points, So that every Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is a point of accumulation of Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev

Ex.17. Show that completeness of X is essential in uniform bounded ness theorem and cannot be omitted.
Hint.17 : 
Consider the sub space Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev consisting of all Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev  where J depends on x, and let Tn be defined by Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev ClearlyQuestion For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is bounded Question For Practice With Solutions:- Normed Linear Spaces Mathematics Notes | EduRev is not bounded.

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