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

## 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

Hint.2:
Use Triangle inequality d
Similarity,

Ex.3. Find a sequence x which is in lp with p >1  but
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)a metric on the power set of X?
Hint.4:
No. Because D(A,B)=0 ≠ > A=B e.g. Choose  where ai ≠ bclearly

Ex.5. Let ( X, d) be any metric space. Is a Metric space where
Hint.5:
Yes. ;

Ex.6. Let (X1, d1) and (X2, d2) be metric Spaces and X = X1 X2. Are  as defined below A metric on X?

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  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 in.
(d) The disk
Hint.8:
use Definition
Ans (a) The integer, (b) R, (c)

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.   be given.Since Y is dense in X and  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
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  which is a Cauchy sequence in (a,b) but does not converge.

Ex.12. Let   be the subspace consisting of all sequence  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  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
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  each of which has only finitely Nonzero terms, and  when y = (ηj). Show that

for j > n is Cauchy but does not converge.
Hint.16:
For every ∈ > 0, ,there is an N s. t. for n > m >N,
But (xn) does not converge to any  Because  so that for n >

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  with metric d(x,y) = |x - y| on R. .Clearly T is 1-1,into & bi continuous so   But R is complete while (-1,1) is an incomplete metric space.

Ex.18. If (X , d) is complete, show that  where  is complete.
Hint.18:

Hence if (xn) is Cauchy is  it is Cauchy in (X, d), and its limit in

Problems on Module-II (Normed and Banach Spaces)

Ex.1. Let (X,   i= 1, 2, ∞ be normed spaces of all ordered pairs
of real numbers where
are defined as

How does unit sphere in these norms look like?
Hint.1:

Ex.2. Show that the discrete metric on a vector space X ≠{0} can not be obtained from a norm.
Hint.2:

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:   where   has 0 value after j>n. Clearly

as well as x(n)∈ γ  but

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.

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

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   which is compact But    is not compact.

Ex.7. Show that  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  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 The null spaces are the the 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.

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.

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) = But 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 we have the strict inequality
Hint.15:
Since

Ex.16. Find the norm of the linear functional f defined on C[-1, 1] by

Hint.16:
For converse, choose x(t) = -1 on [-1,1]. So

Problems on Module III (IPS/Hilbert space)

Ex.1. If x ⊥ y in an IPS X,Show that
Hint.1: Use  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,

Ex.3. If an IPS X is real vector space, show that the condition   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> =   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⊥.

Ex.4. (Apollonius identity): For any elements x, y, z in an IPS X, show that

Hint 4:
Use  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  where α12 are scalars. Consider

.Similarly, one can show that   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.

Ex.7. Let X be the vector space of all ordered pairs of complex numbers. Can we obtain  the norm defined on X by   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 . Can we choose scalars γjk in a completely arbitrary fashion?
Hint.8:
Use
Open it so we get that it depends on

Ex.9. Show that for a sequence (xn ) in an IPS X , the conditions

Hint.9 :
We have

Ex.10. Show that in an IPS X,
for all scalars α.
Hint.10 :
From
condition follows as x ⊥ y.
Conversely,

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,  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].
where  Show that the identity mapping  of Xonto X is continuous. Is it Homeomorphism?
Hint.12 :
Since

Hence I is continuous.
Part-II: Answer No. because X2 is not complete.

Ex.13. Let H be a Hilbert space,   a convex subset, and (xn) is a sequence in M  such that ,  where   Show that (xn) converges in H.
Hint.13 :
(xn) is Cauchy, since from the assumption and the parallelogram equality, we have,

Ex.14. If (ek) is an orthonormal sequence in an IPS X, and x ∈ X, show that x-y with y given by  is orthogonal to the subspace
Hint.14 :

Ex.15. Let (ek) be any orthonormal sequence in an IPS X. Show that for any , x y ∈ X,
Hint.15: Use Cauchy Schwaz’s Inequality & Bessel’s Inequality, we get

Ex.-16. Show that in a Hilbert Space H,convergence of   implies convergence of
Hint.16 :
Let

is a Cauchy. Since H is complete, hence will converge.
converge in H.

Problems On Module IV (On Fundamental theorems)

Ex.1. Let  be a sequence of bounded linear functionals defined as  where . show that (fn)converge strongly to 0 but not uniformly.
Hint.1 : Since
ie
Ex.2. Let  where X is a Banach space and Y a normed space. If (Tn) is strongly  convergent with limit T, then
Hint.2
T linear follows

T is bounded :-  Since
So (Tnx)  is bounded for every x.Since X is complete, so  is bounded by uniform bounded ness theorem. Hence  Taking limit ⇒ T is bounded.

Ex.3. If . Show that (xn)  is point wise convergent on [a,b].
Hint .3 :
A bounded linear functional on

Ex.4. If  in a normed space X. Show that
Hint.4 :
Use Lemma:- ‘’Let Y be a proper closed sub-space of a normed space X and let  be arbitrary point and   then there exists an , dual of X such that  for all y ∈ Y  and
suppose  which is a closed sub space of X. so by the above result ,

hence there exists

Ex.5. Let Tn = Sn, where the operator  is defined by S  Find a bound for
Hint.5 :

For converse, choose x =
(iv)

Ex.6. Let X be a Banach space, Y a normed space and   such that (Tnx) is Cauchy in Y for every x ∈ X. .show that  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 theorem is bounded.

Ex.7. If (xn) in a Banach space X is such that   is bounded for all.Show that   is bounded.
Hint.7 :
Suppose  Thenis bounded for every . So by uniform bounded ness theorem is bounded and .

Ex.8. If a normed space X is reflexive, Show that X' is reflexive.
Hint. 8 :
Let  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  where 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  of norm1. show that
Hint. 9 :
suppose  Then by Lemma: Let X be a normed space and letbe any elementof X. Then there exist a bounded linear functional
would imply the existence of an

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  since Y is closed.
Use the Lemma as given in Ex 4 ( Hint ).
By this Lemma, there is on  which is zero on Y but not zero at x0, which contradicts our assumption.

Ex.11. Prove that Where Tis the adjoint operator of T.
Hint. 11 :

Similarly others.

Ex.12. Prove (ST)= TxSx
Hint. 12 :

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 , ( 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 :   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  is dense in X.
Hint. 16 :
The closure of  is all of X if and if  has no interior points, So that every  is a point of accumulation of

Ex.17. Show that completeness of X is essential in uniform bounded ness theorem and cannot be omitted.
Hint.17 :
Consider the sub space  consisting of all   where J depends on x, and let Tn be defined by  Clearly is bounded  is not bounded.

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