IGNOU MMT-006 Solved Assignment 2024, M.Sc. MACS

IGNOU MMT-006 Solved Assignment 2024 | M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MMT-006 Assignment Question Paper 2024

mmt-006-assignment-question-paper-aab71826-936e-4140-892f-f2b7ca3c6de5

mmt-006-assignment-question-paper-aab71826-936e-4140-892f-f2b7ca3c6de5

  1. a) Let X = { f C [ 0 , 1 ] : f ( 0 ) = 0 } X = { f C [ 0 , 1 ] : f ( 0 ) = 0 } X={f in C[0,1]:f(0)=0}X=\{f \in C[0,1]: f(0)=0\}X={fC[0,1]:f(0)=0}
Y = { g x : 0 1 g ( t ) d t = 0 } Y = g x : 0 1 g ( t ) d t = 0 Y={g in x:int_(0)^(1)g(t)dt=0}Y=\left\{g \in x: \int_0^1 g(t) d t=0\right\}Y={gx:01g(t)dt=0}
Prove that Y Y Y\mathrm{Y}Y is a proper subspace of X X X\mathrm{X}X. Is Y Y Y\mathrm{Y}Y a closed subspace of X X X\mathrm{X}X ? Justify your answer.
b) Let X = L p [ 0 , 1 ] X = L p [ 0 , 1 ] X=L^(p)[0,1]X=L^p[0,1]X=Lp[0,1] and x = x ( t ) = t 2 x = x ( t ) = t 2 x=x(t)=t^(2)x=x(t)=t^2x=x(t)=t2. Find x p x p ||x||_(p)\|x\|_pxp for p = 4 p = 4 p=4p=4p=4 and oo\infty.
c) Let E be a subset of a normed space X , Y = span E X , Y = span E X,Y=span EX, Y=\operatorname{span} EX,Y=spanE and a X a X a in Xa \in XaX. Show that a Y ¯ a Y ¯ a in bar(Y)a \in \bar{Y}aY¯ if and only if f ( a ) = 0 f ( a ) = 0 f(a)=0f(a)=0f(a)=0 whenever f X f X f inX^(‘)f \in X^{\prime}fX and f = 0 f = 0 f=0f=0f=0 everywhere on E E EEE.
2. a) Consider the space c 00 c 00 c_(00)c_{00}c00. For x = ( x 1 , x 2 , , x n , ) c 00 x = x 1 , x 2 , , x n , c 00 x=(x_(1),x_(2),dots,x_(n),dots)inc_(00)x=\left(x_1, x_2, \ldots, x_n, \ldots\right) \in c_{00}x=(x1,x2,,xn,)c00, define f ( x ) = n = 1 x n f ( x ) = n = 1 x n f(x)=sum_(n=1)^(oo)x_(n)f(x)=\sum_{n=1}^{\infty} x_nf(x)=n=1xn. Show that f f fff is a linear functional which is not continuous w.r.t the norm x = sup n | x n | x = sup n x n ||x||=s u p _(n)|x_(n)|\|x\|=\sup _n\left|x_n\right|x=supn|xn|.
b) Consider the space C 1 [ 0 , 1 ] C 1 [ 0 , 1 ] C^(1)[0,1]\mathrm{C}^1[0,1]C1[0,1] of all C 1 C 1 C^(1)\mathrm{C}^1C1 functions on [0,1] endowed with the uniform norm induced from the space C [ 0 , 1 ] C [ 0 , 1 ] C[0,1]\mathrm{C}[0,1]C[0,1], and consider the differential operator D : ( C 1 [ 0 , 1 ] , ) ( C [ 0 , 1 ] , ) D : C 1 [ 0 , 1 ] , C [ 0 , 1 ] , D:(C^(1)[0,1],||*||_(oo))rarr(C[0,1],||*||_(oo))D:\left(C^1[0,1],\|\cdot\|_{\infty}\right) \rightarrow\left(C[0,1],\|\cdot\|_{\infty}\right)D:(C1[0,1],)(C[0,1],) defined by D f = f D f = f Df=f^(‘)D f=f^{\prime}Df=f. Prove that D D DDD is linear, with closed graph, but not continuous. Can we conclude from here that C 1 [ 0 , 1 ] C 1 [ 0 , 1 ] C^(1)[0,1]C^1[0,1]C1[0,1] is not a Banach space? Justify your answer.
3. a) When is a normed linear space called separable? Show that a normed linear space is separable if its dual is separable [You should state all the proposition or theorems or corollaries used for proving the theorem]. Is the converse true? Give justification for your answer. [Whenever an example is given, you should justify that the example satisfies the requirements.]
b) Let X X X\mathrm{X}X be a Banach space, Y Y Y\mathrm{Y}Y be a normed linear space and F F F\mathcal{F}F be a subset of B ( X , Y ) B ( X , Y ) B(X,Y)\mathrm{B}(\mathrm{X}, \mathrm{Y})B(X,Y). If F F F\mathcal{F}F is not uniformly bounded, then there exists a dense subset D D D\mathrm{D}D of X X X\mathrm{X}X such that for every x D , { F ( x ) : F F } x D , { F ( x ) : F F } xinD,{F(x):FinF}\mathrm{x} \in \mathrm{D},\{\mathrm{F}(\mathrm{x}): \mathrm{F} \in \mathcal{F}\}xD,{F(x):FF} is not bounded in Y Y Y\mathrm{Y}Y.
4. a) Read the proof of the closed graph theorem carefully and explain where and how we have used the following facts in the proof.
i) X X X\mathrm{X}X is a Banach space.
ii) Y Y Y\mathrm{Y}Y is a Banach space.
iii) F is a closed map.
iv) Which property of continuity is being established to conclude that F F F\mathrm{F}F is continuous.
b) Which of the following maps are open? Give reasons for your answer.
i) T : R 3 R 2 T : R 3 R 2 quadT:R^(3)rarrR^(2)\quad \mathrm{T}: \mathbb{R}^3 \rightarrow \mathbb{R}^2T:R3R2 given by T ( x , y , z ) = ( x , z ) T ( x , y , z ) = ( x , z ) T(x,y,z)=(x,z)\mathrm{T}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\mathrm{x}, \mathrm{z})T(x,y,z)=(x,z).
ii) T : R 3 R 3 T : R 3 R 3 quadT:R^(3)rarrR^(3)\quad \mathrm{T}: \mathbb{R}^3 \rightarrow \mathbb{R}^3T:R3R3 given by T ( x , y , z ) = ( x , y , 0 ) T ( x , y , z ) = ( x , y , 0 ) T(x,y,z)=(x,y,0)\mathrm{T}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\mathrm{x}, \mathrm{y}, 0)T(x,y,z)=(x,y,0).
5. a) Let f : C [ 0 , 1 ] R f : C [ 0 , 1 ] R f:C[0,1]rarrRf: C[0,1] \rightarrow \mathbb{R}f:C[0,1]R be given by f ( x ) = x ( 1 ) x C [ 0 , 1 ] f ( x ) = x ( 1 ) x C [ 0 , 1 ] f(x)=x(1)AA x in C[0,1]f(x)=x(1) \forall x \in C[0,1]f(x)=x(1)xC[0,1]. Show that f f fff is continuous w.r.t the supnorm and f f fff is not continuous w.r.t the p-norm.
b) Let X X X\mathrm{X}X be an inner product space and x , y X x , y X x,yinX\mathrm{x}, \mathrm{y} \in \mathrm{X}x,yX. Prove that x y x y x_|_y\mathrm{x} \perp \mathrm{y}xy if and only if k x + y 2 = k x 2 + y 2 , k K k x + y 2 = k x 2 + y 2 , k K ||kx+y||^(2)=||kx||^(2)+||y^(2)||,kinK\|\mathrm{kx}+\mathrm{y}\|^2=\|\mathrm{kx}\|^2+\left\|\mathrm{y}^2\right\|, \mathrm{k} \in \mathrm{K}kx+y2=kx2+y2,kK.
6. a) Let H = R 3 H = R 3 H=R^(3)\mathrm{H}=\mathrm{R}^3H=R3 and F F F\mathrm{F}F be the set of all x = ( x 1 , x 2 , x 3 ) x = x 1 , x 2 , x 3 x=(x_(1),x_(2),x_(3))\mathbf{x}=\left(\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3\right)x=(x1,x2,x3) in H H H\mathrm{H}H such that x 1 = 0 x 1 = 0 x_(1)=0\mathrm{x}_1=0x1=0. Find F F F^(_|_)\mathrm{F}^{\perp}F. Verify that every x H x H xinH\mathbf{x} \in \mathrm{H}xH can be expressed as x = y + z x = y + z x=y+z\mathbf{x}=\mathbf{y}+\mathbf{z}x=y+z where y F y F yinF\mathbf{y} \in \mathrm{F}yF and z F z F zinF^(_|_)\mathbf{z} \in \mathrm{F}^{\perp}zF.
b) Given an example of an Hilbert space H H H\mathrm{H}H and an operator A A A\mathrm{A}A on H H H\mathrm{H}H such that σ e ( A ) σ e ( A ) sigma_(e)(A)\sigma_{\mathrm{e}}(\mathrm{A})σe(A) is empty. Justify your choice of example.
c) Let A A AAA be a normal operator on a Hilbert space X X XXX. Show that σ ( A ) σ a ( A ) σ ( A ) σ a ( A ) sigma(A)subsigma _(a)(A)\sigma(A) \subset \sigma_a(A)σ(A)σa(A) where σ a ( A ) σ a ( A ) sigma _(a)(A)\sigma_a(A)σa(A) denotes the approximate eigen spectrum of A A AAA and σ ( A ) σ ( A ) sigma(A)\sigma(A)σ(A) denotes the spectrum of A.
7. a) Let X = c 00 X = c 00 X=c_(00)X=c_{00}X=c00 with p p ||*||_(p)\|\cdot\|_pp. Give an example of a Cauchy sequence in X X XXX that do not converge in X. Justify your choice of example.
b) Give one example of each of the following. Also justify your choice of example.
i) A self-adjoint operator on 2 2 ℓ^(2)\ell^22.
ii) A normal operator on a Hilbert space which is not unitary.
c) Let X X X\mathrm{X}X be a normed space and Y Y Y\mathrm{Y}Y be proper subspace of X X X\mathrm{X}X. Show that the interior Y 0 Y 0 Y^(0)\mathrm{Y}^0Y0 of Y Y Y\mathrm{Y}Y is empty.
8. a) Let X , Y X , Y X,YX, YX,Y be normed spaces and suppose BL(X,Y) and CL(X,Y) denote, respectively, the space of bounded linear operators from X X X\mathrm{X}X to Y Y Y\mathrm{Y}Y and the space of compact linear operators from X to Y. Show that CL(X,Y) is linear subspace of BL(X,Y). Also, Show that if Y Y Y\mathrm{Y}Y is a Banach space, then C L ( X , Y ) C L ( X , Y ) CL(X,Y)\mathrm{CL}(\mathrm{X}, \mathrm{Y})CL(X,Y) is a closed subspace of B L ( X , Y ) B L ( X , Y ) BL(X,Y)\mathrm{BL}(\mathrm{X}, \mathrm{Y})BL(X,Y).
b) Define a Hilbert-Schmidt operator on a Hilbert space H H H\mathrm{H}H and give an example. Is every Hilbert-sehmidt operator a compact operator? Justify your answer.
9. a) Let { A n } A n {A_(n)}\left\{A_n\right\}{An} be a sequence of unitary operators in B L ( H ) B L ( H ) BL(H)B L(H)BL(H). Prove that if A n A 0 , A B L ( H ) A n A 0 , A B L ( H ) ||A_(n)-A||rarr0,A in BL(H)\left\|A_n-A\right\| \rightarrow 0, A \in B L(H)AnA0,ABL(H), then A A AAA is unitary.
b) Define the spectral radius of a bounded linear operator A B L ( X ) A B L ( X ) A in BL(X)A \in B L(X)ABL(X). Find the spectral radius of A A A\mathrm{A}A in B L ( R 3 ) B L R 3 BL(R^(3))\mathrm{BL}\left(\mathbb{R}^3\right)BL(R3), where A A A\mathrm{A}A is given by the matrix
[ 0 1 0 1 0 0 0 0 1 ] 0      1      0 1      0      0 0      0      1 [[0,1,0],[-1,0,0],[0,0,-1]]\left[\begin{array}{rrr} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{array}\right][010100001]
with respect to the standard basis of R 3 R 3 R^(3)\mathbb{R}^3R3.
c) Let X X X\mathrm{X}X be a Banach space and Y Y Y\mathrm{Y}Y be a closed subspace of X X X\mathrm{X}X. Let π : X X / Y π : X X / Y pi:XrarrX//Y\pi: \mathrm{X} \rightarrow \mathrm{X} / \mathrm{Y}π:XX/Y be canonical quotient map. Show that π π pi\piπ is open.
10. State giving reasons, if the following statement are true or false.
a) A closed map on a normed space need not be an open map.
b) c 00 c 00 c_(00)\mathrm{c}_{00}c00 is a closed subspace of ℓ^(oo)\ell^{\infty}.
c) The dual of a finite dimensional space is finite dimensional.
d) If T 1 T 1 T_(1)T_1T1 and T 2 T 2 T_(2)T_2T2 are positive operators on a Hilbert space H H HHH, then T 1 + T 2 T 1 + T 2 T_(1)+T_(2)T_1+T_2T1+T2 is a positive operator on H H H\mathrm{H}H.
e) On a normed space X X X\mathrm{X}X, the norm function : X C : X C ||*||:XrarrC\|\cdot\|: \mathrm{X} \rightarrow \mathbb{C}:XC is a linear map.
\(cos\left(\theta +\phi \right)=cos\:\theta \:cos\:\phi -sin\:\theta \:sin\:\phi \)

MMT-006 Sample Solution 2024

mmt-006-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

mmt-006-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

  1. a) Let X = { f C [ 0 , 1 ] : f ( 0 ) = 0 } X = { f C [ 0 , 1 ] : f ( 0 ) = 0 } X={f in C[0,1]:f(0)=0}X=\{f \in C[0,1]: f(0)=0\}X={fC[0,1]:f(0)=0}
Y = { g x : 0 1 g ( t ) d t = 0 } Y = g x : 0 1 g ( t ) d t = 0 Y={g in x:int_(0)^(1)g(t)dt=0}Y=\left\{g \in x: \int_0^1 g(t) d t=0\right\}Y={gx:01g(t)dt=0}
Prove that Y Y Y\mathrm{Y}Y is a proper subspace of X X X\mathrm{X}X. Is Y Y Y\mathrm{Y}Y a closed subspace of X X X\mathrm{X}X ? Justify your answer.
Answer:
To prove that Y Y YYY is a proper subspace of X X XXX and determine whether Y Y YYY is a closed subspace of X X XXX, we need to examine the definitions and properties of these sets within the context of functional analysis.

X X XXX and Y Y YYY Defined

  • X = { f C [ 0 , 1 ] : f ( 0 ) = 0 } X = { f C [ 0 , 1 ] : f ( 0 ) = 0 } X={f in C[0,1]:f(0)=0}X = \{f \in C[0,1]: f(0) = 0\}X={fC[0,1]:f(0)=0} is the set of all continuous functions on the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1] that vanish at 0 0 000.
  • Y = { g X : 0 1 g ( t ) d t = 0 } Y = g X : 0 1 g ( t ) d t = 0 Y={g in X:int_(0)^(1)g(t)dt=0}Y = \left\{g \in X: \int_0^1 g(t) dt = 0\right\}Y={gX:01g(t)dt=0} is the set of all functions in X X XXX whose integral over [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1] is 0 0 000.

Proving Y Y YYY is a Proper Subspace of X X XXX

To show that Y Y YYY is a proper subspace of X X XXX, we must verify that Y Y YYY satisfies the following criteria for being a subspace:
  1. Non-emptiness: Y Y YYY contains the zero function, which is the function g ( t ) = 0 g ( t ) = 0 g(t)=0g(t) = 0g(t)=0 for all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0,1]t[0,1]. This function clearly belongs to X X XXX and satisfies the integral condition, so Y Y YYY is non-empty.
  2. Closed under addition: If g 1 , g 2 Y g 1 , g 2 Y g_(1),g_(2)in Yg_1, g_2 \in Yg1,g2Y, then g 1 + g 2 Y g 1 + g 2 Y g_(1)+g_(2)in Yg_1 + g_2 \in Yg1+g2Y. This is because the integral of the sum is the sum of the integrals, each of which is 0 0 000, so their sum is also 0 0 000.
  3. Closed under scalar multiplication: If g Y g Y g in Yg \in YgY and α α alpha\alphaα is a scalar, then α g Y α g Y alpha g in Y\alpha g \in YαgY. This follows because 0 1 α g ( t ) d t = α 0 1 g ( t ) d t = α 0 = 0 0 1 α g ( t ) d t = α 0 1 g ( t ) d t = α 0 = 0 int_(0)^(1)alpha g(t)dt=alphaint_(0)^(1)g(t)dt=alpha*0=0\int_0^1 \alpha g(t) dt = \alpha \int_0^1 g(t) dt = \alpha \cdot 0 = 001αg(t)dt=α01g(t)dt=α0=0.
Since Y Y YYY satisfies these criteria, it is a subspace of X X XXX. It is a proper subspace because there exist functions in X X XXX that do not satisfy the integral condition, such as f ( t ) = t f ( t ) = t f(t)=tf(t) = tf(t)=t, which is in X X XXX but not in Y Y YYY since 0 1 t d t = 1 / 2 0 0 1 t d t = 1 / 2 0 int_(0)^(1)tdt=1//2!=0\int_0^1 t dt = 1/2 \neq 001tdt=1/20.

Is Y Y YYY a Closed Subspace of X X XXX?

A subspace Y Y YYY is closed in X X XXX if it contains all its limit points; that is, if a sequence of functions { g n } { g n } {g_(n)}\{g_n\}{gn} in Y Y YYY converges uniformly to a function g g ggg, then g g ggg must also be in Y Y YYY.
To show Y Y YYY is closed, consider a sequence { g n } Y { g n } Y {g_(n)}sub Y\{g_n\} \subset Y{gn}Y that converges uniformly to g X g X g in Xg \in XgX. We need to show that g Y g Y g in Yg \in YgY, meaning 0 1 g ( t ) d t = 0 0 1 g ( t ) d t = 0 int_(0)^(1)g(t)dt=0\int_0^1 g(t) dt = 001g(t)dt=0.
Uniform convergence of { g n } { g n } {g_(n)}\{g_n\}{gn} to g g ggg implies that for every ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists an N N NNN such that for all n N n N n >= Nn \geq NnN, we have | g n ( t ) g ( t ) | < ϵ | g n ( t ) g ( t ) | < ϵ |g_(n)(t)-g(t)| < epsilon|g_n(t) – g(t)| < \epsilon|gn(t)g(t)|<ϵ for all t [ 0 , 1 ] t [ 0 , 1 ] t in[0,1]t \in [0,1]t[0,1]. By the properties of integrals and the limit of a sequence of functions,
lim n 0 1 g n ( t ) d t = 0 1 lim n g n ( t ) d t = 0 1 g ( t ) d t . lim n 0 1 g n ( t ) d t = 0 1 lim n g n ( t ) d t = 0 1 g ( t ) d t . lim_(n rarr oo)int_(0)^(1)g_(n)(t)dt=int_(0)^(1)lim_(n rarr oo)g_(n)(t)dt=int_(0)^(1)g(t)dt.\lim_{n \to \infty} \int_0^1 g_n(t) dt = \int_0^1 \lim_{n \to \infty} g_n(t) dt = \int_0^1 g(t) dt.limn01gn(t)dt=01limngn(t)dt=01g(t)dt.
Since each g n Y g n Y g_(n)in Yg_n \in YgnY, we have 0 1 g n ( t ) d t = 0 0 1 g n ( t ) d t = 0 int_(0)^(1)g_(n)(t)dt=0\int_0^1 g_n(t) dt = 001gn(t)dt=0. Thus, the limit of these integrals as n n n rarr oon \to \inftyn is also 0 0 000, which means 0 1 g ( t ) d t = 0 0 1 g ( t ) d t = 0 int_(0)^(1)g(t)dt=0\int_0^1 g(t) dt = 001g(t)dt=0, and hence g Y g Y g in Yg \in YgY.
Therefore, Y Y YYY is a closed subspace of X X XXX because it contains all its limit points under uniform convergence.
b) Let X = L p [ 0 , 1 ] X = L p [ 0 , 1 ] X=L^(p)[0,1]X=L^p[0,1]X=Lp[0,1] and x = x ( t ) = t 2 x = x ( t ) = t 2 x=x(t)=t^(2)x=x(t)=t^2x=x(t)=t2. Find x p x p ||x||_(p)\|x\|_pxp for p = 4 p = 4 p=4p=4p=4 and oo\infty.
Answer:
To find the L p L p L^(p)L^pLp norm of x ( t ) = t 2 x ( t ) = t 2 x(t)=t^(2)x(t) = t^2x(t)=t2 on the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1] for p = 4 p = 4 p=4p=4p=4 and p = p = p=oop=\inftyp=, we’ll use the definitions of the L p L p L^(p)L^pLp norms.

For p = 4 p = 4 p=4p=4p=4

The L p L p L^(p)L^pLp norm for p = 4 p = 4 p=4p=4p=4 is defined as
x 4 = ( 0 1 | t 2 | 4 d t ) 1 / 4 x 4 = 0 1 | t 2 | 4 d t 1 / 4 ||x||_(4)=(int_(0)^(1)|t^(2)|^(4)dt)^(1//4)\|x\|_4 = \left( \int_0^1 |t^2|^4 dt \right)^{1/4}x4=(01|t2|4dt)1/4
This simplifies to
x 4 = ( 0 1 t 8 d t ) 1 / 4 x 4 = 0 1 t 8 d t 1 / 4 ||x||_(4)=(int_(0)^(1)t^(8)dt)^(1//4)\|x\|_4 = \left( \int_0^1 t^8 dt \right)^{1/4}x4=(01t8dt)1/4
To compute this integral, we use the formula for the integral of t n t n t^(n)t^ntn, which is t n + 1 n + 1 t n + 1 n + 1 (t^(n+1))/(n+1)\frac{t^{n+1}}{n+1}tn+1n+1 for n 1 n 1 n!=-1n \neq -1n1, from 0 0 000 to 1 1 111:
0 1 t 8 d t = t 9 9 | 0 1 = 1 9 0 1 t 8 d t = t 9 9 | 0 1 = 1 9 int_(0)^(1)t^(8)dt=(t^(9))/(9)|_(0)^(1)=(1)/(9)\int_0^1 t^8 dt = \frac{t^9}{9} \Big|_0^1 = \frac{1}{9}