# 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=\left\{f\in C\left[0,1\right]:f\left(0\right)=0\right\}$$X=\left\{f\in C\left[0,1\right]:f\left(0\right)=0\right\}$X={f in C[0,1]:f(0)=0}X=\{f \in C[0,1]: f(0)=0\}$X=\left\{f\in C\left[0,1\right]:f\left(0\right)=0\right\}$
$Y=\left\{g\in x:{\int }_{0}^{1}g\left(t\right)dt=0\right\}$$Y=\left\{g\in x:{\int }_{0}^{1} g\left(t\right)dt=0\right\}$Y={g in x:int_(0)^(1)g(t)dt=0}Y=\left\{g \in x: \int_0^1 g(t) d t=0\right\}$Y=\left\{g\in x:{\int }_{0}^{1}g\left(t\right)dt=0\right\}$
Prove that $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ is a proper subspace of $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$. Is $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ a closed subspace of $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ ? Justify your answer.
b) Let $X={L}^{p}\left[0,1\right]$$X={L}^{p}\left[0,1\right]$X=L^(p)[0,1]X=L^p[0,1]$X={L}^{p}\left[0,1\right]$ and $x=x\left(t\right)={t}^{2}$$x=x\left(t\right)={t}^{2}$x=x(t)=t^(2)x=x(t)=t^2$x=x\left(t\right)={t}^{2}$. Find $\parallel x{\parallel }_{p}$$\parallel x{\parallel }_{p}$||x||_(p)\|x\|_p$\parallel x{\parallel }_{p}$ for $p=4$$p=4$p=4p=4$p=4$ and $\mathrm{\infty }$$\mathrm{\infty }$oo\infty$\mathrm{\infty }$.
c) Let E be a subset of a normed space $X,Y=\mathrm{span}E$$X,Y=\mathrm{span}E$X,Y=span EX, Y=\operatorname{span} E$X,Y=\mathrm{span}E$ and $a\in X$$a\in X$a in Xa \in X$a\in X$. Show that $a\in \overline{Y}$$a\in \overline{Y}$a in bar(Y)a \in \bar{Y}$a\in \overline{Y}$ if and only if $f\left(a\right)=0$$f\left(a\right)=0$f(a)=0f(a)=0$f\left(a\right)=0$ whenever $f\in {X}^{\mathrm{\prime }}$$f\in {X}^{\mathrm{\prime }}$f inX^(‘)f \in X^{\prime}$f\in {X}^{\mathrm{\prime }}$ and $f=0$$f=0$f=0f=0$f=0$ everywhere on $E$$E$EE$E$.
2. a) Consider the space ${c}_{00}$${c}_{00}$c_(00)c_{00}${c}_{00}$. For $x=\left({x}_{1},{x}_{2},\dots ,{x}_{n},\dots \right)\in {c}_{00}$$x=\left({x}_{1},{x}_{2},\dots ,{x}_{n},\dots \right)\in {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=\left({x}_{1},{x}_{2},\dots ,{x}_{n},\dots \right)\in {c}_{00}$, define $f\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{x}_{n}$$f\left(x\right)=\sum _{n=1}^{\mathrm{\infty }} {x}_{n}$f(x)=sum_(n=1)^(oo)x_(n)f(x)=\sum_{n=1}^{\infty} x_n$f\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{x}_{n}$. Show that $f$$f$ff$f$ is a linear functional which is not continuous w.r.t the norm $\parallel x\parallel =\underset{n}{sup}|{x}_{n}|$$\parallel x\parallel =\underset{n}{sup} \left|{x}_{n}\right|$||x||=s u p _(n)|x_(n)|\|x\|=\sup _n\left|x_n\right|$\parallel x\parallel =\underset{n}{sup}|{x}_{n}|$.
b) Consider the space ${\mathrm{C}}^{1}\left[0,1\right]$${\mathrm{C}}^{1}\left[0,1\right]$C^(1)[0,1]\mathrm{C}^1[0,1]${\mathrm{C}}^{1}\left[0,1\right]$ of all ${\mathrm{C}}^{1}$${\mathrm{C}}^{1}$C^(1)\mathrm{C}^1${\mathrm{C}}^{1}$ functions on [0,1] endowed with the uniform norm induced from the space $\mathrm{C}\left[0,1\right]$$\mathrm{C}\left[0,1\right]$C[0,1]\mathrm{C}[0,1]$\mathrm{C}\left[0,1\right]$, and consider the differential operator $D:\left({C}^{1}\left[0,1\right],\parallel \cdot {\parallel }_{\mathrm{\infty }}\right)\to \left(C\left[0,1\right],\parallel \cdot {\parallel }_{\mathrm{\infty }}\right)$$D:\left({C}^{1}\left[0,1\right],\parallel \cdot {\parallel }_{\mathrm{\infty }}\right)\to \left(C\left[0,1\right],\parallel \cdot {\parallel }_{\mathrm{\infty }}\right)$D:(C^(1)[0,1],||*||_(oo))rarr(C[0,1],||*||_(oo))D:\left(C^1[0,1],\|\cdot\|_{\infty}\right) \rightarrow\left(C[0,1],\|\cdot\|_{\infty}\right)$D:\left({C}^{1}\left[0,1\right],\parallel \cdot {\parallel }_{\mathrm{\infty }}\right)\to \left(C\left[0,1\right],\parallel \cdot {\parallel }_{\mathrm{\infty }}\right)$ defined by $Df={f}^{\mathrm{\prime }}$$Df={f}^{\mathrm{\prime }}$Df=f^(‘)D f=f^{\prime}$Df={f}^{\mathrm{\prime }}$. Prove that $D$$D$DD$D$ is linear, with closed graph, but not continuous. Can we conclude from here that ${C}^{1}\left[0,1\right]$${C}^{1}\left[0,1\right]$C^(1)[0,1]C^1[0,1]${C}^{1}\left[0,1\right]$ 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 $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ be a Banach space, $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ be a normed linear space and $\mathcal{F}$$\mathcal{F}$F\mathcal{F}$\mathcal{F}$ be a subset of $\mathrm{B}\left(\mathrm{X},\mathrm{Y}\right)$$\mathrm{B}\left(\mathrm{X},\mathrm{Y}\right)$B(X,Y)\mathrm{B}(\mathrm{X}, \mathrm{Y})$\mathrm{B}\left(\mathrm{X},\mathrm{Y}\right)$. If $\mathcal{F}$$\mathcal{F}$F\mathcal{F}$\mathcal{F}$ is not uniformly bounded, then there exists a dense subset $\mathrm{D}$$\mathrm{D}$D\mathrm{D}$\mathrm{D}$ of $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ such that for every $\mathrm{x}\in \mathrm{D},\left\{\mathrm{F}\left(\mathrm{x}\right):\mathrm{F}\in \mathcal{F}\right\}$$\mathrm{x}\in \mathrm{D},\left\{\mathrm{F}\left(\mathrm{x}\right):\mathrm{F}\in \mathcal{F}\right\}$xinD,{F(x):FinF}\mathrm{x} \in \mathrm{D},\{\mathrm{F}(\mathrm{x}): \mathrm{F} \in \mathcal{F}\}$\mathrm{x}\in \mathrm{D},\left\{\mathrm{F}\left(\mathrm{x}\right):\mathrm{F}\in \mathcal{F}\right\}$ is not bounded in $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$.
4. a) Read the proof of the closed graph theorem carefully and explain where and how we have used the following facts in the proof.
i) $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ is a Banach space.
ii) $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ is a Banach space.
iii) F is a closed map.
iv) Which property of continuity is being established to conclude that $\mathrm{F}$$\mathrm{F}$F\mathrm{F}$\mathrm{F}$ is continuous.
b) Which of the following maps are open? Give reasons for your answer.
i) $\phantom{\rule{1em}{0ex}}\mathrm{T}:{\mathbb{R}}^{3}\to {\mathbb{R}}^{2}$$\phantom{\rule{1em}{0ex}}\mathrm{T}:{\mathbb{R}}^{3}\to {\mathbb{R}}^{2}$quadT:R^(3)rarrR^(2)\quad \mathrm{T}: \mathbb{R}^3 \rightarrow \mathbb{R}^2$\phantom{\rule{1em}{0ex}}\mathrm{T}:{\mathbb{R}}^{3}\to {\mathbb{R}}^{2}$ given by $\mathrm{T}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{x},\mathrm{z}\right)$$\mathrm{T}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{x},\mathrm{z}\right)$T(x,y,z)=(x,z)\mathrm{T}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\mathrm{x}, \mathrm{z})$\mathrm{T}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{x},\mathrm{z}\right)$.
ii) $\phantom{\rule{1em}{0ex}}\mathrm{T}:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$$\phantom{\rule{1em}{0ex}}\mathrm{T}:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$quadT:R^(3)rarrR^(3)\quad \mathrm{T}: \mathbb{R}^3 \rightarrow \mathbb{R}^3$\phantom{\rule{1em}{0ex}}\mathrm{T}:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$ given by $\mathrm{T}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{x},\mathrm{y},0\right)$$\mathrm{T}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{x},\mathrm{y},0\right)$T(x,y,z)=(x,y,0)\mathrm{T}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(\mathrm{x}, \mathrm{y}, 0)$\mathrm{T}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{x},\mathrm{y},0\right)$.
5. a) Let $f:C\left[0,1\right]\to \mathbb{R}$$f:C\left[0,1\right]\to \mathbb{R}$f:C[0,1]rarrRf: C[0,1] \rightarrow \mathbb{R}$f:C\left[0,1\right]\to \mathbb{R}$ be given by $f\left(x\right)=x\left(1\right)\mathrm{\forall }x\in C\left[0,1\right]$$f\left(x\right)=x\left(1\right)\mathrm{\forall }x\in C\left[0,1\right]$f(x)=x(1)AA x in C[0,1]f(x)=x(1) \forall x \in C[0,1]$f\left(x\right)=x\left(1\right)\mathrm{\forall }x\in C\left[0,1\right]$. Show that $f$$f$ff$f$ is continuous w.r.t the supnorm and $f$$f$ff$f$ is not continuous w.r.t the p-norm.
b) Let $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ be an inner product space and $\mathrm{x},\mathrm{y}\in \mathrm{X}$$\mathrm{x},\mathrm{y}\in \mathrm{X}$x,yinX\mathrm{x}, \mathrm{y} \in \mathrm{X}$\mathrm{x},\mathrm{y}\in \mathrm{X}$. Prove that $\mathrm{x}\perp \mathrm{y}$$\mathrm{x}\perp \mathrm{y}$x_|_y\mathrm{x} \perp \mathrm{y}$\mathrm{x}\perp \mathrm{y}$ if and only if $\parallel \mathrm{k}\mathrm{x}+\mathrm{y}{\parallel }^{2}=\parallel \mathrm{k}\mathrm{x}{\parallel }^{2}+\parallel {\mathrm{y}}^{2}\parallel ,\mathrm{k}\in \mathrm{K}$$\parallel \mathrm{k}\mathrm{x}+\mathrm{y}{\parallel }^{2}=\parallel \mathrm{k}\mathrm{x}{\parallel }^{2}+∥{\mathrm{y}}^{2}∥,\mathrm{k}\in \mathrm{K}$||kx+y||^(2)=||kx||^(2)+||y^(2)||,kinK\|\mathrm{kx}+\mathrm{y}\|^2=\|\mathrm{kx}\|^2+\left\|\mathrm{y}^2\right\|, \mathrm{k} \in \mathrm{K}$\parallel \mathrm{k}\mathrm{x}+\mathrm{y}{\parallel }^{2}=\parallel \mathrm{k}\mathrm{x}{\parallel }^{2}+\parallel {\mathrm{y}}^{2}\parallel ,\mathrm{k}\in \mathrm{K}$.
6. a) Let $\mathrm{H}={\mathrm{R}}^{3}$$\mathrm{H}={\mathrm{R}}^{3}$H=R^(3)\mathrm{H}=\mathrm{R}^3$\mathrm{H}={\mathrm{R}}^{3}$ and $\mathrm{F}$$\mathrm{F}$F\mathrm{F}$\mathrm{F}$ be the set of all $\mathbf{x}=\left({\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3}\right)$$\mathbf{x}=\left({\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3}\right)$x=(x_(1),x_(2),x_(3))\mathbf{x}=\left(\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3\right)$\mathbf{x}=\left({\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3}\right)$ in $\mathrm{H}$$\mathrm{H}$H\mathrm{H}$\mathrm{H}$ such that ${\mathrm{x}}_{1}=0$${\mathrm{x}}_{1}=0$x_(1)=0\mathrm{x}_1=0${\mathrm{x}}_{1}=0$. Find ${\mathrm{F}}^{\perp }$${\mathrm{F}}^{\perp }$F^(_|_)\mathrm{F}^{\perp}${\mathrm{F}}^{\perp }$. Verify that every $\mathbf{x}\in \mathrm{H}$$\mathbf{x}\in \mathrm{H}$xinH\mathbf{x} \in \mathrm{H}$\mathbf{x}\in \mathrm{H}$ can be expressed as $\mathbf{x}=\mathbf{y}+\mathbf{z}$$\mathbf{x}=\mathbf{y}+\mathbf{z}$x=y+z\mathbf{x}=\mathbf{y}+\mathbf{z}$\mathbf{x}=\mathbf{y}+\mathbf{z}$ where $\mathbf{y}\in \mathrm{F}$$\mathbf{y}\in \mathrm{F}$yinF\mathbf{y} \in \mathrm{F}$\mathbf{y}\in \mathrm{F}$ and $\mathbf{z}\in {\mathrm{F}}^{\perp }$$\mathbf{z}\in {\mathrm{F}}^{\perp }$zinF^(_|_)\mathbf{z} \in \mathrm{F}^{\perp}$\mathbf{z}\in {\mathrm{F}}^{\perp }$.
b) Given an example of an Hilbert space $\mathrm{H}$$\mathrm{H}$H\mathrm{H}$\mathrm{H}$ and an operator $\mathrm{A}$$\mathrm{A}$A\mathrm{A}$\mathrm{A}$ on $\mathrm{H}$$\mathrm{H}$H\mathrm{H}$\mathrm{H}$ such that ${\sigma }_{\mathrm{e}}\left(\mathrm{A}\right)$${\sigma }_{\mathrm{e}}\left(\mathrm{A}\right)$sigma_(e)(A)\sigma_{\mathrm{e}}(\mathrm{A})${\sigma }_{\mathrm{e}}\left(\mathrm{A}\right)$ is empty. Justify your choice of example.
c) Let $A$$A$AA$A$ be a normal operator on a Hilbert space $X$$X$XX$X$. Show that $\sigma \left(A\right)\subset {\sigma }_{a}\left(A\right)$$\sigma \left(A\right)\subset {\sigma }_{a}\left(A\right)$sigma(A)subsigma _(a)(A)\sigma(A) \subset \sigma_a(A)$\sigma \left(A\right)\subset {\sigma }_{a}\left(A\right)$ where ${\sigma }_{a}\left(A\right)$${\sigma }_{a}\left(A\right)$sigma _(a)(A)\sigma_a(A)${\sigma }_{a}\left(A\right)$ denotes the approximate eigen spectrum of $A$$A$AA$A$ and $\sigma \left(A\right)$$\sigma \left(A\right)$sigma(A)\sigma(A)$\sigma \left(A\right)$ denotes the spectrum of A.
7. a) Let $X={c}_{00}$$X={c}_{00}$X=c_(00)X=c_{00}$X={c}_{00}$ with $\parallel \cdot {\parallel }_{p}$$\parallel \cdot {\parallel }_{p}$||*||_(p)\|\cdot\|_p$\parallel \cdot {\parallel }_{p}$. Give an example of a Cauchy sequence in $X$$X$XX$X$ that do not converge in X. Justify your choice of example.
b) Give one example of each of the following. Also justify your choice of example.
i) A self-adjoint operator on ${\ell }^{2}$${\ell }^{2}$ℓ^(2)\ell^2${\ell }^{2}$.
ii) A normal operator on a Hilbert space which is not unitary.
c) Let $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ be a normed space and $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ be proper subspace of $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$. Show that the interior ${\mathrm{Y}}^{0}$${\mathrm{Y}}^{0}$Y^(0)\mathrm{Y}^0${\mathrm{Y}}^{0}$ of $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ is empty.
8. a) Let $X,Y$$X,Y$X,YX, Y$X,Y$ be normed spaces and suppose BL(X,Y) and CL(X,Y) denote, respectively, the space of bounded linear operators from $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ to $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{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 $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ is a Banach space, then $\mathrm{C}\mathrm{L}\left(\mathrm{X},\mathrm{Y}\right)$$\mathrm{C}\mathrm{L}\left(\mathrm{X},\mathrm{Y}\right)$CL(X,Y)\mathrm{CL}(\mathrm{X}, \mathrm{Y})$\mathrm{C}\mathrm{L}\left(\mathrm{X},\mathrm{Y}\right)$ is a closed subspace of $\mathrm{B}\mathrm{L}\left(\mathrm{X},\mathrm{Y}\right)$$\mathrm{B}\mathrm{L}\left(\mathrm{X},\mathrm{Y}\right)$BL(X,Y)\mathrm{BL}(\mathrm{X}, \mathrm{Y})$\mathrm{B}\mathrm{L}\left(\mathrm{X},\mathrm{Y}\right)$.
b) Define a Hilbert-Schmidt operator on a Hilbert space $\mathrm{H}$$\mathrm{H}$H\mathrm{H}$\mathrm{H}$ and give an example. Is every Hilbert-sehmidt operator a compact operator? Justify your answer.
9. a) Let $\left\{{A}_{n}\right\}$$\left\{{A}_{n}\right\}${A_(n)}\left\{A_n\right\}$\left\{{A}_{n}\right\}$ be a sequence of unitary operators in $BL\left(H\right)$$BL\left(H\right)$BL(H)B L(H)$BL\left(H\right)$. Prove that if $\parallel {A}_{n}-A\parallel \to 0,A\in BL\left(H\right)$$∥{A}_{n}-A∥\to 0,A\in BL\left(H\right)$||A_(n)-A||rarr0,A in BL(H)\left\|A_n-A\right\| \rightarrow 0, A \in B L(H)$\parallel {A}_{n}-A\parallel \to 0,A\in BL\left(H\right)$, then $A$$A$AA$A$ is unitary.
b) Define the spectral radius of a bounded linear operator $A\in BL\left(X\right)$$A\in BL\left(X\right)$A in BL(X)A \in B L(X)$A\in BL\left(X\right)$. Find the spectral radius of $\mathrm{A}$$\mathrm{A}$A\mathrm{A}$\mathrm{A}$ in $\mathrm{B}\mathrm{L}\left({\mathbb{R}}^{3}\right)$$\mathrm{B}\mathrm{L}\left({\mathbb{R}}^{3}\right)$BL(R^(3))\mathrm{BL}\left(\mathbb{R}^3\right)$\mathrm{B}\mathrm{L}\left({\mathbb{R}}^{3}\right)$, where $\mathrm{A}$$\mathrm{A}$A\mathrm{A}$\mathrm{A}$ is given by the matrix
$\left[\begin{array}{rrr}0& 1& 0\\ -1& 0& 0\\ 0& 0& -1\end{array}\right]$$\left[\begin{array}{r}0 1 0\\ -1 0 0\\ 0 0 -1\end{array}\right]$[[0,1,0],[-1,0,0],[0,0,-1]]\left[\begin{array}{rrr} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1 \end{array}\right]$\left[\begin{array}{rrr}0& 1& 0\\ -1& 0& 0\\ 0& 0& -1\end{array}\right]$
with respect to the standard basis of ${\mathbb{R}}^{3}$${\mathbb{R}}^{3}$R^(3)\mathbb{R}^3${\mathbb{R}}^{3}$.
c) Let $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ be a Banach space and $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ be a closed subspace of $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$. Let $\pi :\mathrm{X}\to \mathrm{X}/\mathrm{Y}$$\pi :\mathrm{X}\to \mathrm{X}/\mathrm{Y}$pi:XrarrX//Y\pi: \mathrm{X} \rightarrow \mathrm{X} / \mathrm{Y}$\pi :\mathrm{X}\to \mathrm{X}/\mathrm{Y}$ be canonical quotient map. Show that $\pi$$\pi$pi\pi$\pi$ is open.
10. State giving reasons, if the following statement are true or false.
a) A closed map on a normed space need not be an open map.
b) ${\mathrm{c}}_{00}$${\mathrm{c}}_{00}$c_(00)\mathrm{c}_{00}${\mathrm{c}}_{00}$ is a closed subspace of ${\ell }^{\mathrm{\infty }}$${\ell }^{\mathrm{\infty }}$ℓ^(oo)\ell^{\infty}${\ell }^{\mathrm{\infty }}$.
c) The dual of a finite dimensional space is finite dimensional.
d) If ${T}_{1}$${T}_{1}$T_(1)T_1${T}_{1}$ and ${T}_{2}$${T}_{2}$T_(2)T_2${T}_{2}$ are positive operators on a Hilbert space $H$$H$HH$H$, then ${T}_{1}+{T}_{2}$${T}_{1}+{T}_{2}$T_(1)+T_(2)T_1+T_2${T}_{1}+{T}_{2}$ is a positive operator on $\mathrm{H}$$\mathrm{H}$H\mathrm{H}$\mathrm{H}$.
e) On a normed space $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$, the norm function $\parallel \cdot \parallel :\mathrm{X}\to \mathbb{C}$$\parallel \cdot \parallel :\mathrm{X}\to \mathbb{C}$||*||:XrarrC\|\cdot\|: \mathrm{X} \rightarrow \mathbb{C}$\parallel \cdot \parallel :\mathrm{X}\to \mathbb{C}$ 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=\left\{f\in C\left[0,1\right]:f\left(0\right)=0\right\}$$X=\left\{f\in C\left[0,1\right]:f\left(0\right)=0\right\}$X={f in C[0,1]:f(0)=0}X=\{f \in C[0,1]: f(0)=0\}$X=\left\{f\in C\left[0,1\right]:f\left(0\right)=0\right\}$
$Y=\left\{g\in x:{\int }_{0}^{1}g\left(t\right)dt=0\right\}$$Y=\left\{g\in x:{\int }_{0}^{1} g\left(t\right)dt=0\right\}$Y={g in x:int_(0)^(1)g(t)dt=0}Y=\left\{g \in x: \int_0^1 g(t) d t=0\right\}$Y=\left\{g\in x:{\int }_{0}^{1}g\left(t\right)dt=0\right\}$
Prove that $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ is a proper subspace of $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$. Is $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ a closed subspace of $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ ? Justify your answer.
To prove that $Y$$Y$YY$Y$ is a proper subspace of $X$$X$XX$X$ and determine whether $Y$$Y$YY$Y$ is a closed subspace of $X$$X$XX$X$, we need to examine the definitions and properties of these sets within the context of functional analysis.

### $X$$X$XX$X$ and $Y$$Y$YY$Y$ Defined

• $X=\left\{f\in C\left[0,1\right]:f\left(0\right)=0\right\}$$X=\left\{f\in C\left[0,1\right]:f\left(0\right)=0\right\}$X={f in C[0,1]:f(0)=0}X = \{f \in C[0,1]: f(0) = 0\}$X=\left\{f\in C\left[0,1\right]:f\left(0\right)=0\right\}$ is the set of all continuous functions on the interval $\left[0,1\right]$$\left[0,1\right]$[0,1][0,1]$\left[0,1\right]$ that vanish at $0$$0$00$0$.
• $Y=\left\{g\in X:{\int }_{0}^{1}g\left(t\right)dt=0\right\}$$Y=\left\{g\in X:{\int }_{0}^{1} g\left(t\right)dt=0\right\}$Y={g in X:int_(0)^(1)g(t)dt=0}Y = \left\{g \in X: \int_0^1 g(t) dt = 0\right\}$Y=\left\{g\in X:{\int }_{0}^{1}g\left(t\right)dt=0\right\}$ is the set of all functions in $X$$X$XX$X$ whose integral over $\left[0,1\right]$$\left[0,1\right]$[0,1][0,1]$\left[0,1\right]$ is $0$$0$00$0$.

### Proving $Y$$Y$YY$Y$ is a Proper Subspace of $X$$X$XX$X$

To show that $Y$$Y$YY$Y$ is a proper subspace of $X$$X$XX$X$, we must verify that $Y$$Y$YY$Y$ satisfies the following criteria for being a subspace:
1. Non-emptiness: $Y$$Y$YY$Y$ contains the zero function, which is the function $g\left(t\right)=0$$g\left(t\right)=0$g(t)=0g(t) = 0$g\left(t\right)=0$ for all $t\in \left[0,1\right]$$t\in \left[0,1\right]$t in[0,1]t \in [0,1]$t\in \left[0,1\right]$. This function clearly belongs to $X$$X$XX$X$ and satisfies the integral condition, so $Y$$Y$YY$Y$ is non-empty.
2. Closed under addition: If ${g}_{1},{g}_{2}\in Y$${g}_{1},{g}_{2}\in Y$g_(1),g_(2)in Yg_1, g_2 \in Y${g}_{1},{g}_{2}\in Y$, then ${g}_{1}+{g}_{2}\in Y$${g}_{1}+{g}_{2}\in Y$g_(1)+g_(2)in Yg_1 + g_2 \in Y${g}_{1}+{g}_{2}\in Y$. This is because the integral of the sum is the sum of the integrals, each of which is $0$$0$00$0$, so their sum is also $0$$0$00$0$.
3. Closed under scalar multiplication: If $g\in Y$$g\in Y$g in Yg \in Y$g\in Y$ and $\alpha$$\alpha$alpha\alpha$\alpha$ is a scalar, then $\alpha g\in Y$$\alpha g\in Y$alpha g in Y\alpha g \in Y$\alpha g\in Y$. This follows because ${\int }_{0}^{1}\alpha g\left(t\right)dt=\alpha {\int }_{0}^{1}g\left(t\right)dt=\alpha \cdot 0=0$${\int }_{0}^{1} \alpha g\left(t\right)dt=\alpha {\int }_{0}^{1} g\left(t\right)dt=\alpha \cdot 0=0$int_(0)^(1)alpha g(t)dt=alphaint_(0)^(1)g(t)dt=alpha*0=0\int_0^1 \alpha g(t) dt = \alpha \int_0^1 g(t) dt = \alpha \cdot 0 = 0${\int }_{0}^{1}\alpha g\left(t\right)dt=\alpha {\int }_{0}^{1}g\left(t\right)dt=\alpha \cdot 0=0$.
Since $Y$$Y$YY$Y$ satisfies these criteria, it is a subspace of $X$$X$XX$X$. It is a proper subspace because there exist functions in $X$$X$XX$X$ that do not satisfy the integral condition, such as $f\left(t\right)=t$$f\left(t\right)=t$f(t)=tf(t) = t$f\left(t\right)=t$, which is in $X$$X$XX$X$ but not in $Y$$Y$YY$Y$ since ${\int }_{0}^{1}tdt=1/2\ne 0$${\int }_{0}^{1} tdt=1/2\ne 0$int_(0)^(1)tdt=1//2!=0\int_0^1 t dt = 1/2 \neq 0${\int }_{0}^{1}tdt=1/2\ne 0$.

### Is $Y$$Y$YY$Y$ a Closed Subspace of $X$$X$XX$X$?

A subspace $Y$$Y$YY$Y$ is closed in $X$$X$XX$X$ if it contains all its limit points; that is, if a sequence of functions $\left\{{g}_{n}\right\}$$\left\{{g}_{n}\right\}${g_(n)}\{g_n\}$\left\{{g}_{n}\right\}$ in $Y$$Y$YY$Y$ converges uniformly to a function $g$$g$gg$g$, then $g$$g$gg$g$ must also be in $Y$$Y$YY$Y$.
To show $Y$$Y$YY$Y$ is closed, consider a sequence $\left\{{g}_{n}\right\}\subset Y$$\left\{{g}_{n}\right\}\subset Y${g_(n)}sub Y\{g_n\} \subset Y$\left\{{g}_{n}\right\}\subset Y$ that converges uniformly to $g\in X$$g\in X$g in Xg \in X$g\in X$. We need to show that $g\in Y$$g\in Y$g in Yg \in Y$g\in Y$, meaning ${\int }_{0}^{1}g\left(t\right)dt=0$${\int }_{0}^{1} g\left(t\right)dt=0$int_(0)^(1)g(t)dt=0\int_0^1 g(t) dt = 0${\int }_{0}^{1}g\left(t\right)dt=0$.
Uniform convergence of $\left\{{g}_{n}\right\}$$\left\{{g}_{n}\right\}${g_(n)}\{g_n\}$\left\{{g}_{n}\right\}$ to $g$$g$gg$g$ implies that for every $ϵ>0$$ϵ>0$epsilon > 0\epsilon > 0$ϵ>0$, there exists an $N$$N$NN$N$ such that for all $n\ge N$$n\ge N$n >= Nn \geq N$n\ge N$, we have $|{g}_{n}\left(t\right)-g\left(t\right)|<ϵ$$|{g}_{n}\left(t\right)-g\left(t\right)|<ϵ$|g_(n)(t)-g(t)| < epsilon|g_n(t) – g(t)| < \epsilon$|{g}_{n}\left(t\right)-g\left(t\right)|<ϵ$ for all $t\in \left[0,1\right]$$t\in \left[0,1\right]$t in[0,1]t \in [0,1]$t\in \left[0,1\right]$. By the properties of integrals and the limit of a sequence of functions,
$\underset{n\to \mathrm{\infty }}{lim}{\int }_{0}^{1}{g}_{n}\left(t\right)dt={\int }_{0}^{1}\underset{n\to \mathrm{\infty }}{lim}{g}_{n}\left(t\right)dt={\int }_{0}^{1}g\left(t\right)dt.$$\underset{n\to \mathrm{\infty }}{lim} {\int }_{0}^{1} {g}_{n}\left(t\right)dt={\int }_{0}^{1} \underset{n\to \mathrm{\infty }}{lim} {g}_{n}\left(t\right)dt={\int }_{0}^{1} g\left(t\right)dt.$lim_(n rarr oo)int_(0)^(1)g_(n)(t)dt=int_(0)^(1)lim_(n rarr oo)g_(n)(t)dt=int_(0)^(1)g(t)dt.\lim_{n \to \infty} \int_0^1 g_n(t) dt = \int_0^1 \lim_{n \to \infty} g_n(t) dt = \int_0^1 g(t) dt.$\underset{n\to \mathrm{\infty }}{lim}{\int }_{0}^{1}{g}_{n}\left(t\right)dt={\int }_{0}^{1}\underset{n\to \mathrm{\infty }}{lim}{g}_{n}\left(t\right)dt={\int }_{0}^{1}g\left(t\right)dt.$
Since each ${g}_{n}\in Y$${g}_{n}\in Y$g_(n)in Yg_n \in Y${g}_{n}\in Y$, we have ${\int }_{0}^{1}{g}_{n}\left(t\right)dt=0$${\int }_{0}^{1} {g}_{n}\left(t\right)dt=0$int_(0)^(1)g_(n)(t)dt=0\int_0^1 g_n(t) dt = 0${\int }_{0}^{1}{g}_{n}\left(t\right)dt=0$. Thus, the limit of these integrals as $n\to \mathrm{\infty }$$n\to \mathrm{\infty }$n rarr oon \to \infty$n\to \mathrm{\infty }$ is also $0$$0$00$0$, which means ${\int }_{0}^{1}g\left(t\right)dt=0$${\int }_{0}^{1} g\left(t\right)dt=0$int_(0)^(1)g(t)dt=0\int_0^1 g(t) dt = 0${\int }_{0}^{1}g\left(t\right)dt=0$, and hence $g\in Y$$g\in Y$g in Yg \in Y$g\in Y$.
Therefore, $Y$$Y$YY$Y$ is a closed subspace of $X$$X$XX$X$ because it contains all its limit points under uniform convergence.
b) Let $X={L}^{p}\left[0,1\right]$$X={L}^{p}\left[0,1\right]$X=L^(p)[0,1]X=L^p[0,1]$X={L}^{p}\left[0,1\right]$ and $x=x\left(t\right)={t}^{2}$$x=x\left(t\right)={t}^{2}$x=x(t)=t^(2)x=x(t)=t^2$x=x\left(t\right)={t}^{2}$. Find $\parallel x{\parallel }_{p}$$\parallel x{\parallel }_{p}$||x||_(p)\|x\|_p$\parallel x{\parallel }_{p}$ for $p=4$$p=4$p=4p=4$p=4$ and $\mathrm{\infty }$$\mathrm{\infty }$oo\infty$\mathrm{\infty }$.
To find the ${L}^{p}$${L}^{p}$L^(p)L^p${L}^{p}$ norm of $x\left(t\right)={t}^{2}$$x\left(t\right)={t}^{2}$x(t)=t^(2)x(t) = t^2$x\left(t\right)={t}^{2}$ on the interval $\left[0,1\right]$$\left[0,1\right]$[0,1][0,1]$\left[0,1\right]$ for $p=4$$p=4$p=4p=4$p=4$ and $p=\mathrm{\infty }$$p=\mathrm{\infty }$p=oop=\infty$p=\mathrm{\infty }$, we’ll use the definitions of the ${L}^{p}$${L}^{p}$L^(p)L^p${L}^{p}$ norms.
### For $p=4$$p=4$p=4p=4$p=4$
The ${L}^{p}$${L}^{p}$L^(p)L^p${L}^{p}$ norm for $p=4$$p=4$p=4p=4$p=4$ is defined as
$\parallel x{\parallel }_{4}={\left({\int }_{0}^{1}|{t}^{2}{|}^{4}dt\right)}^{1/4}$$\parallel x{\parallel }_{4}={\left({\int }_{0}^{1} |{t}^{2}{|}^{4}dt\right)}^{1/4}$||x||_(4)=(int_(0)^(1)|t^(2)|^(4)dt)^(1//4)\|x\|_4 = \left( \int_0^1 |t^2|^4 dt \right)^{1/4}$\parallel x{\parallel }_{4}={\left({\int }_{0}^{1}|{t}^{2}{|}^{4}dt\right)}^{1/4}$
$\parallel x{\parallel }_{4}={\left({\int }_{0}^{1}{t}^{8}dt\right)}^{1/4}$$\parallel x{\parallel }_{4}={\left({\int }_{0}^{1} {t}^{8}dt\right)}^{1/4}$||x||_(4)=(int_(0)^(1)t^(8)dt)^(1//4)\|x\|_4 = \left( \int_0^1 t^8 dt \right)^{1/4}$\parallel x{\parallel }_{4}={\left({\int }_{0}^{1}{t}^{8}dt\right)}^{1/4}$
To compute this integral, we use the formula for the integral of ${t}^{n}$${t}^{n}$t^(n)t^n${t}^{n}$, which is $\frac{{t}^{n+1}}{n+1}$$\frac{{t}^{n+1}}{n+1}$(t^(n+1))/(n+1)\frac{t^{n+1}}{n+1}$\frac{{t}^{n+1}}{n+1}$ for $n\ne -1$$n\ne -1$n!=-1n \neq -1$n\ne -1$, from $0$$0$00$0$ to $1$$1$11$1$:
${\int }_{0}^{1}{t}^{8}dt=\frac{{t}^{9}}{9}{|}_{0}^{1}=\frac{1}{9}$${\int }_{0}^{1} {t}^{8}dt=\frac{{t}^{9}}{9}{|}_{0}^{1}=\frac{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}