# IGNOU MMT-006 Solved Assignment 2023 | M.Sc. MACS

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

365.00

Access via our Android App Only

Details For MMT-006 Solved Assignment

## IGNOU MMT-006 Assignment Question Paper 2023

mmt-006-qp-f7c5c3cd-c8c6-4a67-bc59-92f7593d91ab
1. State whether the following statements are true or false. Justify with a short proof or a counter example.
a) The space ${l}^{3}$${l}^{3}$l^(3)l^{3}${l}^{3}$ is a Hilbert space
b) Any non zero bounded linear functional on a Banach space is an open map.
c) Every bounded linear map on a complex Banach space has an eigen value.
d) The image of a Cauchy sequence under a bounded linear map is also a Couchy sequence.
e) If $\mathrm{A}$$\mathrm{A}$A\mathrm{A}$\mathrm{A}$ is a bounded linear operator on a Hilbert space such that ${\mathrm{A}\mathrm{A}}^{\ast }=\mathrm{I}$${\mathrm{A}\mathrm{A}}^{\ast }=\mathrm{I}$AA^(**)=I\mathrm{AA}^{*}=\mathrm{I}${\mathrm{A}\mathrm{A}}^{\ast }=\mathrm{I}$, then ${\mathrm{A}}^{\ast }\text{}\mathrm{A}=\mathrm{I}$${\mathrm{A}}^{\ast }\text{}\mathrm{A}=\mathrm{I}$A^(**)A=I\mathrm{A}^{*} \mathrm{~A}=\mathrm{I}.
2. a) Characterise all bounded linear functionals on a Hilbert space.
b) Show that the map $\mathrm{T}:{\mathbf{R}}^{3}\to {\mathbf{R}}^{2}$$\mathrm{T}:{\mathbf{R}}^{3}\to {\mathbf{R}}^{2}$T:R^(3)rarrR^(2)\mathrm{T}: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}$\mathrm{T}:{\mathbf{R}}^{3}\to {\mathbf{R}}^{2}$ given by $\mathrm{T}\left({\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3}\right)=\left({\mathrm{x}}_{1}+{\mathrm{x}}_{2},{\mathrm{x}}_{3}\right)$$\mathrm{T}\left({\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3}\right)=\left({\mathrm{x}}_{1}+{\mathrm{x}}_{2},{\mathrm{x}}_{3}\right)$T(x_(1),x_(2),x_(3))=(x_(1)+x_(2),x_(3))\mathrm{T}\left(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}\right)=\left(\mathrm{x}_{1}+\mathrm{x}_{2}, \mathrm{x}_{3}\right)$\mathrm{T}\left({\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3}\right)=\left({\mathrm{x}}_{1}+{\mathrm{x}}_{2},{\mathrm{x}}_{3}\right)$ is an open map.
c) Check whether a finite dimensional normed linear space is reflexive? Justify your answer.
1. a) Show how a real linear functional $u$$u$uu$u$ on a complex linear normed space gives rise to a complex linear functional $\mathrm{f}$$\mathrm{f}$f\mathrm{f}$\mathrm{f}$. What is the relation between the boundedness of $\mathrm{u}$$\mathrm{u}$u\mathrm{u}$\mathrm{u}$ and that of $\mathrm{f}$$\mathrm{f}$f\mathrm{f}$\mathrm{f}$ ?
b) In a Hilbert space. Prove that ${\mathrm{x}}_{\mathrm{n}}\to \mathrm{x}$${\mathrm{x}}_{\mathrm{n}}\to \mathrm{x}$x_(n)rarrx\mathrm{x}_{\mathrm{n}} \rightarrow \mathrm{x}${\mathrm{x}}_{\mathrm{n}}\to \mathrm{x}$ provided $\parallel {\mathrm{x}}_{\mathrm{n}}\parallel \to \parallel \mathrm{x}\parallel$$∥{\mathrm{x}}_{\mathrm{n}}∥\to \parallel \mathrm{x}\parallel$||x_(n)||rarr||x||\left\|\mathrm{x}_{\mathrm{n}}\right\| \rightarrow\|\mathrm{x}\|$\parallel {\mathrm{x}}_{\mathrm{n}}\parallel \to \parallel \mathrm{x}\parallel$ and $⟨{\mathrm{x}}_{\mathrm{n}},\mathrm{x}⟩\to ⟨\mathrm{x},\mathrm{x}⟩$$⟨{\mathrm{x}}_{\mathrm{n}},\mathrm{x}⟩\to ⟨\mathrm{x},\mathrm{x}⟩$(:x_(n),x:)rarr(:x,x:)\left\langle\mathrm{x}_{\mathrm{n}}, \mathrm{x}\right\rangle \rightarrow\langle\mathrm{x}, \mathrm{x}\rangle$⟨{\mathrm{x}}_{\mathrm{n}},\mathrm{x}⟩\to ⟨\mathrm{x},\mathrm{x}⟩$.
c) Are Hahn-Banach extensions always unique? Justify.
1. a) State the principle of uniform boundedness. Use it to show that a set $E$$E$EE$E$ in a normed space $X$$X$XX$X$ is bounded if $f\left(E\right)$$f\left(E\right)$f(E)f(E)$f\left(E\right)$ is bounded in $K$$K$KK$K$ for every $f\in {X}^{\mathrm{\prime }}$$f\in {X}^{\mathrm{\prime }}$f inX^(‘)f \in X^{\prime}$f\in {X}^{\mathrm{\prime }}$.
b) If $\mathrm{H}$$\mathrm{H}$H\mathrm{H}$\mathrm{H}$ is a Hilbert space and $\mathrm{S}\mathrm{C}\mathrm{H}$$\mathrm{S}\mathrm{C}\mathrm{H}$SCH\mathrm{SCH}$\mathrm{S}\mathrm{C}\mathrm{H}$, show that ${\mathrm{S}}^{\perp }={\mathrm{S}}^{\perp \perp \perp }$${\mathrm{S}}^{\perp }={\mathrm{S}}^{\perp \perp \perp }$S^(_|_)=S^(⊥⊥⊥)\mathrm{S}^{\perp}=\mathrm{S}^{\perp \perp \perp}${\mathrm{S}}^{\perp }={\mathrm{S}}^{\perp \perp \perp }$. When $\mathrm{S}$$\mathrm{S}$S\mathrm{S}$\mathrm{S}$ is the same as ${\mathrm{S}}^{\perp \perp }$${\mathrm{S}}^{\perp \perp }$S^(⊥⊥)\mathrm{S}^{\perp \perp}${\mathrm{S}}^{\perp \perp }$ ? Justify.
5 a) Show that $\mathrm{Q}$$\mathrm{Q}$Q\mathrm{Q}$\mathrm{Q}$ defined on $\left(\mathrm{C}\left[0,1\right],\parallel \cdot {\parallel }_{\mathrm{\infty }}\right)$$\left(\mathrm{C}\left[0,1\right],\parallel \cdot {\parallel }_{\mathrm{\infty }}\right)$(C[0,1],||*||_(oo))\left(\mathrm{C}[0,1],\|\cdot\|_{\infty}\right)$\left(\mathrm{C}\left[0,1\right],\parallel \cdot {\parallel }_{\mathrm{\infty }}\right)$ by $\mathrm{Q}\left(\mathrm{x}\right)={\int }_{0}^{1}\mathrm{t}\mathrm{x}\left(\mathrm{t}\right)\mathrm{d}\mathrm{t}$$\mathrm{Q}\left(\mathrm{x}\right)={\int }_{0}^{1} \mathrm{t}\mathrm{x}\left(\mathrm{t}\right)\mathrm{d}\mathrm{t}$Q(x)=int_(0)^(1)tx(t)dt\mathrm{Q}(\mathrm{x})=\int_{0}^{1} \mathrm{t} \mathrm{x}(\mathrm{t}) \mathrm{dt}$\mathrm{Q}\left(\mathrm{x}\right)={\int }_{0}^{1}\mathrm{t}\mathrm{x}\left(\mathrm{t}\right)\mathrm{d}\mathrm{t}$ is a bounded linear functional. Calculate $\parallel \mathrm{Q}\parallel$$\parallel \mathrm{Q}\parallel$||Q||\|\mathrm{Q}\|$\parallel \mathrm{Q}\parallel$. b) Let $A$$A$AA$A$ be an operator on a Hilbert space $H$$H$HH$H$. Show if $\parallel Ax\parallel =\parallel {A}^{\ast }x\parallel$$\parallel Ax\parallel =∥{A}^{\ast }x∥$||Ax||=||A^(**)x||\|A x\|=\left\|A^{*} x\right\|$\parallel Ax\parallel =\parallel {A}^{\ast }x\parallel$ for every $x\in H$$x\in H$x in Hx \in H$x\in H$, then $\mathrm{A}$$\mathrm{A}$A\mathrm{A}$\mathrm{A}$ is normal. Is it converse true? Justify.
c) Let $\left\{{\mathrm{u}}_{\mathrm{n}}\right\}$$\left\{{\mathrm{u}}_{\mathrm{n}}\right\}${u_(n)}\left\{\mathrm{u}_{\mathrm{n}}\right\}$\left\{{\mathrm{u}}_{\mathrm{n}}\right\}$ be the sequence in ${l}^{2}$${l}^{2}$l^(2)l^{2}${l}^{2}$ with 1 in the ${\mathrm{n}}^{\text{th}}$n^(“th “)\mathrm{n}^{\text {th }} place and zeroes else where prove that the set $\left\{{\mathrm{u}}_{\mathrm{n}}\right\}$$\left\{{\mathrm{u}}_{\mathrm{n}}\right\}${u_(n)}\left\{\mathrm{u}_{\mathrm{n}}\right\}$\left\{{\mathrm{u}}_{\mathrm{n}}\right\}$ is an orthonormal basis for ${l}^{2}$${l}^{2}$l^(2)l^{2}${l}^{2}$.
6 a) Let $X=C\left[0,1\right]$$X=C\left[0,1\right]$X=C[0,1]X=C[0,1]$X=C\left[0,1\right]$ with Sup norm defined by $\parallel f\parallel =\mathrm{Sup}\left\{|f\left(x\right)|\right\}$$\parallel f\parallel =\mathrm{Sup}\left\{|f\left(x\right)|\right\}$||f||=Sup{|f(x)|}\|f\|=\operatorname{Sup}\{|f(x)|\}$\parallel f\parallel =\mathrm{Sup}\left\{|f\left(x\right)|\right\}$.
$\mathrm{x}\in \left[0,1\right]$$\mathrm{x}\in \left[0,1\right]$xin[0,1]\mathrm{x} \in[0,1]$\mathrm{x}\in \left[0,1\right]$
Let $\mathrm{T}$$\mathrm{T}$T\mathrm{T}$\mathrm{T}$ be a linear map defined on $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ by $\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\left(\frac{1}{2}\right)$$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\left(\frac{1}{2}\right)$T(f)=f((1)/(2))\mathrm{T}(\mathrm{f})=\mathrm{f}\left(\frac{1}{2}\right)$\mathrm{T}\left(\mathrm{f}\right)=\mathrm{f}\left(\frac{1}{2}\right)$.
Show that $\mathrm{T}$$\mathrm{T}$T\mathrm{T}$\mathrm{T}$ is a bounded linear map such that $\parallel \mathrm{T}\parallel =1$$\parallel \mathrm{T}\parallel =1$||T||=1\|\mathrm{T}\|=1$\parallel \mathrm{T}\parallel =1$.
b) Define Eigen Spectrum of a bounded linear operator on a Banach space. Show that the eigen spectrum of the operator $\mathrm{T}$$\mathrm{T}$T\mathrm{T}$\mathrm{T}$ on ${l}^{2}$${l}^{2}$l^(2)l^{2}${l}^{2}$ given by $\mathrm{T}\left({\alpha }_{1},{\alpha }_{2}\dots \right)=\left(0,{\alpha }_{1},{\alpha }_{2}\dots \right)$$\mathrm{T}\left({\alpha }_{1},{\alpha }_{2}\dots \right)=\left(0,{\alpha }_{1},{\alpha }_{2}\dots \right)$T(alpha_(1),alpha_(2)dots)=(0,alpha_(1),alpha_(2)dots)\mathrm{T}\left(\alpha_{1}, \alpha_{2} \ldots\right)=\left(0, \alpha_{1}, \alpha_{2} \ldots\right)$\mathrm{T}\left({\alpha }_{1},{\alpha }_{2}\dots \right)=\left(0,{\alpha }_{1},{\alpha }_{2}\dots \right)$ is empty.
c) Let $\parallel \cdot \parallel$$\parallel \cdot \parallel$||*||\|\cdot\|$\parallel \cdot \parallel$ be a norm on a linear space $X$$X$XX$X$. if $x,y\in X$$x,y\in X$x,y in Xx, y \in X$x,y\in X$ and $\parallel x+y\parallel =\parallel x\parallel +\parallel y\parallel$$\parallel x+y\parallel =\parallel x\parallel +\parallel y\parallel$||x+y||=||x||+||y||\|x+y\|=\|x\|+\|y\|$\parallel x+y\parallel =\parallel x\parallel +\parallel y\parallel$, then show that $\parallel sx+ty\parallel =s\parallel x\parallel +t\parallel y\parallel$$\parallel sx+ty\parallel =s\parallel x\parallel +t\parallel y\parallel$||sx+ty||=s||x||+t||y||\|s x+t y\|=s\|x\|+t\|y\|$\parallel sx+ty\parallel =s\parallel x\parallel +t\parallel y\parallel$ for all $s\ge 0,t\ge 0$$s\ge 0,t\ge 0$s >= 0,t >= 0s \geq 0, t \geq 0$s\ge 0,t\ge 0$.
1. a) Let $X$$X$XX$X$ be an inner product space with the inner product given by $<$$<$<<$<$, $>$$>$>>$>$. For $x\in X$$x\in X$x in Xx \in X$x\in X$, define the function $\parallel \cdot \parallel :X\to K$$\parallel \cdot \parallel :X\to K$||*||:X rarr K\|\cdot\|: X \rightarrow K$\parallel \cdot \parallel :X\to K$ given by $\parallel x\parallel =1/2$$\parallel x\parallel =1/2$||x||=<x,-x > 1//2\|x\|=<x,-x>1 / 2$\parallel x\parallel =1/2$, the non negative square root of $<\mathrm{x},\mathrm{x}>$$<\mathrm{x},\mathrm{x}>$< x,x ><\mathrm{x}, \mathrm{x}>$<\mathrm{x},\mathrm{x}>$. Show that $\parallel \cdot \parallel :\mathrm{X}\to \mathrm{K}$$\parallel \cdot \parallel :\mathrm{X}\to \mathrm{K}$||*||:XrarrK\|\cdot\|: \mathrm{X} \rightarrow \mathrm{K}$\parallel \cdot \parallel :\mathrm{X}\to \mathrm{K}$ defines a norm on $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ and $\mathrm{I}<\left(\mathrm{x},\mathrm{y}\right)>\mathrm{I}\le \parallel \mathrm{x}\parallel \parallel \mathrm{y}\parallel$$\mathrm{I}<\left(\mathrm{x},\mathrm{y}\right)>\mathrm{I}\le \parallel \mathrm{x}\parallel \parallel \mathrm{y}\parallel$I < (x,y) > I <= ||x||||y||\mathrm{I}<(\mathrm{x}, \mathrm{y})>\mathrm{I} \leq\|\mathrm{x}\|\|\mathrm{y}\|$\mathrm{I}<\left(\mathrm{x},\mathrm{y}\right)>\mathrm{I}\le \parallel \mathrm{x}\parallel \parallel \mathrm{y}\parallel$ for all $\mathrm{x},\mathrm{y}\in \mathrm{X}$$\mathrm{x},\mathrm{y}\in \mathrm{X}$x,yinX\mathrm{x}, \mathrm{y} \in \mathrm{X}$\mathrm{x},\mathrm{y}\in \mathrm{X}$. Also show that for all $\mathrm{x},\mathrm{y}\in \mathrm{X}$$\mathrm{x},\mathrm{y}\in \mathrm{X}$x,yinX\mathrm{x}, \mathrm{y} \in \mathrm{X}$\mathrm{x},\mathrm{y}\in \mathrm{X}$, $\parallel x+y{\parallel }^{2}+\parallel x-y{\parallel }^{2}=2\left(\parallel x{\parallel }^{2}+\parallel y{\parallel }^{2}\right)$$\parallel x+y{\parallel }^{2}+\parallel x-y{\parallel }^{2}=2\left(\parallel x{\parallel }^{2}+\parallel y{\parallel }^{2}\right)$||x+y||^(2)+||x-y||^(2)=2(||x||^(2)+||y||^(2))\|x+y\|^{2}+\|x-y\|^{2}=2\left(\|x\|^{2}+\|y\|^{2}\right)$\parallel x+y{\parallel }^{2}+\parallel x-y{\parallel }^{2}=2\left(\parallel x{\parallel }^{2}+\parallel y{\parallel }^{2}\right)$.
b) Let $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ be a vector space. Let $\parallel \cdot {\parallel }^{1}$$\parallel \cdot {\parallel }^{1}$||*||^(1)\|\cdot\|^{1}$\parallel \cdot {\parallel }^{1}$ and $\parallel \cdot {\parallel }^{2}$$\parallel \cdot {\parallel }^{2}$||*||^(2)\|\cdot\|^{2}$\parallel \cdot {\parallel }^{2}$ be two norms on $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$. When are these norms said to be equivalent? Justify your answer.
Let $\mathrm{X}={\mathbb{R}}^{3}$$\mathrm{X}={\mathbb{R}}^{3}$X=R^(3)\mathrm{X}=\mathbb{R}^{3}$\mathrm{X}={\mathbb{R}}^{3}$. For $\mathrm{x}=\left({\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3}\right)$$\mathrm{x}=\left({\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3}\right)$x=(x_(1),x_(2),x_(3))\mathrm{x}=\left(\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}\right)$\mathrm{x}=\left({\mathrm{x}}_{1},{\mathrm{x}}_{2},{\mathrm{x}}_{3}\right)$.
Let $\parallel x{\parallel }^{1}=|{x}_{1}|+|{x}_{2}|+|{x}_{3}|$$\parallel x{\parallel }^{1}=\left|{x}_{1}\right|+\left|{x}_{2}\right|+\left|{x}_{3}\right|$||x||^(1)=|x_(1)|+|x_(2)|+|x_(3)|\|x\|^{1}=\left|x_{1}\right|+\left|x_{2}\right|+\left|x_{3}\right|$\parallel x{\parallel }^{1}=|{x}_{1}|+|{x}_{2}|+|{x}_{3}|$
$\parallel x{\parallel }^{2}=\sqrt{{|{x}_{1}|}^{2}+{|{x}_{2}|}^{2}+{|{x}_{3}|}^{2}}$$\parallel x{\parallel }^{2}=\sqrt{{\left|{x}_{1}\right|}^{2}+{\left|{x}_{2}\right|}^{2}+{\left|{x}_{3}\right|}^{2}}$||x||^(2)=sqrt(|x_(1)|^(2)+|x_(2)|^(2)+|x_(3)|^(2))\|x\|^{2}=\sqrt{\left|x_{1}\right|^{2}+\left|x_{2}\right|^{2}+\left|x_{3}\right|^{2}}$\parallel x{\parallel }^{2}=\sqrt{{|{x}_{1}|}^{2}+{|{x}_{2}|}^{2}+{|{x}_{3}|}^{2}}$
Show that $\parallel \cdot {\parallel }^{1}$$\parallel \cdot {\parallel }^{1}$||*||^(1)\|\cdot\|^{1}$\parallel \cdot {\parallel }^{1}$ and $\parallel \cdot {\parallel }^{1}$$\parallel \cdot {\parallel }^{1}$||*||^(1)\|\cdot\|^{1}$\parallel \cdot {\parallel }^{1}$ are equivalent.
1. a) Let $X={L}^{2}\left[0,2\pi \right]$$X={L}^{2}\left[0,2\pi \right]$X=L^(2)[0,2pi]X=L^{2}[0,2 \pi]$X={L}^{2}\left[0,2\pi \right]$ and ${u}_{n}\left(t\right)=\frac{{e}^{\text{int}}}{\sqrt{2\pi }},t\in \left\{-\pi ,\pi \right\},n\in Z$u_(n)(t)=(e^(“int “))/(sqrt(2pi)),t in{-pi,pi},n in Zu_{n}(t)=\frac{e^{\text {int }}}{\sqrt{2 \pi}}, t \in\{-\pi, \pi\}, n \in Z. Show that the set $\mathrm{E}=\left\{{\mathrm{u}}_{1},{\mathrm{u}}_{2}\dots \right\}$$\mathrm{E}=\left\{{\mathrm{u}}_{1},{\mathrm{u}}_{2}\dots \right\}$E={u_(1),u_(2)dots}\mathrm{E}=\left\{\mathrm{u}_{1}, \mathrm{u}_{2} \ldots\right\}$\mathrm{E}=\left\{{\mathrm{u}}_{1},{\mathrm{u}}_{2}\dots \right\}$ is an orthonormal set in $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$.
b) Let $\mathrm{H}$$\mathrm{H}$H\mathrm{H}$\mathrm{H}$ be a Hilbert space. For any subset $\mathrm{A}$$\mathrm{A}$A\mathrm{A}$\mathrm{A}$ of $\mathrm{H}$$\mathrm{H}$H\mathrm{H}$\mathrm{H}$, define ${\mathrm{A}}^{\perp }$${\mathrm{A}}^{\perp }$A^(_|_)\mathrm{A}^{\perp}${\mathrm{A}}^{\perp }$. If $\mathrm{A}\subseteq \mathrm{B}\subseteq \mathrm{H}$$\mathrm{A}\subseteq \mathrm{B}\subseteq \mathrm{H}$AsubeBsubeH\mathrm{A} \subseteq \mathrm{B} \subseteq \mathrm{H}$\mathrm{A}\subseteq \mathrm{B}\subseteq \mathrm{H}$, then show that:
i) $\phantom{\rule{1em}{0ex}}{\mathrm{B}}^{\perp }\subseteq {\mathrm{A}}^{\perp }$$\phantom{\rule{1em}{0ex}}{\mathrm{B}}^{\perp }\subseteq {\mathrm{A}}^{\perp }$quadB^(_|_)subeA^(_|_)\quad \mathrm{B}^{\perp} \subseteq \mathrm{A}^{\perp}$\phantom{\rule{1em}{0ex}}{\mathrm{B}}^{\perp }\subseteq {\mathrm{A}}^{\perp }$ ii) $\phantom{\rule{1em}{0ex}}\mathrm{A}\subseteq {\mathrm{A}}^{\perp \perp }$$\phantom{\rule{1em}{0ex}}\mathrm{A}\subseteq {\mathrm{A}}^{\perp \perp }$quadAsubeA^(⊥⊥)\quad \mathrm{A} \subseteq \mathrm{A}^{\perp \perp}$\phantom{\rule{1em}{0ex}}\mathrm{A}\subseteq {\mathrm{A}}^{\perp \perp }$
State conditions on $\mathrm{A}$$\mathrm{A}$A\mathrm{A}$\mathrm{A}$ so that ${\mathrm{A}}^{\perp \perp }=\mathrm{A}$${\mathrm{A}}^{\perp \perp }=\mathrm{A}$A^(⊥⊥)=A\mathrm{A}^{\perp \perp}=\mathrm{A}${\mathrm{A}}^{\perp \perp }=\mathrm{A}$.
c) Let $\mathrm{X}={\mathrm{C}}^{\mathrm{\prime }}\left[0,1\right]$$\mathrm{X}={\mathrm{C}}^{\mathrm{\prime }}\left[0,1\right]$X=C^(‘)[0,1]\mathrm{X}=\mathrm{C}^{\prime}[0,1]$\mathrm{X}={\mathrm{C}}^{\mathrm{\prime }}\left[0,1\right]$ and $\mathrm{Y}=\mathrm{C}\left[0,11\right]$$\mathrm{Y}=\mathrm{C}\left[0,11\right]$Y=C[0,11]\mathrm{Y}=\mathrm{C}[0,11]$\mathrm{Y}=\mathrm{C}\left[0,11\right]$ and let $\mathrm{T}:\mathrm{X}\to \mathrm{Y}$$\mathrm{T}:\mathrm{X}\to \mathrm{Y}$T:XrarrY\mathrm{T}: \mathrm{X} \rightarrow \mathrm{Y}$\mathrm{T}:\mathrm{X}\to \mathrm{Y}$ be the linear operator from $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ to $Y$$Y$YY$Y$ given by $T\left(f\right)={f}^{\mathrm{\prime }}$$T\left(f\right)={f}^{\mathrm{\prime }}$T(f)=f^(‘)T(f)=f^{\prime}$T\left(f\right)={f}^{\mathrm{\prime }}$, the derivative of $f$$f$ff$f$ on $\left[0,1\right]$$\left[0,1\right]$[0,1][0,1]$\left[0,1\right]$. Show that $T$$T$TT$T$ is not continuous.
1. a) Let $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ and $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ be Banach spaces and $\mathrm{F}:\mathrm{X}\to \mathrm{Y}$$\mathrm{F}:\mathrm{X}\to \mathrm{Y}$F:XrarrY\mathrm{F}: \mathrm{X} \rightarrow \mathrm{Y}$\mathrm{F}:\mathrm{X}\to \mathrm{Y}$ be a linear map which is continuous and open. Will $\mathrm{F}$$\mathrm{F}$F\mathrm{F}$\mathrm{F}$ always be closed? Will $\mathrm{F}$$\mathrm{F}$F\mathrm{F}$\mathrm{F}$ be always surjective? Give reasons for your answer.
b) Check whether the identity map on an infinite dimensional space is compact.
c) Define $\mathrm{A}:{\mathbb{C}}^{3}\to {\mathbb{C}}^{3}$$\mathrm{A}:{\mathbb{C}}^{3}\to {\mathbb{C}}^{3}$A:C^(3)rarrC^(3)\mathrm{A}: \mathbb{C}^{3} \rightarrow \mathbb{C}^{3}$\mathrm{A}:{\mathbb{C}}^{3}\to {\mathbb{C}}^{3}$ by $\mathrm{A}\left({\mathrm{z}}_{1},{\mathrm{z}}_{2},{\mathrm{z}}_{3}\right)=\left(\mathrm{i}\mathrm{z}{\mathrm{z}}_{1},{\mathrm{e}}^{2\mathrm{i}}{\mathrm{z}}_{2},{\mathrm{z}}_{3}\right)$$\mathrm{A}\left({\mathrm{z}}_{1},{\mathrm{z}}_{2},{\mathrm{z}}_{3}\right)=\left(\mathrm{i}\mathrm{z}{\mathrm{z}}_{1},{\mathrm{e}}^{2\mathrm{i}}{\mathrm{z}}_{2},{\mathrm{z}}_{3}\right)$A(z_(1),z_(2),z_(3))=(izz_(1),e^(2i)z_(2),z_(3))\mathrm{A}\left(\mathrm{z}_{1}, \mathrm{z}_{2}, \mathrm{z}_{3}\right)=\left(\mathrm{iz} \mathrm{z}_{1}, \mathrm{e}^{2 \mathrm{i}} \mathrm{z}_{2}, \mathrm{z}_{3}\right)$\mathrm{A}\left({\mathrm{z}}_{1},{\mathrm{z}}_{2},{\mathrm{z}}_{3}\right)=\left(\mathrm{i}\mathrm{z}{\mathrm{z}}_{1},{\mathrm{e}}^{2\mathrm{i}}{\mathrm{z}}_{2},{\mathrm{z}}_{3}\right)$.
Check whether $A$$A$AA$A$ is i) self adjoint, ii) unitary.
1. a) Let $\mathrm{A}:{\mathrm{X}}_{0}\subseteq \mathrm{X}\to \mathrm{Y}$$\mathrm{A}:{\mathrm{X}}_{0}\subseteq \mathrm{X}\to \mathrm{Y}$A:X_(0)subeXrarrY\mathrm{A}: \mathrm{X}_{0} \subseteq \mathrm{X} \rightarrow \mathrm{Y}$\mathrm{A}:{\mathrm{X}}_{0}\subseteq \mathrm{X}\to \mathrm{Y}$ be a closed operator where $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ and $\mathrm{Y}$$\mathrm{Y}$Y\mathrm{Y}$\mathrm{Y}$ are Banach spaces.
Define $\parallel \mathrm{x}{\parallel }_{\mathrm{A}}=\parallel \mathrm{x}\parallel +\parallel \mathrm{A}\mathrm{x}\parallel ,\mathrm{x}\in {\mathrm{X}}_{0}$$\parallel \mathrm{x}{\parallel }_{\mathrm{A}}=\parallel \mathrm{x}\parallel +\parallel \mathrm{A}\mathrm{x}\parallel ,\mathrm{x}\in {\mathrm{X}}_{0}$||x||_(A)=||x||+||Ax||,xinX_(0)\|\mathrm{x}\|_{\mathrm{A}}=\|\mathrm{x}\|+\|\mathrm{Ax}\|, \mathrm{x} \in \mathrm{X}_{0}$\parallel \mathrm{x}{\parallel }_{\mathrm{A}}=\parallel \mathrm{x}\parallel +\parallel \mathrm{A}\mathrm{x}\parallel ,\mathrm{x}\in {\mathrm{X}}_{0}$.
Then show that the norm $\parallel \cdot {\parallel }_{\mathrm{A}}$$\parallel \cdot {\parallel }_{\mathrm{A}}$||*||_(A)\|\cdot\|_{\mathrm{A}}$\parallel \cdot {\parallel }_{\mathrm{A}}$ is complete.
b) Find a bounded linear functional $\mathrm{f}$$\mathrm{f}$f\mathrm{f}$\mathrm{f}$ on ${\ell }^{3}$${\ell }^{3}$ℓ^(3)\ell^{3}${\ell }^{3}$ such that $\mathrm{f}\left({\mathrm{e}}_{3}\right)=3$$\mathrm{f}\left({\mathrm{e}}_{3}\right)=3$f(e_(3))=3\mathrm{f}\left(\mathrm{e}_{3}\right)=3$\mathrm{f}\left({\mathrm{e}}_{3}\right)=3$ and $\parallel \mathrm{f}\parallel =3$$\parallel \mathrm{f}\parallel =3$||f||=3\|\mathrm{f}\|=3$\parallel \mathrm{f}\parallel =3$.
c) Prove that ${l}^{1}\subset {l}^{2}$${l}^{1}\subset {l}^{2}$l^(1)subl^(2)l^{1} \subset l^{2}${l}^{1}\subset {l}^{2}$. If: $\mathrm{T}:\left({l}^{2},\parallel \cdot {\parallel }_{2}\right)\to \left({l}^{1},\parallel \cdot {\parallel }_{2}\right)$$\mathrm{T}:\left({l}^{2},\parallel \cdot {\parallel }_{2}\right)\to \left({l}^{1},\parallel \cdot {\parallel }_{2}\right)$T:(l^(2),||*||_(2))rarr(l^(1),||*||_(2))\mathrm{T}:\left(l^{2},\|\cdot\|_{2}\right) \rightarrow\left(l^{1},\|\cdot\|_{2}\right)$\mathrm{T}:\left({l}^{2},\parallel \cdot {\parallel }_{2}\right)\to \left({l}^{1},\parallel \cdot {\parallel }_{2}\right)$ is a compact operator, show that: $\mathrm{T}:\left({l}^{2},\parallel \cdot {\parallel }_{2}\right)\to \left({l}^{2},\parallel \cdot {\parallel }_{2}\right)$$\mathrm{T}:\left({l}^{2},\parallel \cdot {\parallel }_{2}\right)\to \left({l}^{2},\parallel \cdot {\parallel }_{2}\right)$T:(l^(2),||*||_(2))rarr(l^(2),||*||_(2))\mathrm{T}:\left(l^{2},\|\cdot\|_{2}\right) \rightarrow\left(l^{2},\|\cdot\|_{2}\right)$\mathrm{T}:\left({l}^{2},\parallel \cdot {\parallel }_{2}\right)\to \left({l}^{2},\parallel \cdot {\parallel }_{2}\right)$ is also compact.
$$2\:cos\:\theta \:cos\:\phi =cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)$$

## MMT-006 Sample Solution 2023

untitled-document-15-c4a14609-db5c-41e1-99f0-81aab3fdac8f

### Question:-01

1. State whether the following statements are true or false. Justify with a short proof or a counter example.
a) The space ${l}^{3}$${l}^{3}$l^(3)l^3${l}^{3}$ is a Hilbert space

### Statement: The space ${l}^{3}$${l}^{3}$l^(3)l^3${l}^{3}$ is a Hilbert space.

#### Justification:

A Hilbert space is a complete inner product space. That is, it is a vector space equipped with an inner product that is complete in the metric induced by the inner product.
The space ${l}^{3}$${l}^{3}$l^(3)l^3${l}^{3}$ consists of all sequences $\left({x}_{1},{x}_{2},{x}_{3},\dots \right)$$\left({x}_{1},{x}_{2},{x}_{3},\dots \right)$(x_(1),x_(2),x_(3),dots)(x_1, x_2, x_3, \ldots)$\left({x}_{1},{x}_{2},{x}_{3},\dots \right)$ such that $\sum _{n=1}^{\mathrm{\infty }}|{x}_{n}{|}^{3}<\mathrm{\infty }$$\sum _{n=1}^{\mathrm{\infty }} |{x}_{n}{|}^{3}<\mathrm{\infty }$sum_(n=1)^(oo)|x_(n)|^(3) < oo\sum_{n=1}^{\infty} |x_n|^3 < \infty$\sum _{n=1}^{\mathrm{\infty }}|{x}_{n}{|}^{3}<\mathrm{\infty }$.
1. Inner Product: We can define an inner product on ${l}^{3}$${l}^{3}$l^(3)l^3${l}^{3}$ as follows:
$⟨x,y⟩=\sum _{n=1}^{\mathrm{\infty }}{x}_{n}\overline{{y}_{n}}$$⟨x,y⟩=\sum _{n=1}^{\mathrm{\infty }} {x}_{n}\overline{{y}_{n}}$(:x,y:)=sum_(n=1)^(oo)x_(n) bar(y_(n))\langle x, y \rangle = \sum_{n=1}^{\infty} x_n \overline{y_n}$⟨x,y⟩=\sum _{n=1}^{\mathrm{\infty }}{x}_{n}\overline{{y}_{n}}$
This definition satisfies the properties of an inner product (conjugate symmetry, linearity, and positive-definiteness).
1. Completeness: The space ${l}^{3}$${l}^{3}$l^(3)l^3${l}^{3}$ is not complete with respect to this inner product. To see this, consider the sequence of sequences ${x}^{\left(k\right)}=\left(1,\frac{1}{2},\dots ,\frac{1}{k},0,0,\dots \right)$${x}^{\left(k\right)}=\left(1,\frac{1}{2},\dots ,\frac{1}{k},0,0,\dots \right)$x^((k))=(1,(1)/(2),dots,(1)/(k),0,0,dots)x^{(k)} = (1, \frac{1}{2}, \ldots, \frac{1}{k}, 0, 0, \ldots)${x}^{\left(k\right)}=\left(1,\frac{1}{2},\dots ,\frac{1}{k},0,0,\dots \right)$. Each ${x}^{\left(k\right)}$${x}^{\left(k\right)}$x^((k))x^{(k)}${x}^{\left(k\right)}$ is in ${l}^{3}$${l}^{3}$l^(3)l^3${l}^{3}$, but its limit as $k\to \mathrm{\infty }$$k\to \mathrm{\infty }$k rarr ook \to \infty$k\to \mathrm{\infty }$, which is the sequence $\left(1,\frac{1}{2},\frac{1}{3},\dots \right)$$\left(1,\frac{1}{2},\frac{1}{3},\dots \right)$(1,(1)/(2),(1)/(3),dots)(1, \frac{1}{2}, \frac{1}{3}, \ldots)$\left(1,\frac{1}{2},\frac{1}{3},\dots \right)$, is not in ${l}^{3}$${l}^{3}$l^(3)l^3${l}^{3}$ because $\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\right)}^{3}=\mathrm{\infty }$$\sum _{n=1}^{\mathrm{\infty }} {\left(\frac{1}{n}\right)}^{3}=\mathrm{\infty }$sum_(n=1)^(oo)((1)/(n))^(3)=oo\sum_{n=1}^{\infty} \left(\frac{1}{n}\right)^3 = \infty$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\right)}^{3}=\mathrm{\infty }$.
Therefore, ${l}^{3}$${l}^{3}$l^(3)l^3${l}^{3}$ is not a Hilbert space because it is not complete with respect to the inner product.

Page Break
b) Any non zero bounded linear functional on a Banach space is an open map.

### Statement: Any non-zero bounded linear functional on a Banach space is an open map.

#### Justification:

A map $T:X\to Y$$T:X\to Y$T:X rarr YT: X \to Y$T:X\to Y$ between topological spaces is said to be open if the image of every open set in $X$$X$XX$X$ under $T$$T$TT$T$ is open in $Y$$Y$YY$Y$.
Let $X$$X$XX$X$ be a Banach space and $f:X\to \mathbb{R}$$f:X\to \mathbb{R}$f:X rarrRf: X \to \mathbb{R}$f:X\to \mathbb{R}$ (or $\mathbb{C}$$\mathbb{C}$C\mathbb{C}$\mathbb{C}$) be a non-zero bounded linear functional. We want to show that $f$$f$ff$f$ is an open map.
1. Boundedness: Since $f$$f$ff$f$ is bounded, there exists a constant $C>0$$C>0$C > 0C > 0$C>0$ such that $|f\left(x\right)|\le C\parallel x\parallel$$|f\left(x\right)|\le C\parallel x\parallel$|f(x)| <= C||x|||f(x)| \leq C \|x\|$|f\left(x\right)|\le C\parallel x\parallel$ for all $x\in X$$x\in X$x in Xx \in X$x\in X$.
2. Non-Zero: $f$$f$ff$f$ is non-zero, which means there exists some ${x}_{0}\in X$${x}_{0}\in X$x_(0)in Xx_0 \in X${x}_{0}\in X$ such that $f\left({x}_{0}\right)\ne 0$$f\left({x}_{0}\right)\ne 0$f(x_(0))!=0f(x_0) \neq 0$f\left({x}_{0}\right)\ne 0$.
3. Open Map: Consider an open ball $B\left(x,ϵ\right)$$B\left(x,ϵ\right)$B(x,epsilon)B(x, \epsilon)$B\left(x,ϵ\right)$ in $X$$X$XX$X$ centered at $x$$x$xx$x$ with radius $ϵ>0$$ϵ>0$epsilon > 0\epsilon > 0$ϵ>0$. We want to show that $f\left(B\left(x,ϵ\right)\right)$$f\left(B\left(x,ϵ\right)\right)$f(B(x,epsilon))f(B(x, \epsilon))$f\left(B\left(x,ϵ\right)\right)$ is an open set in $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ (or $\mathbb{C}$$\mathbb{C}$C\mathbb{C}$\mathbb{C}$).
• Take $y\in f\left(B\left(x,ϵ\right)\right)$$y\in f\left(B\left(x,ϵ\right)\right)$y in f(B(x,epsilon))y \in f(B(x, \epsilon))$y\in f\left(B\left(x,ϵ\right)\right)$. Then there exists $z\in B\left(x,ϵ\right)$$z\in B\left(x,ϵ\right)$z in B(x,epsilon)z \in B(x, \epsilon)$z\in B\left(x,ϵ\right)$ such that $f\left(z\right)=y$$f\left(z\right)=y$f(z)=yf(z) = y$f\left(z\right)=y$.
• Since $z\in B\left(x,ϵ\right)$$z\in B\left(x,ϵ\right)$z in B(x,epsilon)z \in B(x, \epsilon)$z\in B\left(x,ϵ\right)$, we have $\parallel z-x\parallel <ϵ$$\parallel z-x\parallel <ϵ$||z-x|| < epsilon\|z – x\| < \epsilon$\parallel z-x\parallel <ϵ$.
• Using linearity and boundedness of $f$$f$ff$f$, we get $|f\left(z\right)-f\left(x\right)|\le C\parallel z-x\parallel $|f\left(z\right)-f\left(x\right)|\le C\parallel z-x\parallel |f(z)-f(x)| <= C||z-x|| < C epsilon|f(z) – f(x)| \leq C \|z – x\| < C \epsilon$|f\left(z\right)-f\left(x\right)|\le C\parallel z-x\parallel .
• This means $y=f\left(z\right)$$y=f\left(z\right)$y=f(z)y = f(z)$y=f\left(z\right)$ lies in an open interval $\left(f\left(x\right)-Cϵ,f\left(x\right)+Cϵ\right)$$\left(f\left(x\right)-Cϵ,f\left(x\right)+Cϵ\right)$(f(x)-C epsilon,f(x)+C epsilon)(f(x) – C \epsilon, f(x) + C \epsilon)$\left(f\left(x\right)-Cϵ,f\left(x\right)+Cϵ\right)$ which is contained in $f\left(B\left(x,ϵ\right)\right)$$f\left(B\left(x,ϵ\right)\right)$f(B(x,epsilon))f(B(x, \epsilon))$f\left(B\left(x,ϵ\right)\right)$.
Thus, $f\left(B\left(x,ϵ\right)\right)$$f\left(B\left(x,ϵ\right)\right)$f(B(x,epsilon))f(B(x, \epsilon))$f\left(B\left(x,ϵ\right)\right)$ contains an open interval around every point $y$$y$yy$y$ in it, making $f\left(B\left(x,ϵ\right)\right)$$f\left(B\left(x,ϵ\right)\right)$f(B(x,epsilon))f(B(x, \epsilon))$f\left(B\left(x,ϵ\right)\right)$ an open set. Therefore, $f$$f$ff$f$ is an open map.
In conclusion, the statement “Any non-zero bounded linear functional on a Banach space is an open map” is true, and we have justified it with a short proof.

Page Break
c) Every bounded linear map on a complex Banach space has an eigen value.
An eigenvalue $\lambda$$\lambda$lambda\lambda$\lambda$ for a bounded linear map $T:X\to X$$T:X\to X$T:X rarr XT: X \to X$T:X\to X$ on a complex Banach space $X$$X$XX$X$ is a complex number such that there exists a non-zero vector $x\in X$$x\in X$x in Xx \in X$x\in X$ satisfying $T\left(x\right)=\lambda x$$T\left(x\right)=\lambda x$T(x)=lambda xT(x) = \lambda x$T\left(x\right)=\lambda x$.
Consider the right shift operator $T:{\ell }^{2}\to {\ell }^{2}$$T:{\ell }^{2}\to {\ell }^{2}$T:ℓ^(2)rarrℓ^(2)T: \ell^2 \to \ell^2$T:{\ell }^{2}\to {\ell }^{2}$ on the complex Banach space ${\ell }^{2}$${\ell }^{2}$ℓ^(2)\ell^2${\ell }^{2}$ of square-summable sequences. The operator $T$$T$TT$T$ is defined as follows:
$T\left(\left({x}_{1},{x}_{2},{x}_{3},\dots \right)\right)=\left(0,{x}_{1},{x}_{2},{x}_{3},\dots \right)$$T\left(\left({x}_{1},{x}_{2},{x}_{3},\dots \right)\right)=\left(0,{x}_{1},{x}_{2},{x}_{3},\dots \right)$T((x_(1),x_(2),x_(3),dots))=(0,x_(1),x_(2),x_(3),dots)T((x_1, x_2, x_3, \ldots)) = (0, x_1, x_2, x_3, \ldots)$T\left(\left({x}_{1},{x}_{2},{x}_{3},\dots \right)\right)=\left(0,{x}_{1},{x}_{2},{x}_{3},\dots \right)$
Let’s assume, for the sake of contradiction, that $T$$T$TT