Details For MMT-006 Solved Assignment

IGNOU MMT-006 Assignment Question Paper 2024

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

1. a) Let $X=\{f\in C[0,1]:f(0)=0\}$$X=\{f\in C[0,1]:f(0)=0\}$X={f in C[0,1]:f(0)=0}X={f in C[0,1]: f(0)=0}$X=\{f\in C[0,1]:f(0)=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={g in x:int_(0)^(1)g(t)dt=0}Y=left{g in x: int_0^1 g(t) d t=0right}$$Y=\{g\in x:{\int }_0^{1}g(t)dt=0\}$$
Prove that $\mathrm{Y}$$\mathrm{Y}$Ymathrm{Y}$\mathrm{Y}$ is a proper subspace of $\mathrm{X}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$. Is $\mathrm{Y}$$\mathrm{Y}$Ymathrm{Y}$\mathrm{Y}$ a closed subspace of $\mathrm{X}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$ ? Justify your answer.
b) Let $X=L^p[0,1]$$X=L^p[0,1]$X=L^(p)[0,1]X=L^p[0,1]$X=L^p[0,1]$ and $x=x(t)=t^2$$x=x(t)=t^2$x=x(t)=t^(2)x=x(t)=t^2$x=x(t)=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 }$$\mathrm{\infty }$ooinfty$\mathrm{\infty }$.
c) Let E be a subset of a normed space $X,Y=\mathrm{span}E$$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(a)=0$$f(a)=0$f(a)=0f(a)=0$f(a)=0$ whenever $f\in X^{\mathrm{\prime }}$$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}$$c_{00}$c_(00)c_{00}$c_{00}$. For $x=(x_1,x_2,\dots ,x_n,\dots )\in c_{00}$$x=\left(x_1,x_2,\dots ,x_n,\dots \right)\in c_{00}$x=(x_(1),x_(2),dots,x_(n),dots)inc_(00)x=left(x_1, x_2, ldots, x_n, ldotsright) in c_{00}$x=(x_1,x_2,\dots ,x_n,\dots )\in c_{00}$, define $f(x)=\sum _{n=1}^{\mathrm{\infty }}{x}_n$$f(x)=\sum _{n=1}^{\mathrm{\infty }} x_n$f(x)=sum_(n=1)^(oo)x_(n)f(x)=sum_{n=1}^{infty} x_n$f(x)=\sum _{n=1}^{\mathrm{\infty }}{x}_n$. Show that $f$$f$ff$f$ is a linear functional which is not continuous w.r.t the norm $\parallel x\parallel =\underset{n}{sup}\left|x_n\right|$$\parallel x\parallel =\underset{n}{sup} \left|x_n\right|$||x||=s u p _(n)|x_(n)||x|=sup _nleft|x_nright|$\parallel x\parallel =\underset{n}{sup}\left|x_n\right|$.
b) Consider the space ${\mathrm{C}}^1[0,1]$${\mathrm{C}}^1[0,1]$C^(1)[0,1]mathrm{C}^1[0,1]${\mathrm{C}}^1[0,1]$ of all ${\mathrm{C}}^1$${\mathrm{C}}^1$C^(1)mathrm{C}^1${\mathrm{C}}^1$ functions on [0,1] endowed with the uniform norm induced from the space $\mathrm{C}[0,1]$$\mathrm{C}[0,1]$C[0,1]mathrm{C}[0,1]$\mathrm{C}[0,1]$, and consider the differential operator $D:(C^1[0,1],\parallel \cdot {\parallel }_{\mathrm{\infty }})\to (C[0,1],\parallel \cdot {\parallel }_{\mathrm{\infty }})$$D:\left(C^1[0,1],\parallel \cdot {\parallel }_{\mathrm{\infty }}\right)\to \left(C[0,1],\parallel \cdot {\parallel }_{\mathrm{\infty }}\right)$D:(C^(1)[0,1],||*||_(oo))rarr(C[0,1],||*||_(oo))D:left(C^1[0,1],|cdot|_{infty}right) rightarrowleft(C[0,1],|cdot|_{infty}right)$D:(C^1[0,1],\parallel \cdot {\parallel }_{\mathrm{\infty }})\to (C[0,1],\parallel \cdot {\parallel }_{\mathrm{\infty }})$ defined by $Df=f^{\mathrm{\prime }}$$Df=f^{\mathrm{\prime }}$Df=f^(‘)D f=f^{prime}$Df=f^{\mathrm{\prime }}$. Prove that $D$$D$DD$D$ is linear, with closed graph, but not continuous. Can we conclude from here that $C^1[0,1]$$C^1[0,1]$C^(1)[0,1]C^1[0,1]$C^1[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 $\mathrm{X}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$ be a Banach space, $\mathrm{Y}$$\mathrm{Y}$Ymathrm{Y}$\mathrm{Y}$ be a normed linear space and $\mathcal{F}$$\mathcal{F}$Fmathcal{F}$\mathcal{F}$ be a subset of $\mathrm{B}(\mathrm{X},\mathrm{Y})$$\mathrm{B}(\mathrm{X},\mathrm{Y})$B(X,Y)mathrm{B}(mathrm{X}, mathrm{Y})$\mathrm{B}(\mathrm{X},\mathrm{Y})$. If $\mathcal{F}$$\mathcal{F}$Fmathcal{F}$\mathcal{F}$ is not uniformly bounded, then there exists a dense subset $\mathrm{D}$$\mathrm{D}$Dmathrm{D}$\mathrm{D}$ of $\mathrm{X}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$ such that for every $\mathrm{x}\in \mathrm{D},\{\mathrm{F}(\mathrm{x}):\mathrm{F}\in \mathcal{F}\}$$\mathrm{x}\in \mathrm{D},\{\mathrm{F}(\mathrm{x}):\mathrm{F}\in \mathcal{F}\}$xinD,{F(x):FinF}mathrm{x} in mathrm{D},{mathrm{F}(mathrm{x}): mathrm{F} in mathcal{F}}$\mathrm{x}\in \mathrm{D},\{\mathrm{F}(\mathrm{x}):\mathrm{F}\in \mathcal{F}\}$ is not bounded in $\mathrm{Y}$$\mathrm{Y}$Ymathrm{Y}$\mathrm{Y}$.
4. a) Read the proof of the closed graph theorem carefully and explain where and how we have used the following facts in the proof.
i) $\mathrm{X}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$ is a Banach space.
ii) $\mathrm{Y}$$\mathrm{Y}$Ymathrm{Y}$\mathrm{Y}$ is a Banach space.
iii) F is a closed map.
iv) Which property of continuity is being established to conclude that $\mathrm{F}$$\mathrm{F}$Fmathrm{F}$\mathrm{F}$ is continuous.
b) Which of the following maps are open? Give reasons for your answer.
i) $\mathrm{T}:{\mathbb{R}}^3\to {\mathbb{R}}^2$$\mathrm{T}:{\mathbb{R}}^3\to {\mathbb{R}}^2$quadT:R^(3)rarrR^(2)quad mathrm{T}: mathbb{R}^3 rightarrow mathbb{R}^2$\mathrm{T}:{\mathbb{R}}^3\to {\mathbb{R}}^2$ given by $\mathrm{T}(\mathrm{x},\mathrm{y},\mathrm{z})=(\mathrm{x},\mathrm{z})$$\mathrm{T}(\mathrm{x},\mathrm{y},\mathrm{z})=(\mathrm{x},\mathrm{z})$T(x,y,z)=(x,z)mathrm{T}(mathrm{x}, mathrm{y}, mathrm{z})=(mathrm{x}, mathrm{z})$\mathrm{T}(\mathrm{x},\mathrm{y},\mathrm{z})=(\mathrm{x},\mathrm{z})$.
ii) $\mathrm{T}:{\mathbb{R}}^3\to {\mathbb{R}}^3$$\mathrm{T}:{\mathbb{R}}^3\to {\mathbb{R}}^3$quadT:R^(3)rarrR^(3)quad mathrm{T}: mathbb{R}^3 rightarrow mathbb{R}^3$\mathrm{T}:{\mathbb{R}}^3\to {\mathbb{R}}^3$ given by $\mathrm{T}(\mathrm{x},\mathrm{y},\mathrm{z})=(\mathrm{x},\mathrm{y},0)$$\mathrm{T}(\mathrm{x},\mathrm{y},\mathrm{z})=(\mathrm{x},\mathrm{y},0)$T(x,y,z)=(x,y,0)mathrm{T}(mathrm{x}, mathrm{y}, mathrm{z})=(mathrm{x}, mathrm{y}, 0)$\mathrm{T}(\mathrm{x},\mathrm{y},\mathrm{z})=(\mathrm{x},\mathrm{y},0)$.
5. a) Let $f:C[0,1]\to \mathbb{R}$$f:C[0,1]\to \mathbb{R}$f:C[0,1]rarrRf: C[0,1] rightarrow mathbb{R}$f:C[0,1]\to \mathbb{R}$ be given by $f(x)=x(1)\mathrm{\forall }x\in C[0,1]$$f(x)=x(1)\mathrm{\forall }x\in 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)\mathrm{\forall }x\in C[0,1]$. 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}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$ be an inner product space and $\mathrm{x},\mathrm{y}\in \mathrm{X}$$\mathrm{x},\mathrm{y}\in \mathrm{X}$x,yinXmathrm{x}, mathrm{y} in mathrm{X}$\mathrm{x},\mathrm{y}\in \mathrm{X}$. Prove that $\mathrm{x}\perp \mathrm{y}$$\mathrm{x}\perp \mathrm{y}$x_|_ymathrm{x} perp mathrm{y}$\mathrm{x}\perp \mathrm{y}$ if and only if $\parallel \mathrm{k}\mathrm{x}+\mathrm{y}{\parallel }^2=\parallel \mathrm{k}\mathrm{x}{\parallel }^2+\parallel {\mathrm{y}}^2\parallel ,\mathrm{k}\in \mathrm{K}$$\parallel \mathrm{k}\mathrm{x}+\mathrm{y}{\parallel }^2=\parallel \mathrm{k}\mathrm{x}{\parallel }^2+∥{\mathrm{y}}^2∥,\mathrm{k}\in \mathrm{K}$||kx+y||^(2)=||kx||^(2)+||y^(2)||,kinK|mathrm{kx}+mathrm{y}|^2=|mathrm{kx}|^2+left|mathrm{y}^2right|, mathrm{k} in mathrm{K}$\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}$Fmathrm{F}$\mathrm{F}$ be the set of all $\mathbf{x}=({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3)$$\mathbf{x}=\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3\right)$x=(x_(1),x_(2),x_(3))mathbf{x}=left(mathrm{x}_1, mathrm{x}_2, mathrm{x}_3right)$\mathbf{x}=({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3)$ in $\mathrm{H}$$\mathrm{H}$Hmathrm{H}$\mathrm{H}$ such that ${\mathrm{x}}_1=0$${\mathrm{x}}_1=0$x_(1)=0mathrm{x}_1=0${\mathrm{x}}_1=0$. Find ${\mathrm{F}}^{\perp }$${\mathrm{F}}^{\perp }$F^(_|_)mathrm{F}^{perp}${\mathrm{F}}^{\perp }$. Verify that every $\mathbf{x}\in \mathrm{H}$$\mathbf{x}\in \mathrm{H}$xinHmathbf{x} in mathrm{H}$\mathbf{x}\in \mathrm{H}$ can be expressed as $\mathbf{x}=\mathbf{y}+\mathbf{z}$$\mathbf{x}=\mathbf{y}+\mathbf{z}$x=y+zmathbf{x}=mathbf{y}+mathbf{z}$\mathbf{x}=\mathbf{y}+\mathbf{z}$ where $\mathbf{y}\in \mathrm{F}$$\mathbf{y}\in \mathrm{F}$yinFmathbf{y} in mathrm{F}$\mathbf{y}\in \mathrm{F}$ and $\mathbf{z}\in {\mathrm{F}}^{\perp }$$\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}$$\mathrm{H}$Hmathrm{H}$\mathrm{H}$ and an operator $\mathrm{A}$$\mathrm{A}$Amathrm{A}$\mathrm{A}$ on $\mathrm{H}$$\mathrm{H}$Hmathrm{H}$\mathrm{H}$ such that ${\sigma }_{\mathrm{e}}(\mathrm{A})$${\sigma }_{\mathrm{e}}(\mathrm{A})$sigma_(e)(A)sigma_{mathrm{e}}(mathrm{A})${\sigma }_{\mathrm{e}}(\mathrm{A})$ 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 (A)\subset {\sigma }_a(A)$$\sigma (A)\subset {\sigma }_a(A)$sigma(A)subsigma _(a)(A)sigma(A) subset sigma_a(A)$\sigma (A)\subset {\sigma }_a(A)$ where ${\sigma }_a(A)$${\sigma }_a(A)$sigma _(a)(A)sigma_a(A)${\sigma }_a(A)$ denotes the approximate eigen spectrum of $A$$A$AA$A$ and $\sigma (A)$$\sigma (A)$sigma(A)sigma(A)$\sigma (A)$ 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}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$ be a normed space and $\mathrm{Y}$$\mathrm{Y}$Ymathrm{Y}$\mathrm{Y}$ be proper subspace of $\mathrm{X}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$. Show that the interior ${\mathrm{Y}}^0$${\mathrm{Y}}^0$Y^(0)mathrm{Y}^0${\mathrm{Y}}^0$ of $\mathrm{Y}$$\mathrm{Y}$Ymathrm{Y}$\mathrm{Y}$ is empty.
8. a) Let $X,Y$$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}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$ to $\mathrm{Y}$$\mathrm{Y}$Ymathrm{Y}$\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}$$\mathrm{Y}$Ymathrm{Y}$\mathrm{Y}$ is a Banach space, then $\mathrm{C}\mathrm{L}(\mathrm{X},\mathrm{Y})$$\mathrm{C}\mathrm{L}(\mathrm{X},\mathrm{Y})$CL(X,Y)mathrm{CL}(mathrm{X}, mathrm{Y})$\mathrm{C}\mathrm{L}(\mathrm{X},\mathrm{Y})$ is a closed subspace of $\mathrm{B}\mathrm{L}(\mathrm{X},\mathrm{Y})$$\mathrm{B}\mathrm{L}(\mathrm{X},\mathrm{Y})$BL(X,Y)mathrm{BL}(mathrm{X}, mathrm{Y})$\mathrm{B}\mathrm{L}(\mathrm{X},\mathrm{Y})$.
b) Define a Hilbert-Schmidt operator on a Hilbert space $\mathrm{H}$$\mathrm{H}$Hmathrm{H}$\mathrm{H}$ and give an example. Is every Hilbert-sehmidt operator a compact operator? Justify your answer.
9. a) Let $\left\{A_n\right\}$$\left\{A_n\right\}${A_(n)}left{A_nright}$\left\{A_n\right\}$ be a sequence of unitary operators in $BL(H)$$BL(H)$BL(H)B L(H)$BL(H)$. Prove that if $\parallel A_n-A\parallel \to 0,A\in BL(H)$$∥A_n-A∥\to 0,A\in BL(H)$||A_(n)-A||rarr0,A in BL(H)left|A_n-Aright| rightarrow 0, A in B L(H)$\parallel A_n-A\parallel \to 0,A\in BL(H)$, then $A$$A$AA$A$ is unitary.
b) Define the spectral radius of a bounded linear operator $A\in BL(X)$$A\in BL(X)$A in BL(X)A in B L(X)$A\in BL(X)$. Find the spectral radius of $\mathrm{A}$$\mathrm{A}$Amathrm{A}$\mathrm{A}$ in $\mathrm{B}\mathrm{L}\left({\mathbb{R}}^3\right)$$\mathrm{B}\mathrm{L}\left({\mathbb{R}}^3\right)$BL(R^(3))mathrm{BL}left(mathbb{R}^3right)$\mathrm{B}\mathrm{L}\left({\mathbb{R}}^3\right)$, where $\mathrm{A}$$\mathrm{A}$Amathrm{A}$\mathrm{A}$ is given by the matrix
$$\left[\begin{array}{rrr}0& 1& 0\\ -1& 0& 0\\ 0& 0& -1\end{array}\right]$$$$\left[\begin{array}{r}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ -1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-1\end{array}\right]$$[[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}$Xmathrm{X}$\mathrm{X}$ be a Banach space and $\mathrm{Y}$$\mathrm{Y}$Ymathrm{Y}$\mathrm{Y}$ be a closed subspace of $\mathrm{X}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$. Let $\pi :\mathrm{X}\to \mathrm{X}/\mathrm{Y}$$\pi :\mathrm{X}\to \mathrm{X}/\mathrm{Y}$pi:XrarrX//Ypi: mathrm{X} rightarrow mathrm{X} / mathrm{Y}$\pi :\mathrm{X}\to \mathrm{X}/\mathrm{Y}$ be canonical quotient map. Show that $\pi$$\pi$pipi$\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}$$\mathrm{H}$Hmathrm{H}$\mathrm{H}$.
e) On a normed space $\mathrm{X}$$\mathrm{X}$Xmathrm{X}$\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.


[stray-random categories=trigonometry]

MMT-006 Sample Solution 2024

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

1. a) Let $X=\{f\in C[0,1]:f(0)=0\}$$X=\{f\in C[0,1]:f(0)=0\}$X={f in C[0,1]:f(0)=0}X={f in C[0,1]: f(0)=0}$X=\{f\in C[0,1]:f(0)=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={g in x:int_(0)^(1)g(t)dt=0}Y=left{g in x: int_0^1 g(t) d t=0right}$$Y=\{g\in x:{\int }_0^{1}g(t)dt=0\}$$
Prove that $\mathrm{Y}$$\mathrm{Y}$Ymathrm{Y}$\mathrm{Y}$ is a proper subspace of $\mathrm{X}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$. Is $\mathrm{Y}$$\mathrm{Y}$Ymathrm{Y}$\mathrm{Y}$ a closed subspace of $\mathrm{X}$$\mathrm{X}$Xmathrm{X}$\mathrm{X}$ ? Justify your answer.
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=\{f\in C[0,1]:f(0)=0\}$$X=\{f\in C[0,1]:f(0)=0\}$X={f in C[0,1]:f(0)=0}X = {f in C[0,1]: f(0) = 0}$X=\{f\in C[0,1]:f(0)=0\}$ is the set of all continuous functions on the interval $[0,1]$$[0,1]$[0,1][0,1]$[0,1]$ that vanish at $0$$0$00$0$.
• $Y=\{g\in X:{\int }_0^{1}g(t)dt=0\}$$Y=\left\{g\in X:{\int }_0^1 g(t)dt=0\right\}$Y={g in X:int_(0)^(1)g(t)dt=0}Y = left{g in X: int_0^1 g(t) dt = 0right}$Y=\{g\in X:{\int }_0^{1}g(t)dt=0\}$ is the set of all functions in $X$$X$XX$X$ whose integral over $[0,1]$$[0,1]$[0,1][0,1]$[0,1]$ 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$$Y$YY$Y$ contains the zero function, which is the function $g(t)=0$$g(t)=0$g(t)=0g(t) = 0$g(t)=0$ for all $t\in [0,1]$$t\in [0,1]$t in[0,1]t in [0,1]$t\in [0,1]$. This function clearly belongs to $X$$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$$\alpha$alphaalpha$\alpha$ is a scalar, then $\alpha g\in Y$$\alpha g\in Y$alpha g in Yalpha g in Y$\alpha g\in Y$. This follows because ${\int }_0^1\alpha g(t)dt=\alpha {\int }_0^{1}g(t)dt=\alpha \cdot 0=0$${\int }_0^1 \alpha g(t)dt=\alpha {\int }_0^1 g(t)dt=\alpha \cdot 0=0$int_(0)^(1)alpha g(t)dt=alphaint_(0)^(1)g(t)dt=alpha*0=0int_0^1 alpha g(t) dt = alpha int_0^1 g(t) dt = alpha cdot 0 = 0${\int }_0^1\alpha g(t)dt=\alpha {\int }_0^{1}g(t)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$$X$XX$X$ that do not satisfy the integral condition, such as $f(t)=t$$f(t)=t$f(t)=tf(t) = t$f(t)=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$int_(0)^(1)tdt=1//2!=0int_0^1 t dt = 1/2 neq 0${\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 $\{g_n\}$$\{g_n\}${g_(n)}{g_n}$\{g_n\}$ in $Y$$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 $\{g_n\}\subset Y$$\{g_n\}\subset Y${g_(n)}sub Y{g_n} subset Y$\{g_n\}\subset Y$ that converges uniformly to $g\in X$$g\in X$g in Xg in X$g\in X$. We need to show that $g\in Y$$g\in Y$g in Yg in Y$g\in Y$, meaning ${\int }_0^{1}g(t)dt=0$${\int }_0^1 g(t)dt=0$int_(0)^(1)g(t)dt=0int_0^1 g(t) dt = 0${\int }_0^{1}g(t)dt=0$.
Uniform convergence of $\{g_n\}$$\{g_n\}${g_(n)}{g_n}$\{g_n\}$ to $g$$g$gg$g$ implies that for every $ϵ>0$$ϵ>0$epsilon > 0epsilon > 0$ϵ>0$, there exists an $N$$N$NN$N$ such that for all $n\ge N$$n\ge N$n >= Nn geq N$n\ge N$, we have $|g_n(t)-g(t)|<ϵ$$|g_n(t)-g(t)|<ϵ$|g_(n)(t)-g(t)| < epsilon|g_n(t) – g(t)| < epsilon$|g_n(t)-g(t)|<ϵ$ for all $t\in [0,1]$$t\in [0,1]$t in[0,1]t in [0,1]$t\in [0,1]$. By the properties of integrals and the limit of a sequence of functions,
$$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^1{g}_n(t)dt={\int }_0^1\underset{n\to \mathrm{\infty }}{lim}{g}_n(t)dt={\int }_0^{1}g(t)dt.$$$$\underset{n\to \mathrm{\infty }}{lim} {\int }_0^1 g_n(t)dt={\int }_0^1 \underset{n\to \mathrm{\infty }}{lim} g_n(t)dt={\int }_0^1 g(t)dt.$$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.$$\underset{n\to \mathrm{\infty }}{lim}{\int }_0^1{g}_n(t)dt={\int }_0^1\underset{n\to \mathrm{\infty }}{lim}{g}_n(t)dt={\int }_0^{1}g(t)dt.$$
Since each $g_n\in Y$$g_n\in Y$g_(n)in Yg_n in Y$g_n\in Y$, we have ${\int }_0^1{g}_n(t)dt=0$${\int }_0^1 g_n(t)dt=0$int_(0)^(1)g_(n)(t)dt=0int_0^1 g_n(t) dt = 0${\int }_0^1{g}_n(t)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(t)dt=0$${\int }_0^1 g(t)dt=0$int_(0)^(1)g(t)dt=0int_0^1 g(t) dt = 0${\int }_0^{1}g(t)dt=0$, and hence $g\in Y$$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[0,1]$$X=L^p[0,1]$X=L^(p)[0,1]X=L^p[0,1]$X=L^p[0,1]$ and $x=x(t)=t^2$$x=x(t)=t^2$x=x(t)=t^(2)x=x(t)=t^2$x=x(t)=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 }$$\mathrm{\infty }$ooinfty$\mathrm{\infty }$.
Answer:
To find the $L^p$$L^p$L^(p)L^p$L^p$ norm of $x(t)=t^2$$x(t)=t^2$x(t)=t^(2)x(t) = t^2$x(t)=t^2$ on the interval $[0,1]$$[0,1]$[0,1][0,1]$[0,1]$ 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}$$\parallel x{\parallel }_4={\left({\int }_0^1|t^2{|}^{4}dt\right)}^{1/4}$$
This simplifies to
$$\parallel x{\parallel }_4={\left({\int }_0^1{t}^{8}dt\right)}^{1/4}$$$$\parallel x{\parallel }_4={\left({\int }_0^1 t^{8}dt\right)}^{1/4}$$||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}$${\int }_0^1{t}^{8}dt=\frac{t^9}{9}{|}_0^1=\frac{1}{9}$$
Therefore,
$$\parallel x{\parallel }_4={\left(\frac{1}{9}\right)}^{1/4}$$$$\parallel x{\parallel }_4={\left(\frac{1}{9}\right)}^{1/4}$$||x||_(4)=((1)/(9))^(1//4)|x|_4 = left( frac{1}{9} right)^{1/4}$$\parallel x{\parallel }_4={\left(\frac{1}{9}\right)}^{1/4}$$

For $p=\mathrm{\infty }$$p=\mathrm{\infty }$p=oop=infty$p=\mathrm{\infty }$

The $L^{\mathrm{\infty }}$$L^{\mathrm{\infty }}$L^( oo)L^infty$L^{\mathrm{\infty }}$ norm, or the supremum norm, is defined as the essential supremum of $|x(t)|$$|x(t)|$|x(t)||x(t)|$|x(t)|$ over the interval. For $x(t)=t^2$$x(t)=t^2$x(t)=t^(2)x(t) = t^2$x(t)=t^2$ on $[0,1]$$[0,1]$[0,1][0,1]$[0,1]$, this is simply the maximum value of $t^2$$t^2$t^(2)t^2$t^2$ on the interval, which occurs at $t=1$$t=1$t=1t=1$t=1$. Therefore,
$$\parallel x{\parallel }_{\mathrm{\infty }}=\underset{t\in [0,1]}{max}|t^2|=1^2=1$$$$\parallel x{\parallel }_{\mathrm{\infty }}=\underset{t\in [0,1]}{max} |t^2|=1^2=1$$||x||_(oo)=max_(t in[0,1])|t^(2)|=1^(2)=1|x|_infty = max_{t in [0,1]} |t^2| = 1^2 = 1$$\parallel x{\parallel }_{\mathrm{\infty }}=\underset{t\in [0,1]}{max}|t^2|=1^2=1$$

Calculation

Let’s calculate $\parallel x{\parallel }_4$$\parallel x{\parallel }_4$||x||_(4)|x|_4$\parallel x{\parallel }_4$ explicitly:
$$\parallel x{\parallel }_4={\left(\frac{1}{9}\right)}^{1/4}$$$$\parallel x{\parallel }_4={\left(\frac{1}{9}\right)}^{1/4}$$||x||_(4)=((1)/(9))^(1//4)|x|_4 = left( frac{1}{9} right)^{1/4}$$\parallel x{\parallel }_4={\left(\frac{1}{9}\right)}^{1/4}$$
This calculation yields:
$$\parallel x{\parallel }_4={\left(\frac{1}{9}\right)}^{1/4}=\frac{1}{3^{1/2}}=\frac{1}{\sqrt{3}}$$$$\parallel x{\parallel }_4={\left(\frac{1}{9}\right)}^{1/4}=\frac{1}{3^{1/2}}=\frac{1}{\sqrt{3}}$$||x||_(4)=((1)/(9))^(1//4)=(1)/(3^(1//2))=(1)/(sqrt3)|x|_4 = left( frac{1}{9} right)^{1/4} = frac{1}{3^{1/2}} = frac{1}{sqrt{3}}$$\parallel x{\parallel }_4={\left(\frac{1}{9}\right)}^{1/4}=\frac{1}{3^{1/2}}=\frac{1}{\sqrt{3}}$$
So, summarizing:

• For $p=4$$p=4$p=4p=4$p=4$, $\parallel x{\parallel }_4=\frac{1}{\sqrt{3}}$$\parallel x{\parallel }_4=\frac{1}{\sqrt{3}}$||x||_(4)=(1)/(sqrt3)|x|_4 = frac{1}{sqrt{3}}$\parallel x{\parallel }_4=\frac{1}{\sqrt{3}}$.
• For $p=\mathrm{\infty }$$p=\mathrm{\infty }$p=oop=infty$p=\mathrm{\infty }$, $\parallel x{\parallel }_{\mathrm{\infty }}=1$$\parallel x{\parallel }_{\mathrm{\infty }}=1$||x||_(oo)=1|x|_infty = 1$\parallel x{\parallel }_{\mathrm{\infty }}=1$.

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}$$a\in \overline{Y}$a in bar(Y)a in bar{Y}$a\in \overline{Y}$ if and only if $f(a)=0$$f(a)=0$f(a)=0f(a)=0$f(a)=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$$E$EE$E$.
Answer:
To show that $a\in \overline{Y}$$a\in \overline{Y}$a in bar(Y)a in bar{Y}$a\in \overline{Y}$ if and only if $f(a)=0$$f(a)=0$f(a)=0f(a) = 0$f(a)=0$ for every $f\in X^{\prime }$$f\in X^{\prime }$f inX^(‘)f in X’$f\in X^{\prime }$ (the dual space of $X$$X$XX$X$) where $f=0$$f=0$f=0f = 0$f=0$ on $E$$E$EE$E$, we’ll use some fundamental properties of normed spaces, their duals, and the concept of closure.

Definitions:

• $X$$X$XX$X$ is a normed space, and $X^{\prime }$$X^{\prime }$X^(‘)X’$X^{\prime }$ is its dual space, consisting of all continuous linear functionals on $X$$X$XX$X$.
• $E$$E$EE$E$ is a subset of $X$$X$XX$X$, and $Y=\mathrm{span}E$$Y=\mathrm{span}E$Y=span EY = operatorname{span}E$Y=\mathrm{span}E$ is the subspace spanned by $E$$E$EE$E$.
• $\overline{Y}$$\overline{Y}$bar(Y)bar{Y}$\overline{Y}$ denotes the closure of $Y$$Y$YY$Y$ in $X$$X$XX$X$, which includes all limits of convergent sequences in $Y$$Y$YY$Y$.

($\Rightarrow$$\Rightarrow$=>Rightarrow$\Rightarrow$) If $a\in \overline{Y}$$a\in \overline{Y}$a in bar(Y)a in bar{Y}$a\in \overline{Y}$, then $f(a)=0$$f(a)=0$f(a)=0f(a) = 0$f(a)=0$ for every $f\in X^{\prime }$$f\in X^{\prime }$f inX^(‘)f in X’$f\in X^{\prime }$ where $f=0$$f=0$f=0f = 0$f=0$ on $E$$E$EE$E$:

Assume $a\in \overline{Y}$$a\in \overline{Y}$a in bar(Y)a in bar{Y}$a\in \overline{Y}$. This means that for every $ϵ>0$$ϵ>0$epsilon > 0epsilon > 0$ϵ>0$, there exists $y\in Y$$y\in Y$y in Yy in Y$y\in Y$ such that $\parallel a-y\parallel <ϵ$$\parallel a-y\parallel <ϵ$||a-y|| < epsilon|a – y| < epsilon$\parallel a-y\parallel <ϵ$, since $a$$a$aa$a$ can be approximated arbitrarily closely by elements of $Y$$Y$YY$Y$.
Let $f\in X^{\prime }$$f\in X^{\prime }$f inX^(‘)f in X’$f\in X^{\prime }$ be such that $f=0$$f=0$f=0f = 0$f=0$ on $E$$E$EE$E$. Since $Y=\mathrm{span}E$$Y=\mathrm{span}E$Y=span EY = operatorname{span}E$Y=\mathrm{span}E$, $f$$f$ff$f$ also vanishes on $Y$$Y$YY$Y$ because any linear combination of elements in $E$$E$EE$E$ (which $f$$f$ff$f$ maps to $0$$0$00$0$) will also be mapped to $0$$0$00$0$ by $f$$f$ff$f$.
Given $a\in \overline{Y}$$a\in \overline{Y}$a in bar(Y)a in bar{Y}$a\in \overline{Y}$, and considering $f$$f$ff$f$ is continuous, we have $f(a)=\underset{y\to a,y\in Y}{lim}f(y)$$f(a)=\underset{y\to a,y\in Y}{lim} f(y)$f(a)=lim_(y rarr a,y in Y)f(y)f(a) = lim_{y to a, y in Y} f(y)$f(a)=\underset{y\to a,y\in Y}{lim}f(y)$. Since $f(y)=0$$f(y)=0$f(y)=0f(y) = 0$f(y)=0$ for all $y\in Y$$y\in Y$y in Yy in Y$y\in Y$, it follows that $f(a)=0$$f(a)=0$f(a)=0f(a) = 0$f(a)=0$.

($\Leftarrow$$\Leftarrow$lArrLeftarrow$\Leftarrow$) If $f(a)=0$$f(a)=0$f(a)=0f(a) = 0$f(a)=0$ for every $f\in X^{\prime }$$f\in X^{\prime }$f inX^(‘)f in X’$f\in X^{\prime }$ where $f=0$$f=0$f=0f = 0$f=0$ on $E$$E$EE$E$, then $a\in \overline{Y}$$a\in \overline{Y}$a in bar(Y)a in bar{Y}$a\in \overline{Y}$:

Assume $f(a)=0$$f(a)=0$f(a)=0f(a) = 0$f(a)=0$ for every $f\in X^{\prime }$$f\in X^{\prime }$f inX^(‘)f in X’$f\in X^{\prime }$ where $f=0$$f=0$f=0f = 0$f=0$ on $E$$E$EE$E$. Suppose, for the sake of contradiction, that $a\notin \overline{Y}$$a\notin \overline{Y}$a!in bar(Y)a notin bar{Y}$a\notin \overline{Y}$. By the Hahn-Banach theorem, there exists a continuous linear functional $f\in X^{\prime }$$f\in X^{\prime }$f inX^(‘)f in X’$f\in X^{\prime }$ such that $f(a)\ne 0$$f(a)\ne 0$f(a)!=0f(a) neq 0$f(a)\ne 0$ and $f=0$$f=0$f=0f = 0$f=0$ on $\overline{Y}$$\overline{Y}$bar(Y)bar{Y}$\overline{Y}$ (since $a$$a$aa$a$ and $\overline{Y}$$\overline{Y}$bar(Y)bar{Y}$\overline{Y}$ can be separated).
However, since $f=0$$f=0$f=0f = 0$f=0$ on $\overline{Y}$$\overline{Y}$bar(Y)bar{Y}$\overline{Y}$, and $E\subseteq Y\subseteq \overline{Y}$$E\subseteq Y\subseteq \overline{Y}$E sube Y sube bar(Y)E subseteq Y subseteq bar{Y}$E\subseteq Y\subseteq \overline{Y}$, it follows that $f=0$$f=0$f=0f = 0$f=0$ on $E$$E$EE$E$. But we assumed that $f(a)=0$$f(a)=0$f(a)=0f(a) = 0$f(a)=0$ for all such $f$$f$ff$f$, contradicting the existence of such an $f$$f$ff$f$ that separates $a$$a$aa$a$ from $\overline{Y}$$\overline{Y}$bar(Y)bar{Y}$\overline{Y}$.
Therefore, our assumption that $a\notin \overline{Y}$$a\notin \overline{Y}$a!in bar(Y)a notin bar{Y}$a\notin \overline{Y}$ must be false, implying $a\in \overline{Y}$$a\in \overline{Y}$a in bar(Y)a in bar{Y}$a\in \overline{Y}$.