Sample Solution

MMT-006 Solved Assignment 2025

1. State whether the following statements True or False? Justify your answers:
   a) The function $\Vert \cdot \Vert$$\Vert \cdot \Vert$||*||\|\cdot\|$\Vert \cdot \Vert$ defined on ${\mathbb{R}}^n$${\mathbb{R}}^n$R^(n)\mathbb{R}^n${\mathbb{R}}^n$ as:

• $\Vert x\Vert =\sum _{j=1}^n|a_j|\ge 0$$\Vert x\Vert =\sum _{j=1}^n |a_j|\ge 0$||x||=sum_(j=1)^(n)|a_(j)| >= 0\|x\| = \sum_{j=1}^n |a_j| \geq 0$\Vert x\Vert =\sum _{j=1}^n|a_j|\ge 0$ since $|a_j|\ge 0$$|a_j|\ge 0$|a_(j)| >= 0|a_j| \geq 0$|a_j|\ge 0$.
• If $\Vert x\Vert =0$$\Vert x\Vert =0$||x||=0\|x\| = 0$\Vert x\Vert =0$, then $\sum _{j=1}^n|a_j|=0$$\sum _{j=1}^n |a_j|=0$sum_(j=1)^(n)|a_(j)|=0\sum_{j=1}^n |a_j| = 0$\sum _{j=1}^n|a_j|=0$, and since each $|a_j|\ge 0$$|a_j|\ge 0$|a_(j)| >= 0|a_j| \geq 0$|a_j|\ge 0$, we must have $|a_j|=0$$|a_j|=0$|a_(j)|=0|a_j| = 0$|a_j|=0$ for all $j$$j$jj$j$, so $a_j=0$$a_j=0$a_(j)=0a_j = 0$a_j=0$, hence $x=(0,0,\dots ,0)$$x=(0,0,\dots ,0)$x=(0,0,dots,0)x = (0, 0, \ldots, 0)$x=(0,0,\dots ,0)$.
• If $x=0$$x=0$x=0x = 0$x=0$, then $\Vert x\Vert =\sum _{j=1}^n|0|=0$$\Vert x\Vert =\sum _{j=1}^n |0|=0$||x||=sum_(j=1)^(n)|0|=0\|x\| = \sum_{j=1}^n |0| = 0$\Vert x\Vert =\sum _{j=1}^n|0|=0$.
  This holds.

• Vector Space: $C_0(\mathbb{R})$$C_0(\mathbb{R})$C_(0)(R)C_0(\mathbb{R})$C_0(\mathbb{R})$ is a vector space under pointwise addition and scalar multiplication, and if $f,g\in C_0(\mathbb{R})$$f,g\in C_0(\mathbb{R})$f,g inC_(0)(R)f, g \in C_0(\mathbb{R})$f,g\in C_0(\mathbb{R})$, then $f+g$$f+g$f+gf + g$f+g$ and $cf$$cf$cfc f$cf$ (for $c\in \mathbb{R}$$c\in \mathbb{R}$c inRc \in \mathbb{R}$c\in \mathbb{R}$ or $\mathbb{C}$$\mathbb{C}$C\mathbb{C}$\mathbb{C}$) vanish at infinity, as $|f(x)|+|g(x)|\to 0$$|f(x)|+|g(x)|\to 0$|f(x)|+|g(x)|rarr0|f(x)| + |g(x)| \to 0$|f(x)|+|g(x)|\to 0$ and $|c||f(x)|\to 0$$|c||f(x)|\to 0$|c||f(x)|rarr0|c| |f(x)| \to 0$|c||f(x)|\to 0$.
• Normed Space: The supremum norm is well-defined, since continuous functions vanishing at infinity are bounded (if $|f(x)|\nrightarrow 0$$|f(x)|\nrightarrow 0$|f(x)|↛0|f(x)| \not\to 0$|f(x)|\nrightarrow 0$, there’s a sequence $x_n\to \mathrm{\infty }$$x_n\to \mathrm{\infty }$x_(n)rarr oox_n \to \infty$x_n\to \mathrm{\infty }$ with $|f(x_n)|\ge ϵ$$|f(x_n)|\ge ϵ$|f(x_(n))| >= epsilon|f(x_n)| \geq \epsilon$|f(x_n)|\ge ϵ$, contradicting the definition).
• Completeness: Take a Cauchy sequence $\{f_n\}$$\{f_n\}${f_(n)}\{ f_n \}$\{f_n\}$ in $C_0(\mathbb{R})$$C_0(\mathbb{R})$C_(0)(R)C_0(\mathbb{R})$C_0(\mathbb{R})$: for every $ϵ>0$$ϵ>0$epsilon > 0\epsilon > 0$ϵ>0$, there exists $N$$N$NN$N$ such that for $m,n>N$$m,n>N$m,n > Nm, n > N$m,n>N$, $\Vert f_n-f_m{\Vert }_{\mathrm{\infty }}=\underset{x}{sup}|f_n(x)-f_m(x)|<ϵ$$\Vert f_n-f_m{\Vert }_{\mathrm{\infty }}=\underset{x}{sup} |f_n(x)-f_m(x)|<ϵ$||f_(n)-f_(m)||_(oo)=s u p _(x)|f_(n)(x)-f_(m)(x)| < epsilon\|f_n – f_m\|_\infty = \sup_x |f_n(x) – f_m(x)| < \epsilon$\Vert f_n-f_m{\Vert }_{\mathrm{\infty }}=\underset{x}{sup}|f_n(x)-f_m(x)|<ϵ$. For each fixed $x$$x$xx$x$, $\{f_n(x)\}$$\{f_n(x)\}${f_(n)(x)}\{ f_n(x) \}$\{f_n(x)\}$ is a Cauchy sequence in $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$ or $\mathbb{C}$$\mathbb{C}$C\mathbb{C}$\mathbb{C}$, which is complete, so $f_n(x)\to f(x)$$f_n(x)\to f(x)$f_(n)(x)rarr f(x)f_n(x) \to f(x)$f_n(x)\to f(x)$ pointwise. Since $f_n$$f_n$f_(n)f_n$f_n$ is Cauchy in the sup norm, $f_n$$f_n$f_(n)f_n$f_n$ converges uniformly to $f$$f$ff$f$ (standard result: uniform limit of continuous functions is continuous). Now, $f\in C_0(\mathbb{R})$$f\in C_0(\mathbb{R})$f inC_(0)(R)f \in C_0(\mathbb{R})$f\in C_0(\mathbb{R})$: given $ϵ>0$$ϵ>0$epsilon > 0\epsilon > 0$ϵ>0$, choose $N$$N$NN$N$ so $\Vert f_n-f_m{\Vert }_{\mathrm{\infty }}<ϵ/2$$\Vert f_n-f_m{\Vert }_{\mathrm{\infty }}<ϵ/2$||f_(n)-f_(m)||_(oo) < epsilon//2\|f_n – f_m\|_\infty < \epsilon/2$\Vert f_n-f_m{\Vert }_{\mathrm{\infty }}<ϵ/2$ for $m,n>N$$m,n>N$m,n > Nm, n > N$m,n>N$, and since $f_N\in C_0$$f_N\in C_0$f_(N)inC_(0)f_N \in C_0$f_N\in C_0$, there’s a compact $K$$K$KK$K$ such that $|f_N(x)|<ϵ/2$$|f_N(x)|<ϵ/2$|f_(N)(x)| < epsilon//2|f_N(x)| < \epsilon/2$|f_N(x)|<ϵ/2$ for $x\notin K$$x\notin K$x!in Kx \notin K$x\notin K$. Then, $|f(x)|\le |f(x)-f_N(x)|+|f_N(x)|<ϵ/2+ϵ/2=ϵ$$|f(x)|\le |f(x)-f_N(x)|+|f_N(x)|<ϵ/2+ϵ/2=ϵ$|f(x)| <= |f(x)-f_(N)(x)|+|f_(N)(x)| < epsilon//2+epsilon//2=epsilon|f(x)| \leq |f(x) – f_N(x)| + |f_N(x)| < \epsilon/2 + \epsilon/2 = \epsilon$|f(x)|\le |f(x)-f_N(x)|+|f_N(x)|<ϵ/2+ϵ/2=ϵ$ outside $K$$K$KK$K$, so $f\in C_0(\mathbb{R})$$f\in C_0(\mathbb{R})$f inC_(0)(R)f \in C_0(\mathbb{R})$f\in C_0(\mathbb{R})$.

• $0=\lambda x_1$$0=\lambda x_1$0=lambdax_(1)0 = \lambda x_1$0=\lambda x_1$,
• $x_1=\lambda x_2$$x_1=\lambda x_2$x_(1)=lambdax_(2)x_1 = \lambda x_2$x_1=\lambda x_2$,
• $x_2=\lambda x_3$$x_2=\lambda x_3$x_(2)=lambdax_(3)x_2 = \lambda x_3$x_2=\lambda x_3$,
• $x_3=\lambda x_4$$x_3=\lambda x_4$x_(3)=lambdax_(4)x_3 = \lambda x_4$x_3=\lambda x_4$,
• and so on.

• If $\lambda \ne 0$$\lambda \ne 0$lambda!=0\lambda \neq 0$\lambda \ne 0$, then $x_1=0$$x_1=0$x_(1)=0x_1 = 0$x_1=0$.
• Then $x_1=\lambda x_2$$x_1=\lambda x_2$x_(1)=lambdax_(2)x_1 = \lambda x_2$x_1=\lambda x_2$ gives $0=\lambda x_2$$0=\lambda x_2$0=lambdax_(2)0 = \lambda x_2$0=\lambda x_2$, so $x_2=0$$x_2=0$x_(2)=0x_2 = 0$x_2=0$ (since $\lambda \ne 0$$\lambda \ne 0$lambda!=0\lambda \neq 0$\lambda \ne 0$).
• Continue: $x_2=\lambda x_3$$x_2=\lambda x_3$x_(2)=lambdax_(3)x_2 = \lambda x_3$x_2=\lambda x_3$ gives $0=\lambda x_3$$0=\lambda x_3$0=lambdax_(3)0 = \lambda x_3$0=\lambda x_3$, so $x_3=0$$x_3=0$x_(3)=0x_3 = 0$x_3=0$, and inductively, $x_n=0$$x_n=0$x_(n)=0x_n = 0$x_n=0$ for all $n$$n$nn$n$.

• In a reflexive Banach space $X$$X$XX$X$, $X\cong X^{\ast \ast }$$X\cong X^{\ast \ast }$X~=X^(****)X \cong X^{**}$X\cong X^{\ast \ast }$ via $J$$J$JJ$J$.
• The dual $X^{\ast }$$X^{\ast }$X^(**)X^*$X^{\ast }$ is a Banach space (complete since $X$$X$XX$X$ is normed), and its dual is $X^{\ast \ast \ast }$$X^{\ast \ast \ast }$X^(******)X^{***}$X^{\ast \ast \ast }$. Reflexivity of $X^{\ast }$$X^{\ast }$X^(**)X^*$X^{\ast }$ means $X^{\ast }\cong X^{\ast \ast \ast }$$X^{\ast }\cong X^{\ast \ast \ast }$X^(**)~=X^(******)X^* \cong X^{***}$X^{\ast }\cong X^{\ast \ast \ast }$.
• Since $X$$X$XX$X$ is reflexive, $X^{\ast \ast }\cong X$$X^{\ast \ast }\cong X$X^(****)~=XX^{**} \cong X$X^{\ast \ast }\cong X$, and $X^{\ast \ast \ast }=(X^{\ast }{)}^{\ast \ast }$$X^{\ast \ast \ast }=(X^{\ast }{)}^{\ast \ast }$X^(******)=(X^(**))^(****)X^{***} = (X^*)^{**}$X^{\ast \ast \ast }=(X^{\ast }{)}^{\ast \ast }$. We need to determine if $X^{\ast }$$X^{\ast }$X^(**)X^*$X^{\ast }$ maps onto $X^{\ast \ast \ast }$$X^{\ast \ast \ast }$X^(******)X^{***}$X^{\ast \ast \ast }$.

• ${\ell }^2$${\ell }^2$ℓ^(2)\ell^2${\ell }^2$ is reflexive, and $({\ell }^2{)}^{\ast }\cong {\ell }^2$$({\ell }^2{)}^{\ast }\cong {\ell }^2$(ℓ^(2))^(**)~=ℓ^(2)(\ell^2)^* \cong \ell^2$({\ell }^2{)}^{\ast }\cong {\ell }^2$ is also reflexive.
• Finite-dimensional spaces are reflexive, and their duals are too.

• A normed linear space $X$$X$XX$X$ of finite dimension, say $\dim X=n$$\dim X=n$dim X=n\dim X = n$\dim X=n$, has a basis $\{e_1,e_2,\dots ,e_n\}$$\{e_1,e_2,\dots ,e_n\}${e_(1),e_(2),dots,e_(n)}\{ e_1, e_2, \ldots, e_n \}$\{e_1,e_2,\dots ,e_n\}$.
• The dual space $X^{\prime }$$X^{\prime }$X^(‘)X’$X^{\prime }$ consists of all linear functionals $f:X\to \mathbb{K}$$f:X\to \mathbb{K}$f:X rarrKf: X \to \mathbb{K}$f:X\to \mathbb{K}$ (where $\mathbb{K}=\mathbb{R}$$\mathbb{K}=\mathbb{R}$K=R\mathbb{K} = \mathbb{R}$\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$$\mathbb{C}$C\mathbb{C}$\mathbb{C}$), and continuity is automatic in finite dimensions since all linear functionals on a finite-dimensional normed space are bounded.
• Define the dual basis $\{e_1^{\ast },e_2^{\ast },\dots ,e_n^{\ast }\}$$\{e_1^{\ast },e_2^{\ast },\dots ,e_n^{\ast }\}${e_(1)^(**),e_(2)^(**),dots,e_(n)^(**)}\{ e_1^*, e_2^*, \ldots, e_n^* \}$\{e_1^{\ast },e_2^{\ast },\dots ,e_n^{\ast }\}$ where $e_i^{\ast }(e_j)={\delta }_{ij}$$e_i^{\ast }(e_j)={\delta }_{ij}$e_(i)^(**)(e_(j))=delta_(ij)e_i^*(e_j) = \delta_{ij}$e_i^{\ast }(e_j)={\delta }_{ij}$ (Kronecker delta). For any $x=\sum _{i=1}^n{a}_i{e}_i$$x=\sum _{i=1}^n a_i{e}_i$x=sum_(i=1)^(n)a_(i)e_(i)x = \sum_{i=1}^n a_i e_i$x=\sum _{i=1}^n{a}_i{e}_i$, we have $e_i^{\ast }(x)=a_i$$e_i^{\ast }(x)=a_i$e_(i)^(**)(x)=a_(i)e_i^*(x) = a_i$e_i^{\ast }(x)=a_i$, and any functional $f\in X^{\prime }$$f\in X^{\prime }$f inX^(‘)f \in X’$f\in X^{\prime }$ is $f=\sum _{i=1}^{n}f(e_i)e_i^{\ast }$$f=\sum _{i=1}^n f(e_i)e_i^{\ast }$f=sum_(i=1)^(n)f(e_(i))e_(i)^(**)f = \sum_{i=1}^n f(e_i) e_i^*$f=\sum _{i=1}^{n}f(e_i)e_i^{\ast }$.

• For $X={\mathbb{R}}^2$$X={\mathbb{R}}^2$X=R^(2)X = \mathbb{R}^2$X={\mathbb{R}}^2$ with Euclidean norm, $X^{\prime }\cong {\mathbb{R}}^2$$X^{\prime }\cong {\mathbb{R}}^2$X^(‘)~=R^(2)X’ \cong \mathbb{R}^2$X^{\prime }\cong {\mathbb{R}}^2$ (e.g., via $f(x,y)=ax+by$$f(x,y)=ax+by$f(x,y)=ax+byf(x, y) = ax + by$f(x,y)=ax+by$), and $\dim X^{\prime }=2$$\dim X^{\prime }=2$dim X^(‘)=2\dim X’ = 2$\dim X^{\prime }=2$.