Sample Solution

1. State whether the following statements are true or false. Justify with a short proof or a counter example.

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

Justification:

1. Inner Product: We can define an inner product on $l^3$$l^3$l^(3)l^3$l^3$ as follows:

2. Completeness: The space $l^3$$l^3$l^(3)l^3$l^3$ is not complete with respect to this inner product. To see this, consider the sequence of sequences $x^{(k)}=(1,\frac{1}{2},\dots ,\frac{1}{k},0,0,\dots )$$x^{(k)}=(1,\frac{1}{2},\dots ,\frac{1}{k},0,0,\dots )$x^((k))=(1,(1)/(2),dots,(1)/(k),0,0,dots)x^{(k)} = (1, \frac{1}{2}, \ldots, \frac{1}{k}, 0, 0, \ldots)$x^{(k)}=(1,\frac{1}{2},\dots ,\frac{1}{k},0,0,\dots )$. Each $x^{(k)}$$x^{(k)}$x^((k))x^{(k)}$x^{(k)}$ is in $l^3$$l^3$l^(3)l^3$l^3$, but its limit as $k\to \mathrm{\infty }$$k\to \mathrm{\infty }$k rarr ook \to \infty$k\to \mathrm{\infty }$, which is the sequence $(1,\frac{1}{2},\frac{1}{3},\dots )$$(1,\frac{1}{2},\frac{1}{3},\dots )$(1,(1)/(2),(1)/(3),dots)(1, \frac{1}{2}, \frac{1}{3}, \ldots)$(1,\frac{1}{2},\frac{1}{3},\dots )$, 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 }$.

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

Justification:

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(x)|\le C\parallel x\parallel$$|f(x)|\le C\parallel x\parallel$|f(x)| <= C||x|||f(x)| \leq C \|x\|$|f(x)|\le C\parallel x\parallel$ for all $x\in X$$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(x_0)\ne 0$$f(x_0)\ne 0$f(x_(0))!=0f(x_0) \neq 0$f(x_0)\ne 0$.
3. Open Map: Consider an open ball $B(x,ϵ)$$B(x,ϵ)$B(x,epsilon)B(x, \epsilon)$B(x,ϵ)$ in $X$$X$XX$X$ centered at $x$$x$xx$x$ with radius $ϵ>0$$ϵ>0$epsilon > 0\epsilon > 0$ϵ>0$. We want to show that $f(B(x,ϵ))$$f(B(x,ϵ))$f(B(x,epsilon))f(B(x, \epsilon))$f(B(x,ϵ))$ 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(B(x,ϵ))$$y\in f(B(x,ϵ))$y in f(B(x,epsilon))y \in f(B(x, \epsilon))$y\in f(B(x,ϵ))$. Then there exists $z\in B(x,ϵ)$$z\in B(x,ϵ)$z in B(x,epsilon)z \in B(x, \epsilon)$z\in B(x,ϵ)$ such that $f(z)=y$$f(z)=y$f(z)=yf(z) = y$f(z)=y$.
   • Since $z\in B(x,ϵ)$$z\in B(x,ϵ)$z in B(x,epsilon)z \in B(x, \epsilon)$z\in B(x,ϵ)$, 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(z)-f(x)|\le C\parallel z-x\parallel <Cϵ$$|f(z)-f(x)|\le C\parallel z-x\parallel <Cϵ$|f(z)-f(x)| <= C||z-x|| < C epsilon|f(z) – f(x)| \leq C \|z – x\| < C \epsilon$|f(z)-f(x)|\le C\parallel z-x\parallel <Cϵ$.
   • This means $y=f(z)$$y=f(z)$y=f(z)y = f(z)$y=f(z)$ lies in an open interval $(f(x)-Cϵ,f(x)+Cϵ)$$(f(x)-Cϵ,f(x)+Cϵ)$(f(x)-C epsilon,f(x)+C epsilon)(f(x) – C \epsilon, f(x) + C \epsilon)$(f(x)-Cϵ,f(x)+Cϵ)$ which is contained in $f(B(x,ϵ))$$f(B(x,ϵ))$f(B(x,epsilon))f(B(x, \epsilon))$f(B(x,ϵ))$.

Statement: Every bounded linear map on a complex Banach space has an eigenvalue.

Justification:

Counterexample:

Statement: The image of a Cauchy sequence under a bounded linear map is also a Cauchy sequence.

Justification:

Proof:

1. Bounded Linear Map: Since $T$$T$TT$T$ is bounded, there exists a constant $M>0$$M>0$M > 0M > 0$M>0$ such that $\parallel T(x)\parallel \le M\parallel x\parallel$$\parallel T(x)\parallel \le M\parallel x\parallel$||T(x)|| <= M||x||\| T(x) \| \leq M \| x \|$\parallel T(x)\parallel \le M\parallel x\parallel$ for all $x\in X$$x\in X$x in Xx \in X$x\in X$.
2. Cauchy Sequence: Given any $ϵ>0$$ϵ>0$epsilon > 0\epsilon > 0$ϵ>0$, there exists $N$$N$NN$N$ such that for all $m,n\ge N$$m,n\ge N$m,n >= Nm, n \geq N$m,n\ge N$, $\parallel x_m-x_n\parallel <\frac{ϵ}{M}$$\parallel x_m-x_n\parallel <\frac{ϵ}{M}$||x_(m)-x_(n)|| < (epsilon )/(M)\| x_m – x_n \| < \frac{\epsilon}{M}$\parallel x_m-x_n\parallel <\frac{ϵ}{M}$.
3. Image under $T$$T$TT$T$: We have
   $$\parallel T(x_m)-T(x_n)\parallel =\parallel T(x_m-x_n)\parallel$$$$\parallel T(x_m)-T(x_n)\parallel =\parallel T(x_m-x_n)\parallel$$||T(x_(m))-T(x_(n))||=||T(x_(m)-x_(n))||\| T(x_m) – T(x_n) \| = \| T(x_m – x_n) \|$$\parallel T(x_m)-T(x_n)\parallel =\parallel T(x_m-x_n)\parallel$$
   Using the boundedness of $T$$T$TT$T$,
   $$\parallel T(x_m)-T(x_n)\parallel \le M\parallel x_m-x_n\parallel <M\frac{ϵ}{M}=ϵ$$$$\parallel T(x_m)-T(x_n)\parallel \le M\parallel x_m-x_n\parallel <M\frac{ϵ}{M}=ϵ$$||T(x_(m))-T(x_(n))|| <= M||x_(m)-x_(n)|| < M(epsilon )/(M)=epsilon\| T(x_m) – T(x_n) \| \leq M \| x_m – x_n \| < M \frac{\epsilon}{M} = \epsilon$$\parallel T(x_m)-T(x_n)\parallel \le M\parallel x_m-x_n\parallel <M\frac{ϵ}{M}=ϵ$$
   for all $m,n\ge N$$m,n\ge N$m,n >= Nm, n \geq N$m,n\ge N$.

Statement: 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 }\mathrm{A}=\mathrm{I}$${\mathrm{A}}^{\ast }\mathrm{A}=\mathrm{I}$A^(**)A=I\mathrm{A^*A} = \mathrm{I}${\mathrm{A}}^{\ast }\mathrm{A}=\mathrm{I}$.

Justification:

Proof:

1. Given Condition: We are given that $\mathrm{A}$$\mathrm{A}$A\mathrm{A}$\mathrm{A}$ is a bounded linear operator on a Hilbert space $H$$H$HH$H$ and $\mathrm{A}{\mathrm{A}}^{\ast }=\mathrm{I}$$\mathrm{A}{\mathrm{A}}^{\ast }=\mathrm{I}$AA^(**)=I\mathrm{AA^*} = \mathrm{I}$\mathrm{A}{\mathrm{A}}^{\ast }=\mathrm{I}$.
2. Identity Operator: $\mathrm{I}$$\mathrm{I}$I\mathrm{I}$\mathrm{I}$ is the identity operator on $H$$H$HH$H$.
3. To Prove: We need to show that ${\mathrm{A}}^{\ast }\mathrm{A}=\mathrm{I}$${\mathrm{A}}^{\ast }\mathrm{A}=\mathrm{I}$A^(**)A=I\mathrm{A^*A} = \mathrm{I}${\mathrm{A}}^{\ast }\mathrm{A}=\mathrm{I}$.
4. Step 1: Take any $x\in H$$x\in H$x in Hx \in H$x\in H$ and consider ${\mathrm{A}}^{\ast }x$${\mathrm{A}}^{\ast }x$A^(**)x\mathrm{A^*}x${\mathrm{A}}^{\ast }x$. We have
   $$\mathrm{A}({\mathrm{A}}^{\ast }x)=(\mathrm{A}{\mathrm{A}}^{\ast })x=\mathrm{I}x=x$$$$\mathrm{A}({\mathrm{A}}^{\ast }x)=(\mathrm{A}{\mathrm{A}}^{\ast })x=\mathrm{I}x=x$$A(A^(**)x)=(AA^(**))x=Ix=x\mathrm{A} (\mathrm{A^*} x) = (\mathrm{AA^*}) x = \mathrm{I} x = x$$\mathrm{A}({\mathrm{A}}^{\ast }x)=(\mathrm{A}{\mathrm{A}}^{\ast })x=\mathrm{I}x=x$$
   This shows that ${\mathrm{A}}^{\ast }x$${\mathrm{A}}^{\ast }x$A^(**)x\mathrm{A^*} x${\mathrm{A}}^{\ast }x$ is in the range of $\mathrm{A}$$\mathrm{A}$A\mathrm{A}$\mathrm{A}$.
5. Step 2: Now consider ${\mathrm{A}}^{\ast }(\mathrm{A}x)$${\mathrm{A}}^{\ast }(\mathrm{A}x)$A^(**)(Ax)\mathrm{A^*} (\mathrm{A} x)${\mathrm{A}}^{\ast }(\mathrm{A}x)$. We have
   $${\mathrm{A}}^{\ast }(\mathrm{A}x)=({\mathrm{A}}^{\ast }\mathrm{A})x$$$${\mathrm{A}}^{\ast }(\mathrm{A}x)=({\mathrm{A}}^{\ast }\mathrm{A})x$$A^(**)(Ax)=(A^(**)A)x\mathrm{A^*} (\mathrm{A} x) = (\mathrm{A^* A}) x$${\mathrm{A}}^{\ast }(\mathrm{A}x)=({\mathrm{A}}^{\ast }\mathrm{A})x$$
   Since ${\mathrm{A}}^{\ast }x$${\mathrm{A}}^{\ast }x$A^(**)x\mathrm{A^*} x${\mathrm{A}}^{\ast }x$ is in the range of $\mathrm{A}$$\mathrm{A}$A\mathrm{A}$\mathrm{A}$, we can write ${\mathrm{A}}^{\ast }(\mathrm{A}x)=x$${\mathrm{A}}^{\ast }(\mathrm{A}x)=x$A^(**)(Ax)=x\mathrm{A^*} (\mathrm{A} x) = x${\mathrm{A}}^{\ast }(\mathrm{A}x)=x$.
6. Step 3: Combining Steps 1 and 2, we get
   $${\mathrm{A}}^{\ast }(\mathrm{A}x)=x⟹({\mathrm{A}}^{\ast }\mathrm{A})x=x⟹({\mathrm{A}}^{\ast }\mathrm{A}-\mathrm{I})x=0$$$${\mathrm{A}}^{\ast }(\mathrm{A}x)=x⟹({\mathrm{A}}^{\ast }\mathrm{A})x=x⟹({\mathrm{A}}^{\ast }\mathrm{A}-\mathrm{I})x=0$$A^(**)(Ax)=xLongrightarrow(A^(**)A)x=xLongrightarrow(A^(**)A-I)x=0\mathrm{A^*} (\mathrm{A} x) = x \implies (\mathrm{A^* A}) x = x \implies (\mathrm{A^* A} – \mathrm{I}) x = 0$${\mathrm{A}}^{\ast }(\mathrm{A}x)=x⟹({\mathrm{A}}^{\ast }\mathrm{A})x=x⟹({\mathrm{A}}^{\ast }\mathrm{A}-\mathrm{I})x=0$$
   for all $x\in H$$x\in H$x in Hx \in H$x\in H$.
7. Step 4: Since $({\mathrm{A}}^{\ast }\mathrm{A}-\mathrm{I})x=0$$({\mathrm{A}}^{\ast }\mathrm{A}-\mathrm{I})x=0$(A^(**)A-I)x=0(\mathrm{A^* A} – \mathrm{I}) x = 0$({\mathrm{A}}^{\ast }\mathrm{A}-\mathrm{I})x=0$ for all $x\in H$$x\in H$x in Hx \in H$x\in H$, it follows that ${\mathrm{A}}^{\ast }\mathrm{A}=\mathrm{I}$${\mathrm{A}}^{\ast }\mathrm{A}=\mathrm{I}$A^(**)A=I\mathrm{A^* A} = \mathrm{I}${\mathrm{A}}^{\ast }\mathrm{A}=\mathrm{I}$.

2. a) Characterise all bounded linear functionals on a Hilbert space.

Riesz Representation Theorem

Properties

1. Linearity: $f(ax+by)=af(x)+bf(y)$$f(ax+by)=af(x)+bf(y)$f(ax+by)=af(x)+bf(y)f(ax + by) = a f(x) + b f(y)$f(ax+by)=af(x)+bf(y)$ for all $x,y\in \mathcal{H}$$x,y\in \mathcal{H}$x,y inHx, y \in \mathcal{H}$x,y\in \mathcal{H}$ and $a,b\in \mathbb{C}$$a,b\in \mathbb{C}$a,b inCa, b \in \mathbb{C}$a,b\in \mathbb{C}$ (or $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$).
2. Boundedness: $\parallel f\parallel <\mathrm{\infty }$$\parallel f\parallel <\mathrm{\infty }$||f|| < oo\| f \| < \infty$\parallel f\parallel <\mathrm{\infty }$.
3. Uniqueness: The vector $y$$y$yy$y$ in the Riesz representation is unique.
4. Norm Preservation: $\parallel f\parallel =\parallel y\parallel$$\parallel f\parallel =\parallel y\parallel$||f||=||y||\| f \| = \| y \|$\parallel f\parallel =\parallel y\parallel$, where $y$$y$yy$y$ is the vector in the Riesz representation.
5. Continuity: Bounded linear functionals are continuous.
6. Dual Space: The set of all bounded linear functionals on $\mathcal{H}$$\mathcal{H}$H\mathcal{H}$\mathcal{H}$ forms the dual space ${\mathcal{H}}^{\ast }$${\mathcal{H}}^{\ast }$H^(**)\mathcal{H}^*${\mathcal{H}}^{\ast }$.

Examples

1. Finite-Dimensional Spaces: In finite-dimensional Hilbert spaces, every linear functional is bounded.
2. Sequence Spaces: In ${\ell }^2$${\ell }^2$ℓ^(2)\ell^2${\ell }^2$, the space of square-summable sequences, the bounded linear functional corresponding to a sequence $(a_n)$$(a_n)$(a_(n))(a_n)$(a_n)$ is $f(x)=\sum _{n=1}^{\mathrm{\infty }}{a}_n{x}_n$$f(x)=\sum _{n=1}^{\mathrm{\infty }} a_n{x}_n$f(x)=sum_(n=1)^(oo)a_(n)x_(n)f(x) = \sum_{n=1}^{\infty} a_n x_n$f(x)=\sum _{n=1}^{\mathrm{\infty }}{a}_n{x}_n$.
3. Function Spaces: In $L^2([0,1])$$L^2([0,1])$L^(2)([0,1])L^2([0, 1])$L^2([0,1])$, the space of square-integrable functions on $[0,1]$$[0,1]$[0,1][0, 1]$[0,1]$, the bounded linear functional corresponding to a function $g(x)$$g(x)$g(x)g(x)$g(x)$ is $f(f)={\int }_0^{1}f(x)g(x)dx$$f(f)={\int }_0^1 f(x)g(x)dx$f(f)=int_(0)^(1)f(x)g(x)dxf(f) = \int_{0}^{1} f(x) g(x) \, dx$f(f)={\int }_0^{1}f(x)g(x)dx$.

Step 1: Take an Arbitrary Open Set in ${\mathbf{R}}^3$${\mathbf{R}}^3$R^(3)\mathbf{R}^3${\mathbf{R}}^3$

Step 2: Pre-image of $\mathbf{y}$$\mathbf{y}$y\mathbf{y}$\mathbf{y}$

Step 3: Open Ball Around $\mathbf{x}$$\mathbf{x}$x\mathbf{x}$\mathbf{x}$

Step 4: Image of the Open Ball

Step 5: Show the Open Ball in ${\mathbf{R}}^2$${\mathbf{R}}^2$R^(2)\mathbf{R}^2${\mathbf{R}}^2$

Conclusion

Finite-Dimensional Case

Isomorphism Between $X$$X$XX$X$ and $X^{\ast \ast }$$X^{\ast \ast }$X^(****)X^{**}$X^{\ast \ast }$

Isometry

Conclusion