# Abstract Nonsense

## Halmos Chapter one Section 13 and 14: Linear Functionals and Bracket Notation

1.

Problem: Consider the set $\mathbb{C}$ of complex numbers as a vector space over $\mathbb{R}$. Suppose that for each $\zeta=\xi_1+i\xi_2$ in $\mathbb{C}$ (where $\xi_1,\xi_2\in\mathbb{R}$) the function $y$ is given by

a) $y(\zeta)=\xi_1$

b) $y(\zeta)=\xi_2$

c) $y(\zeta)=\xi_1^2$

d) $y(\zeta)=\xi_1-i\xi_2$

e) $y(\zeta)=\sqrt{\xi_1^2+\xi_2^2}$

In which cases are these linear functionals?

Proof:

a) Clearly $y:\mathbb{C}\to\mathbb{R}$ so it is indeed a map from the vector space to it’s underlying field, and so it’s a functional. To see it’s linear we merely notice that if $\zeta=x_1+i\xi_2$ and $\zeta'=\xi'_1+\xi'_2$ then

$\alpha\zeta+\beta\zeta'=(\alpha\xi_1+\beta\xi'_1)+i(\alpha\xi_2+\beta\xi'_2)$

and so

$y\left(\alpha\zeta+\beta\zeta'\right)=\alpha\xi_1+\beta\xi'_1=\alpha y(\zeta)+\beta y(\zeta')$

so that it is indeed linear.

b) Similarly, we can see that $y:\mathbb{C}\to\mathbb{R}$ so it is indeed a function and if $\zeta,\zeta'$ are defined above we get again that

$\alpha\zeta+\beta\zeta'=(\alpha\xi_1+\beta\xi'_1)+i(\alpha\xi_2+\beta\xi'_2)$

and so

$y(\zeta+\zeta')=\alpha\xi_2+\beta\xi'_2=\alpha y(\zeta)+\beta y(\zeta')$

so it is linear.

c) This is not linear, notice that $y(2\cdot 2)=16\ne 2\cdot y(2)=4$

d) This is not a functional since $y:\mathbb{C}\to\mathbb{C}$ surjectively. More simply, notice that $y(i)=-i\notin\mathbb{R}$

e) This is not linear. Notice that

$y(2+4i)=\sqrt{2^2+4^2i}=\sqrt{20}\ne 2+4=\sqrt{2^2}+\sqrt{4^2}=y(2)+y(4i)$

and so $y$ is not linear.

2.

Problem: Suppose that for each $\zeta=(\xi_1,\xi_2,\xi_3)\in\mathbb{C}^3$ the function $y$ is defined by

a) $y(\zeta)=\xi_1+\xi_2$

b) $y(\zeta)=\xi_1-\xi_3^2$

c) $y(\zeta)=\xi_1+1$

d) $\xi_1-2\xi_2+3\xi_3$

Which of these are linear functions for $\mathbb{C}^3$?

Proof: Assuming we’re taking the underlying field to be $\mathbb{C}$

a) This is a functional since clearly $y:\mathbb{C}^3\to\mathbb{C}$. Now, suppose that $\zeta=(\xi_1,\xi_2,\xi_3)$ and $\zeta'=\left(\xi'_1,\xi'_2,\xi'_3\right)$ then clearly

$\alpha\zeta+\beta\zeta'=\left(\alpha\xi_1+\beta\xi'_1,\alpha\xi_2+\beta\xi'_2,\alpha\xi_3+\beta\xi'_3\right)$

and so

$y(\zeta+\zeta')=\alpha \xi+\beta\xi'_1+\alpha\xi_2+\beta\xi'_2=\alpha(\xi_1+\xi_2)+\beta(\xi'_1+\xi'_2)=\alpha y(\zeta)+\beta y(\zeta')$

so that $y$ is clearly linear.

b) This is not linear, since $y(2(0,0,2))=-(4)^2=-16\ne 2y((0,0,2))=2\cdot -(2^2)=-8$

c) This is not linear since $y((0,0,0))=1\ne 0$

d) Clearly this is a functional since $y:\mathbb{C}^3\to\mathbb{C}$. Furthermore, let $\zeta,\zeta'$ be as in a) then we see once again that

$\alpha\zeta+\beta\zeta'=\left(\alpha\xi_1+\beta\xi'_1,\alpha\xi_2+\beta\xi'_2,\alpha\xi_3+\beta\xi'_3\right)$

and so

$y\left(\alpha\zeta+\beta\zeta'\right)=(\alpha\xi_1+\beta\xi'_1)-2(\alpha\xi_2+\beta\xi'_2)+3(\alpha\xi_3+\beta\xi'_3)$

which upon expansion and regrouping is

$\alpha(\xi_1-2\xi_2+3\xi_3)+\beta(\xi'_1-2\xi'_2+3\xi'_3)=\alpha y(\zeta)+\beta y(\zeta')$

and so $y$ is linear.

3.

Problem: Suppose that for each $p\in\mathbb{R}[x]$ define the function $y$ by

a) $\displaystyle y(p)=\int_{-1}^2 p(x)dx$

b) $\displaystyle y(p)=\int_0^2 p(x)^2dx$

c) $\displaystyle y(p)=\int_0^1 x^2p(x)dx$

d) $\displaystyle y(p)=\int_0^1 y\left(x^2\right)dx$

e) $y(p)=p'$

f) $y(p)=p''(1)$

Which of these are linear functionals?

Proof:

a) Clearly $y:\mathbb{R}[x]\to\mathbb{R}$ and so it’s a functional. Furthermore, using elementary properties of integrals we see that

$\displaystyle y\left(\alpha p+\beta q\right)=\int_{-1}^{2}\left(\alpha p(x)+\beta q(x)\right)dx=\alpha \int_{-1}^{2} p(x) dx+\beta \int_{-1}^{2} q(x)dx$

but clearly this is equal to $\alpha y(p)+\beta y(q)$

b) This is not linear since

$\displaystyle y(2x)=\int_{-1}^{2}(2x)^2 dx=4\int_{-1}^{2}x^2 dx \ne 2\int_{-1}^{2}x^2 dx=2 y(x)$

c) This is clearly a functional since $y:\mathbb{R}[x]\to\mathbb{R}$ and

$\displaystyle y(\alpha p+\beta q)=\int_0^1 x^2(\alpha p(x)+\beta q(x))dx=\alpha \int_0^1 x^2p(x) dx+\beta \int_0^1 x^2q(x)dx$

but this is clearly equal to $\alpha y(p)+\beta y(q)$

d) This is clearly a functional since $y:\mathbb{R}[x]\to\mathbb{R}$ and we see that

$\displaystyle y\left(\alpha p+\beta q\right)=\int_0^1\left(\alpha p\left(x^2\right)+\beta q\left(x^2\right)\right)dx=\alpha\int_0^1 p\left(x^2\right) dx+\beta\int_0^1 q\left(x^2\right) dx$

but clearly this is equal to $\alpha y(p)+\beta y(q)$

e) This is not a functional since $y\left(x^2\right)=2x\notin\mathbb{R}$

d) This is clearly a linear function since $y:\mathbb{R}[x]\to\mathbb{R}$. Also,

$y\left(\alpha p+\beta q\right)=\left(\alpha p(x)+\beta q(x)\right)''\mid_{x=1}=\left(\alpha p''(x)+\beta q''(x)\right)\mid_{x=1}=\alpha p''(1)+\beta q''(1)$

but this is evidently equal to $\alpha y(p)+\beta y(q)$

4.

Problem: If $\left\{\alpha_n\right\}_{n\in\mathbb{N}}$ is an arbitrary sequence of complex numbers, and if $p\in\mathbb{C}[x]$ where $\displaystyle p(z)=\sum_{j=0}^{n}\xi_j x^j$ write $\displaystyle y(p)=\sum_{j=0}^{n} \xi_j \alpha ^j$. Prove that $y\in\mathbb{C}[x]^*$ (where the asterisk  indicates the dual space) and that every element of $\mathbb{C}[x]^*$ can be obtained in this manner by a suitable choice of the $\alpha$‘s

Proof: Clearly $y:\mathbb{C}[x]\to\mathbb{C}$ and we see that if $\displaystyle p(z)=\sum_{j=0}^{n}\xi_j z^j$ and $\displaystyle q(z)=\sum_{j=0}^{m}\gamma_j z^j$ where we may assume WLOG that $n\leqslant m$ then

$\displaystyle \alpha p(z)+\beta q(z)=\sum_{j=0}^{m}\left(\alpha \xi_j+\beta \gamma_j\right)z^j$

where we take $\xi_j=0,\text{ }j>n$. Thus,

$\displaystyle y\left(\alpha p(z)+\beta q(z)\right)=\sum_{j=0}^{m}\left(\alpha \xi_j+\beta \gamma_j\right)\alpha_j=\alpha \sum_{j=0}^{m}\xi_j \alpha_j+\beta\sum_{j=0}^{m}\gamma_j \alpha_j$

but by how we defined the $\xi_j$ we may rewrite this as

$\displaystyle \alpha\sum_{j=0}^{n}\xi_j \alpha_j+\beta\sum_{j=0}^{m}\gamma_j \alpha_j=\alpha y(p)+\beta y(q)$

from where it follows that $y\in\mathbb{C}[x]^*$.

The second part follows as an immediate corollary from the following  technical lemma

Lemma: Let $\mathcal{V}$ be a vector space over the field $F$ and let $\mathfrak{B}$ be a basis for $\mathcal{V}$. Then, any $\varphi \in\mathcal{V}^*$ is completely determined on $\mathfrak{B}$, in the sense that if $\eta\in\mathcal{V}^{*}$ is such that $\varphi(x)=\eta(x),\text{ }\forall x\in\mathfrak{B}$ then $\varphi=\eta$

Proof: Let $x\in\mathcal{V}$ then by assumption there exists a unique representation

$\displaystyle x=\sum_{j=1}^{n}\alpha_j x_j,\text{ }\alpha_j\in F,\text{ and }x_j\in\mathfrak{B}$

thus

$\displaystyle \varphi(x)=\varphi\left(\sum_{j=1}^{n}\alpha_j x_j\right)=\sum_{j=1}^{n}\alpha_j \varphi\left(x_j\right)$

but by assumption $\varphi(x_j)=\eta(x_j),\text{ }j=1,\cdots,n$ so that the above says

$\displaystyle \varphi(x)=\sum_{j=1}^{n}\alpha_j\varphi(x_j)=\sum_{j=1}^{n}\alpha_j\eta(x_j)=\eta\left(\sum_{j=1}^{n}\alpha_j x_j\right)=\eta(x)$

$\blacksquare$.

Well, actually the result isn’t really a corollary, but if $\varphi,\mathcal{V},\mathfrak{B}=\left\{x_\alpha\right\}_{\beta\in\mathcal{B}}$ are defined as above then we can see that

$\displaystyle \varphi(x)=\varphi\left(\sum_{\beta\in\mathcal{B}}\alpha_\beta x_\beta\right)=\sum_{\beta\in\mathcal{B}}\alpha_\beta\varphi(x_\beta)$

where it’s understood (by the unique representation of a vector in $\mathcal{V}$ by a  finite linear combination of elements of $\mathfrak{B}$) that $\alpha_\beta=0$ for all but finitely many $\beta\in\mathcal{B}$. In other words, for our example that sequence $\left\{\alpha_n\right\}_{n\in\mathbb{N}\cup\{0\}}$ may be taken to be $\alpha_n=\varphi\left(x^n\right)$

5.

Problem: If $\varphi\in\mathcal{V}^{*}-\{\bold{0}(x)\}$ for a vector space $\mathcal{V}$ over a field $F$ is it true that $\varphi$ is surjective?

Proof: Yes, since $\varphi\ne\bold{0}(x)$ there is some $x_0\in \mathcal{V}$ such that $\varphi(x_0)=\alpha_0\ne 0$. Then, for any $\alpha\in F$ we have that $\varphi(\alpha\alpha_0^{-1}x_0)=\alpha\alpha_0^{-1}\varphi(x_0)=\alpha$.

6.

Problem: Let $V$ be a vector space and $\varphi,\psi\in\text{Hom}\left(\mathscr{V},F\right)$ be such that

$\psi(z)=0\implies\varphi(z)=0$

then, $\varphi=\alpha\psi$ for some $\alpha\in F$

Proof:

Lemma: Let $\eta\in\text{Hom}\left(\mathscr{V},F\right)$ and $v_0\in\mathscr{V}$ be such that $\eta(v_0)\ne 0$. Then,

$\displaystyle \mathscr{V}=\ker\eta\oplus\text{span }\{v_0\}$

Proof: We merely note that if $v\in\mathscr{V}$ then

$\displaystyle v=\frac{\eta(v}{\eta(v_0)}v+\left(v-\frac{\eta(v)}{v_0}v_0\right)$

the first term clearly being in $\text{span }\{v_0\}$ and the second in $\ker\eta$. The conclusion follows by noticing that evidently $\ker\eta\cap\text{span }\{x_0\}=\{\bold{0}\}$. $\blacksquare$

So, now with this we can solve the problem. Clearly if $\psi=\bold{0}$ this is trivial since by assumption this would imply that $\varphi=\bold{0}$. So, assume not, then there exists some $z_0\in\mathscr{V}$ such that $\psi(z_0)\ne 0$. We claim then that

$\displaystyle \varphi=\frac{\varphi(z_0)}{\psi(z_0)}\psi$

To see this let $v\in\mathscr{V}$ be arbitrary. Then, $v=\beta z_0+k$ where $k\in\ker\psi\supseteq\ker\varphi$. Thus,

\begin{aligned}\frac{\varphi(z_0)}{\psi(z_0)}\psi(v) &=\frac{\varphi(z_0)}{\psi(z_0)}\psi\left(\beta z_0+k\right)\\ &=\frac{\varphi(z_0)}{\psi(z_0)}\beta\psi(z_0)\\ &=\beta\varphi(z_0)\\ &= \beta\varphi(z_0)+\varphi(k)\\ &= \varphi(\beta z_0+k)\\ &=\varphi(v)\end{aligned}

and so the conclusion follows.

September 30, 2010 -

1. I really appreciate your work. Do you have any solutions for problem 6 from the same sections: Linear Functionals and Brackets? Thank you in advance!

Comment by popita | November 8, 2010 | Reply

• Thank you. I guess I just forgot to post it. Be sure to respond to this so that I remember to write it up, it’s really late here.

Comment by drexel28 | November 9, 2010 | Reply

• Thank you for your prompt answer! Can you also post this problem 6 from these sections? Thanks!

Comment by ralucatoscano | November 9, 2010

• There you go friend.

Comment by drexel28 | November 10, 2010

2. I really appreciate your work! Do you have any solutions for problem 6 of Section 14: Brackets (page 22)? Thank you in advance!

Comment by ralucatoscano | November 8, 2010 | Reply

3. 1.I really appreciate your work! Do you have any solutions for problem 6 of Section 14: Brackets (page 22)? Thank you in advance!

Comment by ralucatoscano | November 8, 2010 | Reply

4. Resp. Sir/Ma’am,

Myself Deepak, i request how can i download all proof and problems of P R Halmos form your web site.

Thanking you

Comment by Deepak Gawali | August 13, 2011 | Reply