Home » Uncategorized » Solutions to problems posted in July

# Solutions to problems posted in July

1.

a) What is $T+T?.$ Is the representation $z=z_{1}+z_{2},z_{1}\in T,z_{2}\in T$ unique for $z\in T+T?$

b) What is $\bigcup\limits_{n=2}^{\infty}\left\{T+T+...+T\right.$(n terms)$\left.\right\}?$ Do we have uniqueness here?

c) If $X$ is any normed linear space of dimension greater than $1$ prove that $\{x:\left\Vert x\right\Vert \leq 2\}=\{y+z:\left\Vert y\right\Vert =\left\Vert z\right\Vert =1\}.$

We claim that $T+T=\{z:\left\vert z\right\vert \leq 2\}$ and that the representation is unique. Clearly $T+T\subset \{z:\left\vert z\right\vert\leq 2\}.$ Let $0\leq r\leq 2$. Then $r=e^{ia}+e^{-ia}$ where $\cos a=\frac{r}{2}.$ For
any $b\in \mathbb{R}$ we have the representation $re^{ib}=e^{i(a+b)}+e^{i(b-a)}.$ If $r=e^{i\alpha }+e^{i\beta }$ with $\alpha$ and $\beta$ real then $\sin\alpha =-\sin \beta$ and $\cos \alpha +\cos \beta =r.$ It follows that $\cos \alpha =\cos \beta =r/2.$ Thus $\beta =2\pi -\alpha$ and $e^{i\beta }=e^{-i\alpha }.$ Thus the representation is unique. Uniqueness of representation for $re^{ib}$ follows from this.

$\mathbb{C}=\bigcup\limits_{n=2}^{\infty}\left\{T+T+...+T\right.$(n terms)$\left.\right\}$

and the representation $z=z_{1}+z_{2}+...+z_{n}$ with $\left\vert z_{j}\right\vert=1$ for all $j$ is unique if and only if $\left\vert z\right\vert =n.$ Clearly any $z\in \mathbb{C}$ is a sum of complex numbers with modulus at most $2$ and hence it can be written as a sum of elements of $T.$ Now fix $n$ and suppose $z=z_{1}+z_{2}+...+z_{n}$ with $\left\vert z_{j}\right\vert =1$ for all $j.$ Then $\left\vert z\right\vert \leq 1+1+...+1=n$ and if $\left\vert z\right\vert =n$ then $z_{j}=t_{j}z_{1},1\leq j\leq n$ with $t_{j}\geq 0$ for all $j$. But $\left\vert z_{j}\right\vert =1$ for all $j$ so $t_{j}=1$ for all $j$ and $z_{j}=\frac{z}{n}.$ Thus the representation is unique when $\left\vert z\right\vert =n.$ We claim that it is not unique if $\left\vert z\right\vert =2.$ There are infinitely many complex numbers $c$ such that $\left\vert c\right\vert =1$ and $\left\vert z-c\right\vert In fact one such number is $e^{it}$ where $z=re^{it}$ is the polar form of $z.$ By continuity the inequality $\left\vert z-e^{i\alpha }\right\vert   holds for any $\alpha$ sufficiently close to $t$. Any representation of $c$ as a sum of elements of $T$ gives a similar representation for $z.$ Hence the representation is not unique.

c) In the case of complex scalars we can restrict to a one-dimensional subspace and in the case of real scalars we can look at a two dimensional subspace containing a given vector $x$ with $\left\Vert x\right\Vert \leq 2$.

Note that uniqueness fails: let $f_{1}(x)=x,g_{1}(x)=1-x,f_{2}(x)=x^{2}$ and $g_{2}(x)=1-x^{2}$ in $C[0,1]$. Then $f_{1}+g_{1}=1=f_{2}+g_{2}$.

2. If $X_{n}^{\prime }$ are independent identically distributed positive random variables does it follow that $EX_{1}X_{2}...=(EX_{1})(EX_{2})...$ assuming that all the products and expectations exist.

No! Let $X_{n}^{\prime }$ take values $1/2$ and $3/2$ with probabilty $1/2$ each. Then $EX_{n}=1/4+3/4=1$ so $(EX_{1})(EX_{2})...=1$. We claim that $EX_{1}X_{2}...=0$. Since $E(X_{1}X_{2}...X_{n+1}/X_{1},X_{2},...,X_{n})=X_{1}X_{2}...X_{n}EX_{n}=X_{1}X_{2}...X_{n}$ it follows that $\{X_{1}X_{2}...X_{n}\}$ is a non-negative martingale. Hence it converges a.s. We shall show that the limit is $0$ a.s. thereby completing the proof. Let $N_{n}=\#\{k\leq n:X_{k}=3/2\}$. Then $X_{1}X_{2}...X_{n}=(1/2)^{n-N_{n}}(3/2)^{N_{n}}=\frac{3^{N_{n}}}{2^{n}}$. By SLLN applied to $\{I_{X_{k}}=3/2\}$ we see that $\frac{1}{n}N_{n}\rightarrow 1/2$. Hence $\log (\frac{3^{N_{n}}}{2^{n}})=N_{n}\log 3-n\log 2=n(\frac{1}{n}N_{n}\log 3-\log 2)\rightarrow \rightarrow \infty$ a.s. because $1/2\log3-\log 2<0$.

3. Consider the inequalities $m(A+B)\leq m(A)+m(B)$ and $m(A+B)\geq m(A)+m(B)$. Are these true when $A$ and $B$ are open sets in $\mathbb{R}$? Are these true when $A$ and $B$ are compact sets in $\mathbb{R}$?

Suppose $A=K_{1}$ and $B=K_{2}$ are compact. Translate $K_{2}$ by $\sup K_{1}-\inf K_{2}$. Then the inequality does not change and the proof is therefore reduced to the case $\sup K_{1}=\inf K_{2}(=c,$ say$)$. In this case $K_{1}\cup K_{2}\subset K_{1}+K_{2}-c$ ( because $c\in K_{1}\cap K_{2})$ and $m(K_{1})+m(K_{2})=m(K_{1}\cup K_{2})$ (because $K_{1}\backslash \{c\}$ and $K_{2}$ are disjoint) so $m(K_{1})+m(K_{2})\leq m(K_{1}+K_{2}-c)=m(K_{1}+K_{2})$. The reverse inequality fails when $K_{1}=K_{2}=C$ the Cantor set. It is not true that $m(U+V)\leq m(U)+m(V)$ for all open sets: there is an open set $U$ such that $C\subset U$ and $m(U)<1/2$. Since $[0,2]=C+C\subset U+U$ we would have $2\leq m(U+U)\leq 2m(U)$ and choose compact sets $H,K$ such that $H\subset U,K\subset V,m(U) and $m(V). Then $m(U)+m(V).