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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.

The answer to (b) is

\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 <n-1. 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 <n-1  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)<m(H)+\epsilon and m(V)<m(K)+\epsilon . Then m(U)+m(V)<m(H)+m(K)+2\epsilon \leq m(H+K)+2\epsilon \leq m(U+V)+2\epsilon .

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2 thoughts on “Solutions to problems posted in July

  1. Why there is no question from any part of algebra? I am expecting some problems related to algebra in near future.

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