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Consider the following set

\begin{eqnarray*} S &=&\Big\{\lambda\in \mathbb{C};\;\exists (x_n,y_n)\in \mathbb{C}^2\,;\;\;\;|x_n+y_n|=1, \\ &&\phantom{++++++++++}\;\displaystyle\lim_{n\longrightarrow+\infty}|x_n|^2+x_n\overline{y_n}= \lambda\;\;\hbox{and}\;\;\displaystyle\lim_{n\longrightarrow+\infty}|x_n|=\frac{1}{2}\Big\}. \end{eqnarray*}

I want to show that $S$ is not convex.

  • If $(x_n,y_n)=(1/2,1/2)$ then $1/2\in S$.

  • If $(x_n,y_n)=(-1/2,3/2)$ then $-1/2\in S$.

I hope to show that $0\notin S$.

Thank you for your help.

Schüler
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  • From $|x+y|=1$ you get that $[|x^2|+x\overline{y}]+[|y|^2+\overline{x}y]=1$. Therefore, $x\overline{y}\to \lambda-1/4$. If $\lambda$ is real, then $\overline{x}y\to \lambda-1/4$. Hence, $|y|^2\to 5/4-2\lambda$. Now, $x\overline{y}\overline{x}y$ is on one hand $|x|^2|y|^2=\frac{1}{4}(5/4-2\lambda)$ and on the other $(\lambda-1/4)^2$. This gives you a condition $(\lambda-1/4)^2=\frac{1}{4}(5/4-\lambda)$ that $\lambda$ must satisfy and $\lambda=0$ doesn't. – blueInk Mar 24 '18 at 12:23

1 Answers1

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As you have shown $\lambda_1=1/2$ and $\lambda_2=-1/2$ both are in $S$. If $S$ were convex then for every $\alpha\in(0,1)$ we must have $\alpha\lambda_1+(1-\alpha)\lambda_2\in S$. In particular for $\alpha=1/2$ it follows that $0\in S$. By definition of $S$ this implies that there exists $(x_n,y_n)\in \mathbb{C}^2$ such that $|x_n+y_n|=1, \lim_n(|x_n|^2+x_n\overline{y}_n)=0$ and $\lim_n|x_n|=1/2$. From these conditions we get the following implications $$\lim_nx_n\overline{y}_n=-\lim_n|x_n|^2=-1/4\Rightarrow\lim_n|x_n||\overline{y}_n|=|\lim_nx_n\overline{y}_n|=1/4\Rightarrow \lim_n|\overline{y}_n|=1/2$$ since the absolute value function $|\cdot|$ is continuous. On the other hand $$|x_n+y_n|=1\Rightarrow (x_n+y_n)(\overline{x}_n+\overline{y}_n)=1\Rightarrow|x_n|^2+x_n\overline{y}_n+\overline{x}_ny_n+|y_n|^2=1$$ This implies $$\lim_n(\overline{x}_ny_n+|y_n|^2)=1\Rightarrow |\lim_n\overline{x}_ny_n+|y_n|^2|=1\Rightarrow\lim_n|\overline{x}_ny_n|+\lim_n|y_n|^2\geqslant 1$$ by triangle inequality. Using $\lim_n|\overline{x}_n|=\lim_n|x_n|=1/2$ and $\lim_n|y_n|=\lim_n|\overline{y}_n|=1/2$ one gets $$\lim_n|\overline{x}_ny_n|+\lim_n|y_n|^2=\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2^2}=\frac{1}{2}<1$$ which is a contradiction.

Arian
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