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MSMalekan
MSMalekan
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  1. Each $C_\epsilon$ is closed and non-empty ( since $\inf\{\|x-fx\|: x\in K\}=0$).
  2. If $\bigcap_\epsilon C_\epsilon=\emptyset$, then $a>0$; by definition there exists a $y_\epsilon\in C_\epsilon$ such that $\|y_\epsilon\|\leq a(C_\epsilon)+\epsilon$, and $\|y_\epsilon-fy_\epsilon\|\leq\epsilon$, so if $a=0$, then $\lim_\epsilon y_\epsilon=0$, whence $0\in\bigcap_\epsilon C_\epsilon$.
  3. For $u_1,u_2\in C_\epsilon$, $\frac12(u_1+u_2)\in C_{\phi(\epsilon)}$; because of the computation in the paper.
  4. By 3. for $u_1,u_2\in D_\epsilon$, we have $\|u_1\|,\|u_2\|\leq a+\epsilon$ and $\frac12\|u_1+u_2\|\geq a(C_{\phi(\epsilon)})$, so by the lemma in the paper ($u=u_1$, $v=u_2$, and $w=0$), and the computation there, we get $\lim_{\epsilon\rightarrow0}\text{diam}(D_\epsilon)=0$ (this is where $a>0$ is used), so by Cantor's intersection theorem $\bigcap_\epsilon D_\epsilon\neq\emptyset$, and this is the contradication, since $\bigcap_\epsilon D_\epsilon\subset\bigcap_\epsilon C_\epsilon$ and we assume at first that $\bigcap_\epsilon C_\epsilon=\emptyset$.
  1. Each $C_\epsilon$ is closed and non-empty ( since $\inf\{\|x-fx\|: x\in K\}=0$).
  2. If $\bigcap_\epsilon C_\epsilon=\emptyset$, then $a>0$; by definition there exists a $y_\epsilon\in C_\epsilon$ such that $\|y_\epsilon\|\leq a(C_\epsilon)+\epsilon$, and $\|y_\epsilon-fy_\epsilon\|\leq\epsilon$, so if $a=0$, then $\lim_\epsilon y_\epsilon=0$, whence $0\in\bigcap_\epsilon C_\epsilon$.
  3. For $u_1,u_2\in C_\epsilon$, $\frac12(u_1+u_2)\in C_{\phi(\epsilon)}$; because of the computation in the paper.
  4. By 3. for $u_1,u_2\in D_\epsilon$, we have $\|u_1\|,\|u_2\|\leq a+\epsilon$ and $\frac12\|u_1+u_2\|\geq a(C_{\phi(\epsilon)})$, so by the lemma in the paper ($u=u_1$, $v=u_2$, and $w=0$), and the computation there, we get $\lim_{\epsilon\rightarrow0}\text{diam}(D_\epsilon)=0$, so by Cantor's intersection theorem $\bigcap_\epsilon D_\epsilon\neq\emptyset$, and this is the contradication, since $\bigcap_\epsilon D_\epsilon\subset\bigcap_\epsilon C_\epsilon$ and we assume at first that $\bigcap_\epsilon C_\epsilon=\emptyset$.
  1. Each $C_\epsilon$ is closed and non-empty ( since $\inf\{\|x-fx\|: x\in K\}=0$).
  2. If $\bigcap_\epsilon C_\epsilon=\emptyset$, then $a>0$; by definition there exists a $y_\epsilon\in C_\epsilon$ such that $\|y_\epsilon\|\leq a(C_\epsilon)+\epsilon$, and $\|y_\epsilon-fy_\epsilon\|\leq\epsilon$, so if $a=0$, then $\lim_\epsilon y_\epsilon=0$, whence $0\in\bigcap_\epsilon C_\epsilon$.
  3. For $u_1,u_2\in C_\epsilon$, $\frac12(u_1+u_2)\in C_{\phi(\epsilon)}$; because of the computation in the paper.
  4. By 3. for $u_1,u_2\in D_\epsilon$, we have $\|u_1\|,\|u_2\|\leq a+\epsilon$ and $\frac12\|u_1+u_2\|\geq a(C_{\phi(\epsilon)})$, so by the lemma in the paper ($u=u_1$, $v=u_2$, and $w=0$), and the computation there, we get $\lim_{\epsilon\rightarrow0}\text{diam}(D_\epsilon)=0$ (this is where $a>0$ is used), so by Cantor's intersection theorem $\bigcap_\epsilon D_\epsilon\neq\emptyset$, and this is the contradication, since $\bigcap_\epsilon D_\epsilon\subset\bigcap_\epsilon C_\epsilon$ and we assume at first that $\bigcap_\epsilon C_\epsilon=\emptyset$.
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MSMalekan
MSMalekan
  • 2.1k
  • 1
  • 10
  • 19

  1. Each $C_\epsilon$ is closed and non-empty ( since $\inf\{\|x-fx\|: x\in K\}=0$).
  2. If $\bigcap_\epsilon C_\epsilon=\emptyset$, then $a>0$; by definition there exists a $y_\epsilon\in C_\epsilon$ such that $\|y_\epsilon\|\leq a(C_\epsilon)+\epsilon$, and $\|y_\epsilon-fy_\epsilon\|\leq\epsilon$, so if $a=0$, then $\lim_\epsilon y_\epsilon=0$, whence $0\in\bigcap_\epsilon C_\epsilon$.
  3. For $u_1,u_2\in C_\epsilon$, $\frac12(u_1+u_2)\in C_{\phi(\epsilon)}$; because of the computation in the paper.
  4. By 3. for $u_1,u_2\in D_\epsilon$, we have $\|u_1\|,\|u_2\|\leq a+\epsilon$ and $\frac12\|u_1+u_2\|\geq a(C_{\phi(\epsilon)})$, so by the lemma in the paper ($u=u_1$, $v=u_2$, and $w=0$), and the computation there, we get $\lim_{\epsilon\rightarrow0}\text{diam}(D_\epsilon)=0$, so by Cantor's intersection theorem $\bigcap_\epsilon D_\epsilon\neq\emptyset$, and this is the contradication, since $\bigcap_\epsilon D_\epsilon\subset\bigcap_\epsilon C_\epsilon$ and we assume at first that $\bigcap_\epsilon C_\epsilon=\emptyset$.