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Let $X$ and $Y$ be normed linear spaces over a scalar field, and let $T:X\to Y$ be a linear map.

Suppose $T:X\to Y$ is a continuous at the origin, I want to show that $T$ is Lipschitz, i.e. there exists some constant $K\geq 0,$ such that \begin{align} \Vert T(x) \Vert \leq K\Vert x\Vert ,\;\;\forall\;x\in X. \end{align}

HERE IS A PROOF

Suppose for contradiction, that $\forall\, n\in \Bbb{N},\exists\,x_n\in X$ such $x_n\neq 0$ and \begin{align} \Vert T(x_n) \Vert > n\Vert x_n\Vert . \end{align} So, \begin{align} \frac{\Vert T(x_n) \Vert}{n\Vert x_n\Vert} > 1 ,\;\;\forall\, n\in \Bbb{N}. \end{align} Define \begin{align} u_n=\frac{x_n}{n\Vert x_n\Vert} ,\;\;\forall\, n\in \Bbb{N}. \end{align} Then, \begin{align} u_n\to 0, \;\;\text{as} \;\;n\to\infty, \end{align} but \begin{align} \Vert T(u_n) -0\Vert=\frac{\Vert T(x_n) \Vert}{n\Vert x_n\Vert} > 1 ,\;\;\forall\, n\in \Bbb{N}. \end{align} This implies that $T(u_n)\not\to 0, \;\;\text{as} \;\;n\to\infty.$ Contadiction and we're done.

My question: Why must $x_n\neq 0 ,\;\;\forall\, n\in \Bbb{N}$ at negation?

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If you chose $x_n$ such that $\|Tx_n\| >n\|x_n\|$ and $x_n=0$ you get $0>0$! So $x_n \neq 0$ automatically.

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  • $\begingroup$ Oh, I see Sir! Thanks! +1) $\endgroup$ – Omojola Micheal Dec 17 '18 at 9:41
  • $\begingroup$ I believe I can still go directly and use the argument. Since $T(x_n)\to 0,$ then it is bounded. This implies that it is Lipschitz. Right? $\endgroup$ – Omojola Micheal Dec 17 '18 at 9:43

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