# If $T:X\to Y$ is a continuous linear map at the origin, then $T$ is Lipschitz

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?

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.
• 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? – Omojola Micheal Dec 17 '18 at 9:43