For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent

Assume $T$ is bounded, with constant $C$ from [L1]. Then $\mathrm{\parallel }Tx-Ty\mathrm{\parallel }=\mathrm{\parallel }T(x-y)\mathrm{\parallel }\le C\mathrm{\parallel }x-y\mathrm{\parallel }$ for all $x,y\in X$, so $T$ is Lipschitz.

If $T$ is continuous on $X$, then in particular it is continuous at $0$, so $(3)\Rightarrow (2)$.

Assume $T$ is continuous at $0$. Applying [L3] with $\epsilon =1$ gives $\delta >0$ such that $\mathrm{\parallel }u\mathrm{\parallel }<\delta$ implies $\mathrm{\parallel }Tu\mathrm{\parallel }<1$.

Step 1.1 proves $(1)\Rightarrow (4)$, and [L2] gives $(4)\Rightarrow (3)$.

Let $x\in X$ with $x\mathrm{\ne }0$ and put $u:={\displaystyle \frac{\delta }{2\mathrm{\parallel }x\mathrm{\parallel }}}x$. Then $\mathrm{\parallel }u\mathrm{\parallel }=\delta \mathrm{/}2<\delta$, so $\mathrm{\parallel }Tu\mathrm{\parallel }<1$ by step 1.3. By linearity, $Tu={\displaystyle \frac{\delta }{2\mathrm{\parallel }x\mathrm{\parallel }}}Tx$, hence $\mathrm{\parallel }Tx\mathrm{\parallel }<{\displaystyle \frac{2}{\delta }}\mathrm{\parallel }x\mathrm{\parallel }$. The corresponding non-strict inequality $\mathrm{\parallel }Tx\mathrm{\parallel }\le (2\mathrm{/}\delta )\mathrm{\parallel }x\mathrm{\parallel }$ holds also at $x=0$ because $T0=0$, so $T$ is bounded.

Thus $(2)\Rightarrow (1)$. Combining steps 2.1, 1.2, and 2.2 gives all four equivalences.