[proofplan] We prove the two implications separately. If $T$ satisfies a uniform norm estimate, then linearity turns the estimate into a Lipschitz estimate for differences, which gives continuity. Conversely, continuity at $0_X$ controls $T$ on a small ball; rescaling an arbitrary nonzero vector into that ball converts this local control into a global linear bound. [/proofplan] [step:Use the boundedness estimate to prove Lipschitz continuity] Assume that there exists $C \geq 0$ such that $\|T x\|_Y \leq C\|x\|_X$ for every $x \in X$. For arbitrary $x,x' \in X$, linearity gives \begin{align*} T x - T x' = T(x - x'). \end{align*} Applying the assumed bound to $x - x' \in X$ yields \begin{align*} \|T x - T x'\|_Y = \|T(x - x')\|_Y \leq C\|x - x'\|_X. \end{align*} Thus $T: X \to Y$ is Lipschitz continuous with Lipschitz constant $C$, and hence continuous with respect to the norm topologies. [/step] [step:Use continuity at the origin to obtain a uniform bound] Let $0_X \in X$ denote the zero vector of $X$, and let $0_Y \in Y$ denote the zero vector of $Y$. Conversely, assume that $T: X \to Y$ is continuous. Since $T$ is continuous at $0_X \in X$ and $T(0_X)=0_Y$ by linearity, applying continuity with $\varepsilon = 1$ gives a number $\delta > 0$ such that \begin{align*} \|u\|_X < \delta \implies \|T u\|_Y < 1 \end{align*} for every $u \in X$. We prove a global bound. If $x = 0_X$, then $\|T x\|_Y = 0$, so the desired inequality holds. Let $x \in X$ with $x \neq 0_X$, and define \begin{align*} u_x := \frac{\delta}{2\|x\|_X}x \in X. \end{align*} Then \begin{align*} \|u_x\|_X = \left|\frac{\delta}{2\|x\|_X}\right|\|x\|_X = \frac{\delta}{2} < \delta, \end{align*} so $\|T u_x\|_Y < 1$. By linearity, \begin{align*} T u_x = T\left(\frac{\delta}{2\|x\|_X}x\right) = \frac{\delta}{2\|x\|_X}T x. \end{align*} Taking norms gives \begin{align*} \frac{\delta}{2\|x\|_X}\|T x\|_Y = \|T u_x\|_Y < 1. \end{align*} Multiplying by $2\|x\|_X/\delta$ yields \begin{align*} \|T x\|_Y < \frac{2}{\delta}\|x\|_X. \end{align*} Thus the boundedness estimate holds for every $x \in X$ with the constant $C := 2/\delta > 0$. [/step] [step:Conclude the equivalence] The first step proves that every bounded [linear map](/page/Linear%20Map) $T: X \to Y$ is continuous. The second step proves that every continuous linear map $T: X \to Y$ satisfies a uniform bound of the form \begin{align*} \|T x\|_Y \leq C\|x\|_X \end{align*} for all $x \in X$. Therefore $T$ is bounded if and only if $T$ is continuous. [/step]