# Continuity implies boundness in Frechet Spaces?

Consider two Frechet spaces $$(X,d_X),\;(Y,d_Y)$$. Let $$T:X \rightarrow Y$$ be a continues linear operator. Is it true that $$T$$ is bounded(or Lipschitz) in the following sense: $$\exists \alpha > 0 \; \forall x,y \in X \; d_Y(Tx,Ty)\leq \alpha d_X(x,y)$$

By the definition of continuity and translation invariance of the metric we know there exists $$\delta > 0$$ such that $$d_X(x,y)<\delta \implies d_Y(Tx,Ty)<1$$ But I am not sure whether this imples the former condition.

Suppose $$X$$ is a normed linear space and $$Y=X$$ with the metric $$d_Y(x,y)=\frac {\|x-y\|} {1+\|x-y\|}$$. Then $$Y$$ is a Frechet space. Let $$T:(Y,d_Y) \rightarrow (X,\Vert\cdot\Vert)$$ be the idenitity map. $$T$$ is continuous. If the inequality you have stated holds then $$\|x-y\| \leq \alpha$$ for all $$x,y$$, which is false.