# bounded operator sends bounded set to bounded set

How it follows from the title, the (supposet to be easy) exercise is to show that $$\begin{Vmatrix} Tx\end{Vmatrix} \leq C \begin{Vmatrix} x\end{Vmatrix}\iff T\text{ sends bounded set to bounded set}$$ (we are talking about linear operator between normed spaces).

Let $$A: = \{ y: \begin{Vmatrix} x - y\end{Vmatrix} \leq C_x \}$$ then $$\begin{Vmatrix} Tx + Ty\end{Vmatrix} \leq \begin{Vmatrix} Tx\end{Vmatrix} + \begin{Vmatrix} Ty \end{Vmatrix} \leq C_1 \begin{Vmatrix} x \end{Vmatrix} + C_2 \begin{Vmatrix} y\end{Vmatrix}$$ so the image of $A$ is bounded. Now I have a difficultes with the converse, it's the matter of notation, I'm sure it se more than easy using right notation.

Suppose a linear operator $T\colon X\longrightarrow Y$ maps bounded sets in $X$ into bounded sets in $Y$. WLOG, choose any bounded sets that is centred at zero. This means that for any fixed $R>0$, there exists a constant $M_R>0$ such that $\|x\|\le R\implies \|Tx\|\le M_R$. We now take any nonzero $y\in X$ and set $$x=R\dfrac{y}{\|y\|} \implies \|x\| = R.$$ Thus, \begin{align*} \dfrac{R}{\|y\|}\|Ty\| = \left\|T\left(\dfrac{R}{\|y\|}y\right)\right\| = \|Tx\| & \le M_R.\\ \implies \|Ty\| & \le \dfrac{M_R}{R}\|y\|. \end{align*} where we crucially used the linearity of $T$. Rearranging and taking supremum over all $y$ of norm 1 shows that $T$ is bounded.