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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.

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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.

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  • $\begingroup$ Yeah, right, I thought about bounded sets not necesserely into a ball centred in zero. $\endgroup$
    – Invincible
    Apr 25, 2017 at 8:31
  • $\begingroup$ You were right there, but I can always rescale the set to be centred at zero :pI remembered getting points off from my professor because I forgot to mention this when I wrote my solution last time! $\endgroup$
    – Chee Han
    Apr 25, 2017 at 8:32
  • $\begingroup$ And I forgot that we can always use the ball centred in zero when I wrote my question here (generally, asking such kind of questions is quite ashamed ;d) $\endgroup$
    – Invincible
    Apr 25, 2017 at 8:44
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    $\begingroup$ @Vladislav Don't be, in fact you should be proud of yourself; it takes a lot of courage to ask any questions in MSE! $\endgroup$
    – Chee Han
    Apr 25, 2017 at 9:35
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    $\begingroup$ @CheeHan Why do we take supremum over all y of norm equal to exactly 1? Why is the value 1 important? $\endgroup$
    – blahblah
    Nov 1, 2020 at 9:54

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