Bounded linear transformation definition clarification

Question

My professor defines bounded linear transformations (operators) between two norm linear spaces as "linear map that maps bounded sets to bounded sets".
But in many books has defined it as a linear map $T:X\to Y$ such that: for all $x\neq 0$ there is $M\ge 0$ with $$\dfrac{\|Tx\|_Y}{\|x\|_X}\le M.$$ However, I can not see any immediate equivalence between these two definitions.
Are these two equivalent? If it is yes, How can you see it?

Also, I would like to know the definition and the intuition behind the unbounded linear operators.

Answer 1

Clearly, if there is an $M > 0$ such that $\frac{\|Tx\|_Y}{\|x\|_X}\leq M$ for all $x\in X$, then $T$ maps bounded sets to bounded sets. Take $S\subset X$ such that $\|x\|_X\leq B < \infty$ for all $x\in S$. Then, for $y\in T(S)$, $y = Tx$ for some $x\in S$, which implies that $\|y\|_Y\leq M\|x\|_X\leq MB$, so $T(S)$ is bounded.

Similarly, if $T$ maps bounded sets to bounded sets, then we must have some $M\geq 0$ such that $\frac{\|Tx\|_Y}{\|x\|_X}\leq M$. Otherwise, take an increasing sequence $\{M_n\}_{n=1}^{\infty}$ such that $M_n\to \infty$, and for each $n$ choose an $x_n\in X$ such that $\|x_n\|_X = 1$ and $\frac{\|Tx_n\|_Y}{\|x_n\|_X} = \|Tx_n\|_Y > M_n$, which we can do by the linearity of $T$, as $$\frac{\|Tx\|_Y}{\|x\|_X} = \left\|T\frac{x}{\|x\|_X}\right\|_Y$$ is only dependent on the "direction" of $x$, not on its length. Then, $\{x_n\}_{n=1}^{\infty}$ is bounded as it lies within the unit sphere, but $\{Tx_n\}_{n=1}^{\infty}$ is not bounded.

Answer 2

The condition that you have written is the same as $$ \sup_{||x||\leq1} ||Tx||_Y = \sup_{x\in X} \frac{||Tx||_Y}{||x||_X}\leq M$$ because $T $ is linear and $||\lambda x|| = |\lambda|||x||$.

Therefore, if $T$ sends a bounded sets to bounded sets it sends the unit ball in a bounded set (so $M$ exists). Vice versa, if the inequality above holds, take a bounded set $K$, it must be contained in a ball or radius $r>0$, i.e. $K \subset r B_X $, it follows that $T(K)\subset M r B_Y$ (where $B_X$ is the unit ball in the space $X$).