I am confused about the definition of a bounded operator (which is probably a consequence of my unsatisfactory understanding of bounedesness and local boundedness). The definition is (https://en.wikipedia.org/wiki/Bounded_operator):
Let $T:V\to V'$ be a linear map between two normed vector spaces. $T$ is said to be bounded if
$$
\exists M\geq1,\ M\in\mathbb{N}\text{ s.t. } \|Tx\|\leq M\|x\|\ \forall x\in V.
$$
Questions:
- Does it mean that it is a definition of a local boundedeness, or there is also a difference with the latter?
- What is the idea/intuition about the norm of the space of bdd linear operators as the maximum value taken by the ratio of the norm of the image of $x$ and $x$ itself?
- (excuse me for the generality of this question) Is there some clear link with the way the continuous dual space is equipped with a norm?
Because of linearity, $T$ is continuous iff it is continuous at $0$. This is because $$ T(x)-T(y)=T(x-y)-T(0). $$ The linear $T$ is continuous at $0$ iff the inverse image of the open unit ball centered at the origin in $V'$ contains an open ball $B_{r}(0)$ in $V$ for some $r > 0$. Because $T(\alpha x)=\alpha(x)$, then you can see that $\frac{r}{\|x\|+\epsilon}x$ is mapped by $T$ to a vector whose length is less than $1$. That is $\|T(x)\|< (\|x\|+\epsilon)/r$. This is true for all $\epsilon > 0$ and, therefore, $\|Tx\| \le (1/r)\|x\|$. So, if $T$ is continuous and linear, then $T$ is bounded.
Conversely, suppose $T$ is bounded by $M$. If $\{ x_{n}\}$ tends to 0 in $V$, then $T(x_{n})$ tends to 0 because $\|T(x_{n})\|\le M\|x_{n}\|$; so $T$ is continuous at $0$. So boundedness is equivalent to continuity at the origin, which is equivalent to continuity everywhere for a linear $T$.