Let $A \in L(X,Y)$, $\tilde y \in Y^*$, $x \in X$. Then from $\left(A^*(\tilde y)\right)(x) := \tilde y (A(x))$ you get
$$|A^*(\tilde y)(x)|= |\tilde y (A(x))|≤\|\tilde y\|_{Y^*} \|A(x)\|_Y ≤ \|\tilde y\|_{Y^*} \|A\|_{L(X,Y)} \|x\|_X$$
For $A^*$ to be bounded you must consider the following expression
$$\|A^*\|_{L(Y^*,X^*)}=\sup_{\tilde y \in Y^*, \ \|\tilde y\|_{Y^*}≤1} \left\{\|A^*(\tilde y)\|_{X^*}\right\}$$
Note that you have the following inequality, which follows from the first one
$$\|A^*(\tilde y)\|_{X^*}=\sup_{x \in X, \ \|x\|_X≤1}\{|A^*(\tilde y)(x)|\}≤\sup_{x \in X, \ \|x\|_X≤1}\{\|\tilde y\|_{Y^*} \|A\|_{L(X,Y)}\ \|x\|_X \}$$
Putting it together gives you
$$\|A^*\|_{L(Y^*,X^*)}≤\|A\|_{L(X,Y)}$$
Note further that the construction gives you $\|A^{**}\|_{L(X^{**},Y^{**})}≤\|A^*\|_{L(Y^*,X^*)}≤\|A\|_{L(X,Y)}$. If you restrict $A^{**}$ to the subspace given by the embedding of $X$ into $X^{**}$ you get precisely $A$ again (composed with the embedding of $Y$ into $Y^{**}$). So $\|A^{**}\|_{L(X^{**},Y^{**})}≥\|A\|_{L(X,Y)}$ also holds, and the two inequalities are equalities, so also $\|A^*\|=\|A\|$.
Suppose $L$ is a homeomorphism. Then $L$ is bounded, and hence $\| L x \| \leq \|L \| \|x \|$. Similarly, $\|x\| = \| L^{-1} L x \| \leq \| L^{-1} \| \|L x \|$. Setting $m := \|L^{-1} \|^{-1}$ and $M := \|L\|$ yields
$$ m \| x \| \leq \| L x \| \leq M \| x \|. $$
Conversely, suppose there exists $m, M > 0$ such that, for all $x \in V$,
$$ m \| x \| \leq \| L x \| \leq M \| x \|. $$
Then, in particular, it holds that $\| L x \| \leq M \| x \|$ for all $x \in V$, that is, the linear map $L$ is bounded. Furthermore, the inequality $m \| x \| \leq \| L x \|$ shows that $L x \neq 0$ whenever $x \neq 0$, that is, the map $L$ is injective. As you have the surjectivity of $L$ by assumption, it follows that $L$ is bijective. To show that $L^{-1}$ is bounded, note that $\|L^{-1} L x \| = \|x \| \leq m^{-1} \|Lx\|$ for all $x \in X$.
Best Answer
No, there is no difference. If $T:X\to Y$ is linear, then
$T$ is bounded $ \iff T$ is Lipschitz.