Weak-to-weak continuous operator which is not norm-continuous Can one give a "relatively easy" example of a linear mapping $T\colon X\to X$ ($X$ a Banach space) which is 
a) weak-to-weak continuous
b) weak*-to-weak* continuous ($X=Y^*$)
but not norm-to-norm continuous (not bounded). This needs some choice I guess.
 A: I shouldn't have answered in the comments, so here's a full-fledged answer.
In short: there are no such examples.

a) Jonas Meyer proved in his answer here that a weak-weak continuous operator $T: X \to Y$ between normed space is bounded.

To see this, observe that a weak-weak continuous operator is weakly sequentially continuous. If $T$ is unbounded then there is a sequence $(x_n)$ such that $\|x_n\| \to 0$ while $\|Tx_n\| \to \infty$. By the uniform boundedness principle — applied to the sequence of functionals $\phi \mapsto \phi (x_n)$ on the dual space $X^{\ast}$ — a weakly convergent sequence is (norm-)bounded. But this means $Tx_n$ can't converge weakly and thus $T$ can't be weak-weak continuous.

b) A linear map $T: Y^\ast \to X^\ast$ is weak$^{\ast}$-weak$^{\ast}$ continuous if and only if $T = S^{\ast}$ for some bounded operator $S:X \to Y$. In particular $T$ must be bounded.

That an adjoint operator $T = S^\ast$ of a bounded operator $S$ is weak$^{\ast}$-weak$^\ast$ continuous operator is clear: If $y'_i \to y'$ is a weak$^{\ast}$ convergent net in $Y^\ast$ then $\langle Ty'_i - Ty', x\rangle_{X',X} = \langle y'_i - y', Sx \rangle_{Y',Y} \to 0$.
For the other direction note first that a linear functional $\phi$ on $X^{\ast}$ is weak$^{\ast}$-continuous if and only if $\phi = \operatorname{ev}_x$ for some $x \in X$. (Since the weak$^{\ast}$-topology is Hausdorff $x$ is necessarily unique). 
Conversely, assume that $T: Y^{\ast} \to X^{\ast}$ is weak$^\ast$-weak$^\ast$ continuous. Given this, note that $\operatorname{ev}_x T$ is weak$^{\ast}$ continuous on $Y$ thus it is of the form $\operatorname{ev}_{S(x)}$ for a (unique) $S(x) \in Y$. Since $S(x)$ is uniquely determined, it follows that $S$ is linear. Now let us check that $S$ is continuous by applying the closed graph theorem: if $x_n \to x$ and $Sx_n \to y$ (both convergences in norm) then for each $\phi$ in $Y^{\ast}$ we have
$$
\langle \phi, y\rangle_{Y^\ast,Y} =
\lim{\langle \phi, Sx_n \rangle_{Y^\ast,Y}} =
\lim{\langle T\phi, x_n \rangle_{X^{\ast},X}} =
\langle T\phi, x \rangle_{X^\ast,X} =
\langle \phi, Sx \rangle_{Y^\ast,Y}
$$
and thus $y = Sx$, as we wanted. Hence $S$ is bounded and therefore its adjoint $T = S^{\ast}$ is bounded, too.

Two concluding remarks:


*

*In course of the proof of b) we have established the Hellinger-Toeplitz theorem:  

If $T: Y^\ast \to X^\ast$ and $S: X \to Y$ are linear maps such that $\langle T\phi, x \rangle_{X^{\ast},X} = \langle \phi, Sx \rangle_{Y^{\ast},Y}$ for all $\phi \in Y^\ast$ and $x \in X$ then both $S$ and $T$ are bounded. 

In particular, an everywhere defined symmetric operator on a Hilbert space is bounded.

*The only non-trivial facts we used are the uniform boundedness principle and the closed graph theorem. For the former let me mention that Alan Sokal recently gave a beautiful proof, relying on the gliding hump method and thus avoiding Baire.
