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.

  • 1
    $\begingroup$ Ad b) an operator $T: Y^\ast \to X^\ast$ is weak$^\ast$-weak$^{\ast}$ continuous if and only if it is of the form $T = S^{\ast}$ for some bounded operator $S: X \to Y$ in particular $T$ must be bounded. To see this, simply note that a functional on $Y^\ast$ is weak$^{\ast}$ continuous if and only if it is an evaluation functional and apply this to $\operatorname{ev}_{x} \circ T$. $\endgroup$
    – t.b.
    Aug 4, 2011 at 22:34
  • $\begingroup$ Ad a) This is answered here: a weakly continuous operator is weakly sequentially continous, as Jonas proved there, such an operator is necessarily bounded. $\endgroup$
    – t.b.
    Aug 4, 2011 at 22:50
  • $\begingroup$ This answer is modulo Eberlein-Smulian theorem? Weakly sequentially continuous operator must be weakly continuous, right? $\endgroup$ Aug 4, 2011 at 22:55
  • $\begingroup$ No, it's the other way around: a weakly continuous operator is weakly sequntially continous. Then it's just the uniform boundedness principle. $\endgroup$
    – t.b.
    Aug 4, 2011 at 22:59
  • 2
    $\begingroup$ Instead of uniform boundedness, you can use closed graph theorem, if you think it is more elementary. If $T$ is w-w continuous, then its graph is weakly closed, a weakly closed set is strongly closed, so this graph is strongly closed, and therefore (by CGT) T is norm-norm continuous. $\endgroup$
    – GEdgar
    Aug 5, 2011 at 13:52

1 Answer 1


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:

  1. 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.

  2. 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.

  • 3
    $\begingroup$ Here's a simple example of a bounded operator $T$ on a dual space $X^*$ which is not weak$*$-weak$*$ continuous. Take $X = \ell^1 = (c_0)^*$. Let $e_n$ denote the sequence which is $1$ in the $n$th position and $0$ elsewhere. Let $T$ be the rank-1 operator given by $T(\{x_n\}) = \left( \sum x_n \right) \cdot e_1$. Then $\|T\|=1$, but $T e_n = e_1$ for all $n$ even though $e_n \to 0$ weak$*$. $\endgroup$
    – Mike F
    Dec 18, 2012 at 20:00
  • 1
    $\begingroup$ @Mike: that's my favorite example, too :-) It may be worth adding that the summation functional is adjoint to the limit functional on $c$ (convergent sequences). That is, if you view $\ell^1$ as the dual space of $c$ via the pairing $\langle x,y \rangle_{\ell^1,c} = \left(\sum x_n\right) \lim y_n + \sum x_n y_n$ then $T$ is adjoint to $S(y) = (\lim y_n) \cdot e_1$. $\endgroup$
    – t.b.
    Dec 19, 2012 at 8:13

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