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The textbook I am using says that a sequence $(T_n)$ in $\mathcal{B}(X,Y)$ for $X$, $Y$ normed linear spaces converges strongly to $T$ if $\lim_{n\rightarrow\infty}T_nx=Tx$ for every $x\in X$.

The (topological) dual space to $X$ a Banach space is defined as $X^\ast=\mathcal{B}(X,\mathbb{R})$, and $\varphi\in X^\ast$ is the weak-$\ast$ limit of $(\varphi_n)$ if $\varphi_n(x)\rightarrow\varphi(x)$ as $n\rightarrow\infty$ for every $x\in X$.

This seems like the same definition to me. Is the difference that we are considering $X^\ast$ to be a space in its own right, and then by strong convergence in $X^\ast$ we actually mean norm convergence, where $\|\varphi\|=\sup_{x\neq0}\frac{|\varphi(x)|}{\|x\|}$?

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Strong convergence is reserved for usual norm convergece. To differ this notion from strong convergence of operators one usually say that operators converges in the strong operator topology.

And of courese you are right that weak-$^*$ convergence is the same as strong operator convergence in $X^*$. These are just different names for the same thing. Of course when one deals with functionals he uses notion of weak-$^*$ convergence. The term strong operator convergence is used for "real" operators.

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  • $\begingroup$ So, sloppy terminology in this book. As I expected/hoped. Thanks. $\endgroup$
    – gmoss
    Mar 27, 2013 at 22:35

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