Given a normed space $X$ over $\mathbb{R}$ (or $\mathbb{C}$), define its dual space $X^*$ as the set of all continuous & linear functionals. $\forall g \in X^*, \|g\|_{X^*}\triangleq \sup_\limits{\|x\|_{X}=1\\ x\in X}|g(x)|$ (the absolute value is taken since $g(x) \in\mathbb{R}$ (or $\mathbb{C}$). For convenience I will drop the subscripts below.
A sequence $\{f_n\}\in X^*$ strongly converges to $f\in X^*$ means $\|f_n-f\| \to 0$ (denote as $f_n \to f$).
A sequence $\{f_n\}\in X^*$ weakly converges to $f\in X^*$ means $\forall T\in X^{**}, |T(f_n) - T(f)| \to 0$ (denote as $f_n \xrightarrow{w}f$).
A sequence $\{f_n\}\in X^*$ converge to $f\in X^*$ in the weak* topology means $\forall x\in X, |f_n(x)-f(x)| \to 0$ (denote as $f_n\xrightarrow{w*} f$). (This is equivalent to $f_n$ converges to $g$ pointwise, right?)
Now, it's easy to see that if $f_n \to f$, then $f_n\xrightarrow{w}f$ (use the continuity of $T \in T^{**}$).
Also, notice that $f_n \xrightarrow{w} f \implies f_n\xrightarrow{w*}f$ since we can define $\hat{x} \in X^{**}$ by $\hat{x}(g) = g(x),\forall g \in X^*$.
In a word, strong convergence implies weak convergence and weak convergence implies convergence in weak* topology.
The question is
If $f_n \xrightarrow{w*} f$, then $\forall x\in X$, $|f_n(x)-f(x)|\to 0$. This may imply $\|f_n-f\| = \sup_\limits{\|x\|=1\\x\in X}|f_n(x)-f(x)| \to 0$. i.e. $f_n$ converges to $f$ strongly. But this cannot be true by intuition. What's the problem? Can anyone give me a counter example that $f_n \xrightarrow{w} f$ but $f_n$ does not converge to $f$ strongly. Also, I want a counterexample that $f_n\xrightarrow{w*} f$ but $f_n$ does not converge to $f$ weakly.