Does increasing convergence imply convergence in ${\cal L}^\infty$ Is this true?
$$
 0 \le f_n\nearrow f,\ \Vert f\Vert_{{\cal L}^\infty} \le M < \infty \implies f_{n_k}\to f \textrm{ in }{\cal L}^\infty, \text{ for some subsequence }(f_{n_k})_{k\in\mathbb N}.
$$

No. See:


*

*$f\equiv 1$ and $f_n=1_{[-n,n]}$ in $L^\infty(\Bbb R)$;

*$f\equiv 1$ and $f_n(x)=1-x^n$ in $L^\infty[0,1)$.
In both cases, $\lVert f-f_n\rVert_\infty=1$ for all $n$.
Remark: It should be pointed out that, under the hypothesis that the sequence of functions increases, passage to subsequences is completely irrelevant.
