If a sequence of functions converges pointwise, then the supremum converges?

Question

Let $X$ be a separabel Banach space and let $f_n:X\to\mathbb{R}$ be a sequence of bounded functions which converge pointwise to a bounded function $f:X\to\mathbb{R}$. Can I say something about the convergence of the supremum, i.e. does $\sup_{x\in X}f_n(x)$ converge to $\sup_{x\in X}f(x)$? If not are there any additional conditions, which would guarantee the convergence of the supremum?

Answer 1

The condition you are looking for is uniform convergence. If you want a counterexample just think, on the real line, about a sequence of functions with compact support obtained translating by $+n$ a given nonnegative and non identically zero function with compact support.

Answer 2

No, for example consider the functions $f_n:\mathbb R\to \mathbb R$ where $f_n=0$ outside the region $[\sum_{m<n}a_n,\sum_{m\leq n}a_n]$ and on that region rises linearly from $0$ at each end to $a_n$ at the midpoint, where $a_n$ is any sequence of positive reals. These functions tend pointwise to $0$ and are pretty nice (they are continuous uniformly over the whole sequence) but the sequence of suprema is arbitrary.

Answer 3

If $f_n(x) =0$ for $x <0$ as well as for $x >\frac 1n$ and $f_n(x)=4n^{2}x(\frac 1n -x)$ for $0 <x<\frac 1n$ then you can check that $f_n \to f$ point-wise but supremum of $f_n(x)$ is $1$ for each $n$.

Under uniform convergence the conclusion is true.