Let $$ \ell^1(\mathbb{R}) = \{ x=(x_k)_{k=1}^\infty : x_k \in \mathbb{R}, ||x||=\sum_{k=1}^\infty |x_k| < \infty \}$$ and let $$B=\{x \in \ell^1: ||x||\leq1\}$$ be the unit ball in $\ell^1$.
Is there a continuous function on $\ell^1$ that doesn't take its maximum on $B$?
I know that a continuous function on compact sets takes it maximum and minimum, but $B$ here isn't compact anymore since $\ell^1$ is infinite dimensional.
Let $f(x)=\sum (1-\frac 1n) x_n$ where $x=(x_n)$. Then $|f(x)| \leq \sum (1-\frac 1n) |x_n| \leq 1$ for $\|x\| \leq 1$ and equality fails unless $\|x\|=1$ and $(1-\frac 1 n) |x_n|=|x_n|$ for al $n$, a contradiction. But in this case $x_n=0$ for all $n$. Note that the supremum of $f$ on the closed unit ball is at most $1$. Since $f(e_n)=(1-\frac 1 n) \to 1 $ it follows that the surepmum is exactly $1$. Hence $f$ does not attain a maximum value on the closed unit ball.