Non-existence of minimisers in $L_1$ and $L_{\infty}$ In this note, Exercise 11 asks for finding counterexamples to the existence of minimisers in $L_1$ and $L_{\infty}$, which is

For $p = 1$ or $=\infty$, there exists a Banach space $L_p(X,\mu)$ such that: There exists a function $f\in L_p(X,\mu)$ and a non-empty closed convex subset $K\subset L_p(X,\mu)$ which we can't find any $g$ in $K$ satisfying $\|f-g\|_p = dist(f,K)$.

I try to find counterexamples in $l_1(X)$ and $l_{\infty}(X)$ instead. It is necessary that $X$ is infinite (if not, $l_p(X)$ is finite dimensional, i.e. $\mathbb{R}^n$, let any non-empty closed convex subset $K$ and any point $x$, take the disc $D(x; r)$ where $r$ large enough for the intersection with $K$ is non-empty; considering this intersection, we reduce the problem to a compact subset instead, which always has minimiser.) So I try to find counterexamples in $l_1(\mathbb{N})$ and $l_{\infty}(\mathbb{N})$. Here I make no progress.
The kind of questions about finding counterexamples is always hard to me. It is very helpful to me if anyone could show me the ways you find counterexamples in math. Thanks for any help!
 A: What follows is not a complete answer.
First, a general fact. Let $E$ be a Banach space (say real for simplicity) and let $\phi$ be a continuous linear functional on $E$. Set $K:=\ker(\phi)$, and let also $f\in E$. Then one can find $g\in K$ such that $\Vert f-g\Vert={\rm dist}(f,G)$ if and only if the linear functional $\phi$ attains its norm, which means that one can find $u\in E$ such that $\Vert u\Vert =1$ and $\phi(u)=\Vert\phi\Vert$. In fact, is $\phi$ does not attain its norm, then ${\rm dist}(f,\ker(\phi))$ is not attained for any $f\in E\setminus\ker(\phi)$. This is not very difficult to prove.
So, to find $f\in E$ and a closed convex set $K$ such that ${\rm dist}(f,K)$ is not attained, it is enough to find a linear functional $\phi$ which does not attain its norm.
By a famous theorem due to James ("James' theorem"), this can be done if and only if the space $E$ is not reflexive; hence in particular if $E=\ell_1$ of $\ell_\infty$.
In the case of $\ell_1$, it is not difficult to find an explicit example of a linear functional $\phi$ which does not attain its norm. For example, define $\phi$ as follows: if $x=(x_n)_{n\geq 1}\in\ell_1$, then 
$$\phi(x)=\sum_{n=1}^\infty \left(1-\frac1n \right)x_n\, .$$
Then $\Vert\phi\Vert=1$, but as you can easily check, there is no $x\in\ell_1$ such that $\Vert x\Vert=1$ and $\phi(x)=1$.
In the case of $\ell_\infty$, I don't have any concrete example at hand right now. As you now, $\ell_\infty$ is isometric to $\mathcal C(\beta\mathbb N)$, where $\beta\mathbb N$ is the Stone-Čech compactification of $\mathbb N$, so the dual space is the space of all Baire measures on $\beta\mathbb N$. Hence, we have to find a measure $\mu$ on $\beta\mathbb N$ such that $\Vert\mu\Vert=1$ but $\int u\, d\mu$ holds for no $u\in\mathcal C(\beta\mathbb N)$ satisfying $\Vert u\Vert_\infty =1$.
This would be done if one can prove that there exists two disjoint countable sets $A,B\subset\beta\mathbb N$ whose closures are not disjoint. In this case, writing $A=\{ a_n;\; n\geq 1\}$ and $B=\{ b_n;\; n\geq 1\}$ one could define $\mu$ as follows :
$$\mu:=\frac12\left(\sum_{n=1}^\infty 2^{-n}\delta_{a_n} -\sum_{n=1}^\infty 2^{-n}\delta_{b_n}\right). $$
Then $\Vert\mu\Vert=1$, but there is no $u\in\mathcal C(\beta\mathbb N)$ such that $\Vert u\Vert_\infty=1$ and $\int u\, d\mu=1$. Indeed, such a function $u$ would have to satisfy $u(a_n)=1$ and $u(b_n)=-1$ for all $n\geq 1$, and this would not be compatible with the continuity of $u$ since $\overline A\cap\overline B\neq\emptyset$. However, I don't know if it is possible to find the sets $A$ and $B$.
