Geometrical representation of the unit ball? Let $E$  be the vector space of $\mathbb{R}$-valued continuous functions on $[0\ 1]$. With the norm $\| f \| = \max \{\ | f (x) |; 0 \leq x \leq 1\}$, the open ball centered at $f$ and radius $r$ has a simple graphical representation: it is a “parallel to f band”: the distance from all point on the function at each of its two edges is constant and equal to $r$; for example the closed ball of center the constant function $f(x)= 5$ and radius $1$ is the set of all functions in $E$ contained in the closed rectangle of vertices $(0,4),(1,4),(1,6),(0, 6)$.
Is there a similar or analog geometrical representation when the norm on $E$ is given by
$\int_{{0}}^{1}|f(x)|$?
 A: $\newcommand{\Reals}{\mathbf{R}}$Let $B$ denote the unit ball in the space of continuous functions on $[0, 1]$ with respect to the $1$-norm, i.e., the set of continuous, real-valued functions $f$ such that
$$
\|f\|_{1} = \int_{0}^{1} |f(x)|\, dx < 1.
$$
There is no geometric characterization of $B$ in the following sense:
Theorem: If $X$ is a proper subset of $[0, 1] \times \Reals$, then there exists an element of $B$ whose graph is not contained in $X$.
Proof: If $(x_{0}, y_{0})$ is an arbitrary point of $[0, 1] \times \Reals$, the graph of the continuous function $f:\Reals \to \Reals$ defined by
$$
f(x) = \begin{cases}
  y_{0}\bigl[1 - (|y_{0}| + 1) \cdot |x - x_{0}|\bigr] & \text{if $|x - x_{0}| < \dfrac{1}{|y_{0}| + 1}$,} \\
  0 & \text{otherwise}
\end{cases}
$$
is a triangular spike with apex at $(x_{0}, y_{0})$ and enclosing area
$$
\tfrac{1}{2}(\text{base})(\text{height})
  = \frac{|y_{0}|}{|y_{0}| + 1} < 1.
$$
Restricting to $[0, 1]$ gives an element of $B$ whose graph contains the point $(x_{0}, y_{0})$. That is, for each point $p$ of the strip $[0, 1] \times \Reals$, there exists an element of $B$ whose graph contains $p$.
