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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 $\int_{{0}}^{1}|f(x)|$?

Piquito
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    You don't like "the set of all functions $f$ for which the region bounded by the horizontal axis and the graph of $f$ has unit area"...? :) – Andrew D. Hwang May 14 '15 at 19:19
  • $user86418 Are you giving me an answer to the question or a correction for my description of the first ball? (it is very hard to me writing in English) – Piquito May 14 '15 at 19:39
  • Sorry for my ambiguity. It seemed to me that 1. The unit ball has a nice geometric interpretation, given by "the absolute area enclosed by the graph of a function $f$ is at most one"; 2. You probably already knew this; 3. There must consequently be something else specific you were seeking. Anyway, I can't imagine there being a geometric description substantially different from the area condition, certainly nothing analogous to the ball in $L^{\infty}$. The unit ball in $L^{1}$ contains, for example, triangular spikes of arbitrarily large height. :) – Andrew D. Hwang May 14 '15 at 19:48
  • $user86418 I would like you answer the question with your good English. I think you are right, mainly by your remark about "triangular spikes" wich was precisaly the reason of my ask. Is it possible to say "geometrically" that all the product [0, 1] x$\mathbbR$ considering that the very little base of these triangles can have every point of [0, 1]? – Piquito May 14 '15 at 21:28
  • Sorry, I do not know what happened to my comment to user86418. I can not even delete it to write well again. What I said was in good terms and accepted his comment as appropriate and adding a question about if one can take these balls as all the product of the interval [0,1] x R which seems to me as true and as false at the same time. – Piquito May 14 '15 at 21:52

1 Answers1

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$\newcommand{\Reals}{\mathbf{R}}$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 $p$.

Andrew D. Hwang
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  • @Luis: It looks as if a dollar sign got omitted from your comment. There probably is the usual delete button, but it's way off the page to the right, maybe superimposed on (or underneath...) other page elements. :) – Andrew D. Hwang May 14 '15 at 22:07
  • $user86418 Beautiful theorem I did not know. Following the "perspective" of the other ball, "geometrically" expressed and having in account that for all point we can show one of your triangles is it possible to say that the ball is [0, 1] x R? I think the unthinkable: yes and not. What do you think about it, please? – Piquito May 15 '15 at 00:00
  • @Luis: Note first that the unit ball in the $L^{\infty}$-norm isn't the rectangle $[0, 1] \times [-1, 1]$; it's the (infinite-dimensional) space of continuous sections of projection on the first factor. A similar assertion is true for the $L^{1}$-norm; the unit ball isn't the strip $[0, 1] \times (-\infty, \infty)$ itself, but the space of continuous sections whose graphs enclose at most one unit of area. – Andrew D. Hwang May 15 '15 at 10:25
  • I do not agree. Each function is its graphic and the rectangle (in the norm of maximum) is the (non-countable) union of these graphics. It is clear that there are many more functions but I have written geometrical representation in quotes, "geometrical", for the sole purpose of "viewing" the object. All objection to this kind of game it should be attributed to my poor English and not to an error about it. (almost all this comment has been written with the help of Google). – Piquito May 15 '15 at 12:48
  • Two thoughts: 1. If we identify the unit ball in the $L^{1}$-norm with $[0, 1] \times \Reals$, we also have to identify the space of all continuous functions with $[0, 1] \times \Reals$. 2. If you consider measurable functions instead of continuous functions, the function space now has strictly larger cardinality than $[0, 1] \times \Reals$. So, I think you cannot truly identify these function spaces with subsets of the plane. (Sending a function to its graph maps a function space into the power set of $[0, 1] \times \Reals$, a much larger set.) :) – Andrew D. Hwang May 15 '15 at 13:49
  • It was not at all matter of "identifying"! (the first time I learned about a nice example of cardinality superior to the continum was with an example containing yours: "all the functions"; it was in a book of the great Laurent Schwartz. When he has died I recall this example I never forget). Regards. – Piquito May 15 '15 at 14:56