Extreme points of the unit ball of $l^1(\mathbb{Z^+})$ and $L^1[0,1]$

Question

Determine the extreme points of the unit ball of $l^1(\mathbb{Z^+})$ and $L^1[0,1]$.

My attempt: I know the definition but I don't know how to find these extreme points.Please help me to solve this problem.Thanks in advance.

Extreme point:An element $f$ of the convex subset $K$ of $X$ is said to be an extreme point of
$K$ if for no distinct pair $f_1$ and $f_2$ in $K$ is $f=\frac{f_1+f_2}{2}$

Edit: The answer is the extreme points of the unit ball of $L^1[0,1]$ are the delta functions and the unit ball of $l^1(\mathbb{Z^+})$ has no extreme points .

2nd Edit: $l^1(\mathbb{Z^+})$ denote the collection of all complex functions $f$ on $\mathbb{Z^+}$ such that $\sum_{n=0}^{\infty}|f(n)|<\infty$.It is exercise number 9 of first chapter of Banach algebra techniques in operator theory by Douglas.

3rd Edit: My previous answer was wrong,the correct answer is the extreme points of the unit ball of that $l^1(\mathbb{Z^+})$ are the delta functions and the unit ball of $L^1[0,1]$ has no extreme points .Can now someone prove this?

Hint: for $\ell ^1$, consider first what the extreme points are of the unit ball in $\mathbb R^2$ (with the $\ell ^1$ norm). — Jonas Dahlbæk, Sep 02 '14 at 06:26

score 13 · Accepted Answer · edited Jul 11 '25 at 07:53

13

The answer is the extreme points of the unit ball of $L^1[0,1]$ are the delta functions

Wrong. There is no such thing as a delta function in $L^1[0,1]$. There are no extreme points of the unit ball in $L^1$. Reason: given a unit-norm function $f\in L^1$, you can find $b\in [0,1]$ such that both $f\chi_{[0,b]}$ and $f\chi_{[b,1]}$ have norm $1/2$. (Use the intermediate value theorem on the map $t\mapsto \int_{[0,t]} |f|$ defined on $[0,1]$.)

the unit ball of $l^1(\mathbb Z^+)$ has no extreme points

Wrong. One extreme point is $(1,0,0,0,\dots)$ and there are other such.

The difference between the two cases is that the measure space in the second cases has atoms, whereas $[0,1]$ does not. This is why the support of an $L^1$ function can be divided in two subsets of positive measure, but the support of a sequence sometimes cannot be.

edited Jul 11 '25 at 07:53

Dean Miller

7,015
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answered Sep 02 '14 at 12:57

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@Math Check your Math. :) My answer works for both kinds of scalars. – Sep 03 '14 at 06:02
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How do we prove that for $l^1(N)$, only $\pm e_i$, where $e_i$ is the standard unit vector, are the extreme points? – pikachuchameleon Apr 01 '16 at 16:08
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how can we apply the intermediate value thm if f is not continuous? (f doesnt have to be continuous to be in $L^1$) – cesare borgia Feb 02 '17 at 12:21
why is the answer with the delta function wrong? – Its me Mar 27 '24 at 16:34

score 10 · Answer 2 · edited Feb 06 '17 at 15:05

10

I'm answering my question in the comments, show explicitly what @user147263 mentioned above.

$\,$

Pick any $f \in B_1(0,L^1([0,1]))$, the closed unit ball of $L^1([0,1])$.

The function $F : \mathbb R \rightarrow \mathbb R$, $t \mapsto \int_0^t |f(x)|\,dx$, is continuous, since for all $t_0, \, t_1 \in [0,1]$ there holds $$|F(t_0) - F(t_1) | = \left| \int_{t_1}^{t_0} |f(x)| \, dx \right| \leq ||f||_{L^1([0,1])} \, |t_0 - t_1|, $$ hence $||F||_{L(\mathbb R)} \leq ||f||_{L^1([0,1])} < \infty$.

Using the intermediate value theorem we can find $c \in [0,1]$, such that $F(c) = \frac{1}{2} ||f||_{L^1}$.

Let now $g_1 := 2 \, f|_{\chi_{[0,c]}}$, $g_2 := 2\, f|_{\chi_{[c,1]}}$. Then $g_1, \, g_2 \in B_1(0,L^1([0,1]))$ and $g_1 \neq f \neq g_2$, on the other hand $$f = \frac{1}{2} g_1 + \frac{1}{2} g_2.$$ Thus $f$ is not extremal, therefore there are no extremal points in $B_1(0,L^1([0,1]))$.

edited Feb 06 '17 at 15:05

augustin souchy

664

answered Feb 03 '17 at 14:26

cesare borgia

1,052

Great answer! Why would we need $F$ to show that $g_1 \ne f \ne g_2$? – nekodesu Mar 19 '18 at 18:16
In fact, if $f = 0$, then $g_1 = g_2 = f$. But in that case it is trivial that $f$ is not extreme. – J. De Ro May 20 '20 at 13:46
Why is your inequality used to prove the uniform continuity of $F$ true? – Peter Griffin Feb 13 '25 at 20:42

Extreme points of the unit ball of $l^1(\mathbb{Z^+})$ and $L^1[0,1]$

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