2
$\begingroup$

I have the following doubt:

I know that space $l_1$ of sequences is not strictly convex. However, I also know that $l_p$ is strictly convex ($1<p<\infty$) owing to the following reasoning:

If $x,y\in{l_p}$ such that $||x||_p=||y||_p=1$ we have using triangle inequality that $||x+y||_p\leq{||x||_p+||y||_p}=2$ then if $||x+y||_p=2$ using the case equality in triangle inequality we conclude that $x=ty$ with $t>0$ and since $||x||_p=1$ we have that $x=y$.

Where is the mistakes in this proof for $l_1$?

Thanks.

$\endgroup$
4
  • 1
    $\begingroup$ What is "the case equality in triangle inequality"? $\endgroup$ Commented Dec 23, 2018 at 15:42
  • $\begingroup$ $||x+y||=||x||+||y||$ if and only if $x=t·y$ with $t\geq{0}$ $\endgroup$
    – mathlife
    Commented Dec 23, 2018 at 15:44
  • 2
    $\begingroup$ The statement "$\|x+y\|=\|x\|+\|y\|\iff\exists t\geq0:x=ty$" is equivalent to the statement "$\|\cdot\|$ is strictly convex" so your argument uses a circular reasoning. $\endgroup$ Commented Dec 23, 2018 at 15:48
  • $\begingroup$ So how would you conclude that the equality cannot hold? $\endgroup$
    – tornt
    Commented Oct 31, 2022 at 10:53

3 Answers 3

1
$\begingroup$

In $\ell_1$, $\|e_1+e_2\| = 2 = \|e_1\| + \|e_2\|$ where $e_j$ are the standard unit vectors ($(e_j)_n = 1$ iff $j=n$).

$\endgroup$
1
$\begingroup$

You can't conclude that $x$ is a scalar multiple of $y$ from $\|x\|_p = \|y\|_p =1$ and $\|x\|_p + \|y\|_p = 2$ when $p = 1$. E.g., take $x = (1, 0, 0, 0, \dots)$ and $y = (0, 1, 0, 0, \ldots)$. The rule that equality holds in the triangle inequality for $x$ and $y$ iff $y$ is a positive scalar multiple of $x$ is equivalent to the statement that the unit disc is strictly convex.

$\endgroup$
0
$\begingroup$

$x$ and $y$ need not be multiples for equality to hold. Draw the picture in $\mathbb R^2$, to see what is going on. The line segment connecting $e_1$ and $e_2$ lies on the edge of the $\ell_1$ unit ball, and $\|(1,0)+(0,1)\|_1=\|(1,0)\|_1+\|(0,1)\|_1$ Once $p> 1$, the ball gets "rounded", so the interior of the line segment is interior to the ball.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .