# $l_1$ is not strictly convex space

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) 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.

• What is "the case equality in triangle inequality"? Commented Dec 23, 2018 at 15:42
• $||x+y||=||x||+||y||$ if and only if $x=t·y$ with $t\geq{0}$ Commented Dec 23, 2018 at 15:44
• 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. Commented Dec 23, 2018 at 15:48
• So how would you conclude that the equality cannot hold? Commented Oct 31, 2022 at 10:53

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 $$j=n$$).
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.
$$x$$ 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.