Why isn't the "l1"-norm differentiable?

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Discussion Overview

The discussion revolves around the differentiability of the "l1"-norm, specifically addressing why the norm defined as ||x||_0 = |x_1| + |x_2| + ... + |x_N| is not differentiable. Participants explore geometric interpretations and seek clarification on the implications for matrix calculus.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants clarify that the correct formulation of the norm includes absolute values, stating that ||x||_0 = |x_1| + |x_2| + ... + |x_N| is necessary for it to be a norm.
  • One participant mentions that the non-differentiability arises because the unit sphere defined by this norm has "corners," drawing a parallel to the function f(x) = |x|, which is not differentiable at x=0.
  • Another participant expresses appreciation for the geometric explanation but requests further elaboration on the topic in the context of matrix calculus, indicating a need for deeper understanding.
  • There is a reiteration that the non-differentiability at extreme points of the unit sphere, such as (1,0,...,0), contributes to the overall confusion regarding the norm's properties.

Areas of Agreement / Disagreement

Participants generally agree on the necessity of absolute values in defining the norm and recognize the geometric reasoning behind its non-differentiability. However, there remains some confusion regarding its implications in matrix calculus, indicating that the discussion is not fully resolved.

Contextual Notes

Some assumptions about the definitions of norms and differentiability are present, and the discussion highlights the need for clarity in mathematical contexts, particularly regarding matrix calculus.

OhMyMarkov
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Hello everyone!

I've searched a lot for this one, but couldn't find an answer:

If x is in R^N then ||x||_0 = x_1 + x_2 + ... + x_N. Why, then, isn't this norm differentiable?

(Btw, how to make LaTeX work?)

Thank you!
 
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OhMyMarkov said:
Hello everyone!

I've searched a lot for this one, but couldn't find an answer:

If x is in R^N then ||x||_0 = x_1 + x_2 + ... + x_N. Why, then, isn't this norm differentiable?

(Btw, how to make LaTeX work?)

Thank you!
Put your math in between $\$$ signs. For example

$\$$||x||_0 = x_1 + x_2 + ... + x_N$\$ $ gives $ ||x||_0 = x_1 + x_2 + ... + x_N $.
 
OhMyMarkov said:
Hello everyone!

I've searched a lot for this one, but couldn't find an answer:

If x is in R^N then ||x||_0 = x_1 + x_2 + ... + x_N. Why, then, isn't this norm differentiable?
I think you mean $\|x\|_0 = |x_1| + |x_2| + ... + |x_N|$ (without the absolute values it is not a norm). The geometric reason for non-differentiability of the norm is that the unit sphere $\{x:\|x\|_0=1\}$ has "corners". In one dimension, the function $f(x)=|x|$ is not differentiable at $x=0$. In $R^N$, the function $\|x\|_0$ is not differentiable at an extreme point of the unit sphere, such as the point $(1,0,\ldots,0).$
 
Ohh, I missed the part with the absolute values... It's all clear now
 
Opalg said:
I think you mean $\|x\|_0 = |x_1| + |x_2| + ... + |x_N|$ (without the absolute values it is not a norm). The geometric reason for non-differentiability of the norm is that the unit sphere $\{x:\|x\|_0=1\}$ has "corners". In one dimension, the function $f(x)=|x|$ is not differentiable at $x=0$. In $R^N$, the function $\|x\|_0$ is not differentiable at an extreme point of the unit sphere, such as the point $(1,0,\ldots,0).$
Hi Opalg,

I like your explanation from geometric point of view. It is very helpful. But Still it is confusing in terms of matrix calculus. could you elaborate the answer? Please.

Thank you
Venki
 
Last edited:
Venki said:
Hi Opalg,

I like your explanation from geometric point of view. It is very helpful. But Still it is confusing in terms of matrix calculus. could you elaborate the answer? Please.

Thank you
Venki

It is because |x| is not differentiable at the corner x=0.

CB
 

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