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I denote $N(x)$ as the norm-function, although in the denominator it stays $\|x\|$.

$$\lim_{x\to 0} \frac{N(x)-N(0)}{\|x\|} = \lim_{x\to 0} \frac{N(x) - 0}{\|x\|} = \lim_{x\to 0} 1 = 1 \ne 0$$

1) Is it right?

Also, I've seen a similar way showing it, by taking the limit:

$$\lim_{x\to 0} \frac{N(x)-x \cdot v}{\left\|x\right\|}=0.$$

2) What was done here exactly with this $x\cdot v$?

Royi
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AlonAlon
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  • Hint: consider the Euclidean norm in $\mathbb{R}^1$: $|x| = \sqrt{x^2} = |x|$. – Emily Jan 19 '15 at 18:52
  • (1) The scalar $\lim_{h\to0}\frac{f(x+h)-f(x)}{|h|}$ (where $h\to0$ in the vector space) is not the definition of the derivative of $f$ at $x$; the derivative is not a scalar at all. (2) Did you read the answer you linked to? Are you familiar with the derivative as a linear map? – anon Jan 19 '15 at 18:59

1 Answers1

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Ad 1): This is right, and proves that at the origin the directional derivative of $N(\cdot)$ is $=1$ in all directions. Here the directional derivative of a function $f$ at a point $p$ in direction $u\in S^{n-1}$ is defined as $$D_uf(p):=\lim_{t\to 0+}{f(p+tu)-f(p)\over t}\ .$$ The fact that $D_uN(0)=1$ for all $u\in S^{n-1}$ already proves that the function $N(x):=\|x\|$ is not differentiable at $0$, for the following reason:

When a real-valued function $f$ is differentiable at the point $p\in{\mathbb R}^n$ then one has $$D_uf(p)=df(p).u\ ,$$ and here the right hand side is a linear function of $u$. This implies that $$D_{-u}f(p)=-D_u f(p)$$ for all $u\in S^{n-1}$.

Now the origin is a very special point for the function $N(\cdot)$. At all points $p\ne0$ the norm is differentiable, and one has $$\nabla N(p)={p\over\|p\|}\ ,$$ which implies $$D_u N(p)=\nabla N(p)\cdot u={p\cdot u\over\|p\|}\ .$$

Christian Blatter
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