Are there any elementary functions of norms that are still norms?

[askmathquestions]

If $d(x,y)$ is a metric, then it is said $\frac{d}{1+d}$ is also a metric. I don't know the proof of this, I'd appreciate a reference, but it got me wondering:

If $N(x)$ is a norm on a Banach space $x \in X$, then are there functions in a single real (or complex) variable $f$ (besides the identity function) for which $f(N)$ is also a norm? What would be the purpose of different norms of this nature?

 

[pasmith]

If $d(x,y)$ is a metric, then it is said $\frac{d}{1+d}$ is also a metric. I don't know the proof of this, I'd appreciate a reference,

Establishing the triangle inequality comes down to proving that for all positive $a$, $b$ and $c$, if $a + b \geq c$ then $$\frac{a}{1 + a} + \frac{b}{1 + b} \geq \frac{c}{1 + c}.$$ This is done by showing that $x/(1 + x)$ is strictly increasing for $x > 0$ and showing that $$\frac{a}{1 + a} + \frac{b}{1 + b} = \frac{a + b + 2ab}{1 + a + b + ab} \geq \frac{a + b}{1 + a + b}$$ for all positive $a$ and $b$.

If $N(x)$ is a norm on a Banach space $x \in X$, then are there functions in a single real (or complex) variable $f$ (besides the identity function) for which $f(N)$ is also a norm? What would be the purpose of different norms of this nature?

If $N$ is a norm, and you want $f \circ N$ to be a norm, then you must satisfy the linearity requirement: $$f(N(ax)) = f(|a|N(x)) = |a|f(N(x)).$$ Since you also need $f(0) = 0$ the conclusion is that $f(x) = Cx$ for some positive constant $C$. This amounts to changing your unit of measurement, which isn't all that interesting.

 

[askmathquestions]

Thanks for your reply. There's so many types of norms though, norms defined by integration, norms defined by differentiation (both of which are linear operators), norms defined by supremums, norms defined by the eigenvalues of a matrix, so on. None of these operations are simply scalar multiples of each other, you're absolutely sure there's no elementary functions for which $f(N)$ is also a norm?

 

[pasmith]

Thanks for your reply. There's so many types of norms though, norms defined by integration, norms defined by differentiation (both of which are linear operators), norms defined by supremums, norms defined by the eigenvalues of a matrix, so on. None of these operations are simply scalar multiples of each other, you're absolutely sure there's no elementary functions for which $f(N)$ is also a norm?

A norm $N: V \mapsto \mathbb{R}$ on a vector space $V$ has to satisfy the requirements of being a norm:

• For every $v \in V$, $N(v) \geq 0$ with $N(v) = 0$ iff $v = 0$.
• For every $v \in V$ and every scalar $a$, $N(av) = |a|N(v)$.
• For every $v \in V$ and every $w \in V$, $N(v + w) \leq N(v) + N(w)$.

$f : [0, \infty) \to \mathbb{R}$ has to satisfy conditions which make $f \circ N: V \mapsto \mathbb{R}$ a norm, given that $N$ satisfies the above requirements:

• For every $x \in [0, \infty)$, $f(x) \geq 0$ with $f(x) = 0$ iff $x = 0$.
• For every $x \in [0, \infty)$ and every $a \in [0, \infty)$, $f(ax) = af(x)$.

The second requirement is extremely restrictive for real functions, and only functions of the form $x \mapsto Cx$ satisfy it.

An entirely different question is: given two vector spaces $V$ and $W$, a (not necessarily linear) map $L : V \to W$ and a norm $N$ on $W$, under what circumstances is $N \circ L$ a norm on $V$? I suspect you will find that $L$ does actually need to be linear and injective, but that still admits much more possibility than just positive multiples of the identity.

 

[askmathquestions]

Thanks for explaining. I guess the linearity is a very restrictive condition, since the only elementary functions with that property are linear functions, though compositions of integrals and derivatives might still yield a norm.