Let $M:[0,\infty)\to[0,\infty)$ be continuous and convex. Further $M$ satisfies $M(t)=0\Leftrightarrow t=0$.

Let $$\mathcal L_M(\mathbb R):=\left\{f:\mathbb R\to\mathbb R \mathrm{\ measurable\ }|\exists c>0: \int_\mathbb R M\left(\frac{|f(t)|}{c}\right)dt<\infty\right\}$$

Now consider the quotient-space $L_M(\mathbb R):=\mathcal L_M(\mathbb R)/ \{f|f=0\mathrm{\ almost\ everywhere}\}$.

Then $\|f\|_M=\inf\{c>0| \int_\mathbb R M\left(\frac{|f(t)|}{c}\right)dt\leq 1\}$ defines a norm on the quotient-space and $L_M(\mathbb R)$ is a banach space.

I am currently trying to show that $\|\cdot\|$ is indeed a norm, but I am not getting anywhere with it.

First I am trying to show that $\|f\|=0\Leftrightarrow f=0$ for all $f\in L_M(\mathbb R)$.

If $\|f\|=0$, we know that $\inf_{c>0}\{\int_\mathbb R M\left(\frac{|f(t)|}{c}\right)dt\leq 1\}=0$. Since $|f(t)|/c\to\infty$ for $c\to 0$ and the convexity of $M$ I think $f\equiv 0$ is the only possibility that this holds, but I don't think this is a good mathematical argument.

I also have to show $\|\alpha f\|=|\alpha|\|f\|$ and $\|f+g\|\leq\|f\|+\|g\|$, but for these 2 I am completely clueless.

Can anyone hint me in the right direction here?


Positivity. If the set $\{|f|>0\}$ has positive measure, then there is $n$ such that $\{|f|>1/n\}$ has positive measure, say $\delta$. It follows that $\int M(|f|/c) \ge \delta\,M(1/(cn))$. The inequality $\delta\,M(1/(cn))\le 1$ cannot hold for arbitrarily small $c$.

Homogeneity. Observe that $$\int_\mathbb R M\left(\frac{ |\alpha f(t)|}{c }\right)dt\le 1 \tag{1}$$ if and only if the number $c' = c/|\alpha|$ satisfies $$\int_\mathbb R M\left(\frac{ | f(t)|}{c' }\right)dt\le 1$$ Conclude that the set of $c$ that work for $\alpha f$ is related to the set of $c$ that work for $f$.

Triangle inequality. It is easier to approach it via the convexity of the unit ball: namely, show that if $\|f\|\le 1$ and $\|g\|\le 1$, then $\|\lambda f+(1-\lambda )g \|\le 1$. Indeed, for every $c>0$ you have $\int M(|f|/c)\le 1$ and $\int M(|g|/c)\le 1$. By convexity of $M$, $\int M(|\lambda f+(1-\lambda )g|/c)\le 1$. It is an exercise (unrelated to this particular norm) that convexity of unit ball and homogeneity imply the triangle inequality.

  • $\begingroup$ Very helpful, thank you! $\endgroup$
    – dinosaur
    Apr 22 '14 at 13:08

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