Norm in $\mathcal{C}[\Bbb{R},\Bbb{R}]$ [duplicate]

I’m interested in expliciting a norm in the space of continuous functions from $\Bbb{R}$ to itself. It does not need to induce a complete metric.

marked as duplicate by Community♦Sep 15 '18 at 11:26

• "Expliciting?" What is your question? – Sean Roberson Sep 15 '18 at 10:36
• I know you can show the existence of such a norm by means of the Axiom of Choice. However this approach is naturally not constructive. – matboy Sep 15 '18 at 10:47

Continuity implies Riemann integrability. Multiplication preserves continuity, so the product is Riemann integrable. For $a,b\in\mathbb{R}$ with $a<b$, we can define the inner product $\langle\cdot,\cdot\rangle: V\times V\to\mathbb{R}$ by $$\langle f, g \rangle = \int_a^b fg$$ This induces a norm via $\langle f, f\rangle = \| f \|^2$.
EDIT: Only a semi-norm. Not an inner product space. Semi-norm is given by $$\| f \| = \sqrt{\int_a^b f^2}$$ I believe you can quotient out by the kernal of the semi-norm to create a norm but this may be incorrect and probably violates the conditions of the question.
• Unfortunately, this is not a norm in ${\cal C}(\mathbb{R})$, it defines a seminorm. – Rodrigo Dias Sep 15 '18 at 10:44
• That makes it only a seminorm ($||f||=0 \not\Rightarrow f=0$). – Kolja Sep 15 '18 at 10:44
• I can only see counterexamples for $b\leq a$. I should have said $a<b$. In this case, are there still counter examples? – user512116 Sep 15 '18 at 10:48