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This question already has an answer here:

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.

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

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  • $\begingroup$ "Expliciting?" What is your question? $\endgroup$ – Sean Roberson Sep 15 '18 at 10:36
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    $\begingroup$ 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. $\endgroup$ – matboy Sep 15 '18 at 10:47
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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.

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    $\begingroup$ Unfortunately, this is not a norm in ${\cal C}(\mathbb{R})$, it defines a seminorm. $\endgroup$ – Rodrigo Dias Sep 15 '18 at 10:44
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    $\begingroup$ That makes it only a seminorm ($||f||=0 \not\Rightarrow f=0$). $\endgroup$ – Kolja Sep 15 '18 at 10:44
  • $\begingroup$ I can only see counterexamples for $b\leq a$. I should have said $a<b$. In this case, are there still counter examples? $\endgroup$ – user512116 Sep 15 '18 at 10:48
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    $\begingroup$ A function that is 0 on an interval, and has a non zero value somewhere else. Think of bump functions. $\endgroup$ – Kolja Sep 15 '18 at 10:50
  • $\begingroup$ Ah right definitely. $\endgroup$ – user512116 Sep 15 '18 at 10:51

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