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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.

matboy
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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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    Unfortunately, this is not a norm in ${\cal C}(\mathbb{R})$, it defines a seminorm. – Rodrigo Dias Sep 15 '18 at 10:44
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    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? –  Sep 15 '18 at 10:48
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    A function that is 0 on an interval, and has a non zero value somewhere else. Think of bump functions. – Kolja Sep 15 '18 at 10:50
  • Ah right definitely. –  Sep 15 '18 at 10:51
  • Thanks for your answer but, as others have mentionned, this doesn’t define a norm. For any $a,b\in\Bbb{R}$, you can easily find a non-zero continuous function that equals $0$ in the interval $[a,b]$. – matboy Sep 15 '18 at 10:52
  • Yes, there are. For any $a<b$, take $f(x)=0$ for $x\in(a,b)$ and $f(x)=|x-\frac{a+b}{2}|-\frac{b-a}{2}$ for $x\notin(a,b)$, for example. – Rodrigo Dias Sep 15 '18 at 10:52