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If the product is defined as $$(f,g) = \int_{-\infty}^\infty f'(x)\overline{g'(x)} \, dx$$

then is the space of continuously differentiable functions with finite support an inner product space?

I initially thought No because if $f$ is the Cantor function then $(f,f)=0$ however $f\neq 0 $. While $f$ is differentiable almost everywhere and is continuous everywhere.... maybe it is not $C^1$ because we require more than just a.e. continuous differentiability to be in $C^1$?

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    $\begingroup$ It is an inner product space, but not a Hilbert space. Its completion is the Sobolov space $H^1(\mathbb{R})$. $\endgroup$ – Sangchul Lee Sep 5 '17 at 21:23
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We require everywhere differentiability, with a continuous differentiate for functions in $C^1$.

But you're on the right track, exactly the constant functions will have $(f,f)=0$ with respect to this positive semidefinite bilinear form.

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  • $\begingroup$ Thank you for clarifying!! I appreciate you answering the questions as asked! $\endgroup$ – Clclstdnt Sep 5 '17 at 21:30
  • $\begingroup$ This is not true. The only constant function with finite support is the zero function. Hence, $(f,f) \ne 0$ for all $f \ne 0$. $\endgroup$ – gerw Sep 6 '17 at 6:49
  • $\begingroup$ Does the notation $C^1$ mean compactly supported? $\endgroup$ – Berci Sep 6 '17 at 7:45
  • $\begingroup$ It means continuously differentiable. So in this setting we are working with continuously differentiable functions with finite support. $\endgroup$ – Clclstdnt Sep 6 '17 at 22:13
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It is an inner product space, but it is not complete ( so, not a Hilbert space).

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