# Is space of $C^1$ functions with finite support a Hilbert space

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 $C^1$?

• It is an inner product space, but not a Hilbert space. Its completion is the Sobolov space $H^1(\mathbb{R})$. – Sangchul Lee Sep 5 '17 at 21:23

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.
• This is not true. The only constant function with finite support is the zero function. Hence, $(f,f) \ne 0$ for all $f \ne 0$. – gerw Sep 6 '17 at 6:49
• Does the notation $C^1$ mean compactly supported? – Berci Sep 6 '17 at 7:45