Given a certain open set $\Omega$ I'm trying to prove that continuous, and piecewise continuously differentiable functions with bounded support in $\overline{\Omega}$ have weak derivatives in $\Omega$, and hence belong in $H^1(\Omega)$.\

Since they are continuous in $\overline{\Omega}$ and have bounded support is easy to see they belong in $L^2(\Omega)$ since by Weierstrass theorem they are bounded, so $$\int_\Omega|f|^2dx\leq\max\{|f(x)|\colon x\in\Omega\}^2\cdot\text{vol}(\text{supp}(f))$$ Unfortunately, we've barely seen any result whatsoever to prove the existance of weak derivatives (this is not a course in functional analysis), that is, a function $\omega_i\in L^2(\Omega)$ such that $$\int_\Omega f(x)\frac{d\phi}{d x_i}(x)dx=-\int_\Omega \omega_i(x)\phi(x)dx$$ for every $\phi$ which is infinitely differentiable and with compact support, that is, for every $\phi\in\mathcal{C}^\infty_c(\Omega)$. Namely, I only know the mere definition and a result that says that if you can bound $$\left|\int_\Omega f(x)\frac{d\phi}{d x_i}(x)dx\right|\leq C\|\phi\|_{L^2}=C\left(\int_\Omega|\phi(x)|^2\right)^{1/2}$$ for every $\phi\in\mathcal{C}^\infty_c(\Omega)$ (the constant $C$ must not depend on $\phi$) and for all indexes $i$, then $f$ is weakly differentiable.

I thought maybe Cauchy-Schwart could help $$\left|\int_\Omega f(x)\frac{d\phi}{dx_i}dx\right|\leq\left(\int_\Omega |f(x)|^2dx\right)^{1/2}\left(\int_\Omega\left|\frac{d\phi}{dx_i}\right|^2dx\right)^{1/2}=\|f\|\left\|\frac{d\phi}{dx_i}\right\|$$ but I get this bound in terms of the norm of the derivativate of $\phi$, which is not what I need. I realize I haven't used the piecewise continuous differentiability, but I don't know how to apply it to get the result.

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    $\begingroup$ Do you know that Lipschitz functions have a weak derivative (which is in $L^\infty$)? Piecewise differentiability with bounded derivative implies the Lipschitz property. $\endgroup$ – user357151 Jan 21 '17 at 4:24
  • $\begingroup$ Thanks for the comment, this has allowed me to research the required results to compose the proof I needed. $\endgroup$ – F.Webber Jan 21 '17 at 22:18

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