Show that $\delta_0$, Dirac function, defined than $\left<{\delta_0,\phi}\right> = \phi(0)$ is linear.
I trying: Let be $\phi_1,\phi_2$ $\in W^{m,p}(\Omega)$ then $\delta_0(\phi_1+\phi_2)=(\phi_1+\phi_2)(0)$, but I need more steps.
Show that $\delta_0$, Dirac function, defined than $\left<{\delta_0,\phi}\right> = \phi(0)$ is linear.
I trying: Let be $\phi_1,\phi_2$ $\in W^{m,p}(\Omega)$ then $\delta_0(\phi_1+\phi_2)=(\phi_1+\phi_2)(0)$, but I need more steps.
Hint: What is the definition of a linear functional? Just plug this into the definition and see if it works. Under trying, $(\phi_1 + \phi_2)(0)=\phi_1 (0) + \phi_2 (0)$
I am a little confused about your question, so please do not downvote my answer. If $<>$ represents the inner product, use the fact that the Heaviside step function is the antiderivative of the Dirac delta as follows: $< \varphi ,\delta > = \int_{ - \infty }^\infty {\varphi (x)\delta (x)dx = \left[ {\varphi (x)H(x)} \right]_{ - \infty }^\infty } - \int_{ - \infty }^\infty {\varphi '(x)} H(x)dx$ which simplifies to $- \int_0^\infty {\varphi '(x)dx = \left[ { - \varphi (x)} \right]_0^\infty = \varphi (0)}.$ Note that I used the definition of the Dirac delta.