Understanding the proof that $f\in L^1_{loc} =0 $ a.e if its integral with all test function is 0

Question

I know this question has been asked before but there is one specific point about this proof that is bothering me :

So I was wondering what was the point of extending $f$ to the whole on $R^n$ until I reread the definition of mollifier which explicitly defined them to be functions on $R^n$ whose integral over $R^n$ is 1. Ok, so that was how they were defined on these notes, but at the same time, we can easily shift and rescale any mollifiers so that their support is contained in $\Omega$ and if we integrate our mollifier over $\Omega$, we would still get 1. That way we can just work on $\Omega$ instead of extending to the whole of $R^n$. I know, either way, it really doesn't change how long the proof is overall or make this proof any easier or harder, but in all the proofs I have seen so far, we extended $f$ to $R^n$ which really doesn't feel necessary at all.

Answer 1

The convolution you are using is defined for functions defined on the whole $\mathbb R^n$. Just for the record, the convolution product is defined by $$ f*g(x)=\int_{\mathbb R^n} f(y)g(x-y)dy $$ when the integral is well defined for the Lebesgue theory. I know there exists convolution on suitable groups, I am not an expert, but I suspect that if you replace $\mathbb R^n$ by a nasty set you might have problems due to lack of symmetry and/or convexity. So my first answer would be: I am not very smart so I prefer to spend few words to make my problem fit the known theory, which works.

You say that you can shift the mollifier so its support is contained in $\Omega$. This would not solve your problem and worst it would break the mollification procedure. Indeed, I agree with you that for any $x_0 \in \Omega$, there is $\epsilon_0 > 0$ such that forall $0 < \epsilon < \epsilon_0$, if you define $$ \psi_\epsilon := \phi_\epsilon(x_0 + \cdot) $$ you get that $(\psi_\epsilon)_{0 < \epsilon < \epsilon_0}$ is a family of $\mathcal D (\Omega)$. Then observe the following

1. It is yet not evident how to define $f * \psi_\epsilon$ without extending $f$. Indeed formally $$ f * \psi_\epsilon(x) = \int_{\mathbb R^n}\psi_\epsilon(y)f(x-y)dy = \int_{B(x_0,\epsilon)}\psi_\epsilon(y)f(x-y)dy $$ and except with an extension by $0$ I don't know how to give a sens to this integral. This is because the $\epsilon_0$ depends on $x_0$.

2. More importantly, you loose the key property of a mollification kernel: regularization. This can be understood philosophically, the mollification kernel does average of the function which makes it more smooth. But if you shift it you are taking the average with a shift, so you are not seing an average of what is surrounding you. To see it suppose $\Omega = \mathbb R^n$, I claim that the following is false $$ \forall f \in L^1(\Omega),\quad f * \psi_\epsilon \xrightarrow[\epsilon \longrightarrow 0^+]{L^1(\Omega)} f. $$ Denote $\tau$ the operator which shifts the functions of $x_0$, you get that $$ \begin{align} f * \psi_\epsilon(x) &= \int_{\mathbb R^n}\psi_\epsilon(y)f(x-y)dy\\ &= \int_{\mathbb R^n}\phi_\epsilon(x_0+y)f(x-y)dy\\ &= \int_{\mathbb R^n}\phi_\epsilon(y)f(x-y+x_0)dy \\ &= \int_{\mathbb R^n}\phi_\epsilon(y)(\tau f)(x-y)dy \\ &= \phi_\epsilon * (\tau f)(x) \end{align} $$ so that $$ f * \psi_\epsilon = \phi_\epsilon * (\tau f) \xrightarrow[\epsilon \longrightarrow 0^+]{L^1(\Omega)} \tau f $$ and the right hand side can be taken different of $f$.

Notice that in my computations I performed a change of variable that is a translation, if you change the integration domain computations become more delicate.