MHB Understanding the Integral in a Proof: A Closer Look at the Role of Supp and Chi

[evinda]

Hello! (Wave)

I am looking at the following proof:View attachment 5454

Why does it hold that $\int u_j \phi dx= \int (\psi_j \ast u) \phi dx$ ?

 

[evinda]

We have that $X(\frac{x}{j})=1$ for $|x|<j$, right? And since we pick $j$ sufficiently large, it's like if we had $-\infty<j<+\infty$ and so we have $X(\frac{x}{j})=1$ for all the values of $x$, right?

 

[Euge]

You are correct that $\chi(x/j) = 1$ for $|x| < j$. To proceed, let $M > 0$ such that $\operatorname{supp}(\phi)\subset [-M,M]^n$; choose $j > M$. Then $\phi(x) = 0$ for all $|x| \ge j$. Thus, for all $j > M$,

$$\int u_j\phi\, dx = \int \chi\left(\frac{x}{j}\right)(\psi_j * u)\phi(x)\ dx = \int_{|x| < j} \chi\left(\frac{x}{j}\right)(\psi_j * u)\phi(x)\, dx = \int_{|x| < j} (\psi_j * u)\phi\, dx = \int (\psi_j * u)\phi\, dx$$

 

[evinda]

You are correct that $\chi(x/j) = 1$ for $|x| < j$. To proceed, let $M > 0$ such that $\operatorname{supp}(\phi)\subset [-M,M]^n$; choose $j > M$. Then $\phi(x) = 0$ for all $|x| \ge j$. Thus, for all $j > M$,

$$\int u_j\phi\, dx = \int \chi\left(\frac{x}{j}\right)(\psi_j * u)\phi(x)\ dx = \int_{|x| < j} \chi\left(\frac{x}{j}\right)(\psi_j * u)\phi(x)\, dx = \int_{|x| < j} (\psi_j * u)\phi\, dx = \int (\psi_j * u)\phi\, dx$$

I got it... Thanks a lot! (Smirk)