2
$\begingroup$

For $\int_{\mathbb R^n} \mid x \mid \mid g(x) \mid dx< \infty$ and $\int g dx = 0$. Suppose $\omega \in C_0 ^{\infty} (\mathbb R)$ and $\int_{\mathbb R} \omega (x) dx =1$.

For $1 \leq k \leq n$, we define the sequence functions $g_k$ by $g_0 = g$ and if $k \geq 1$: $$g_k (y)= \int_{\mathbb R^k} g(x_1, x_2, ...,x_k,y_{k+1},\cdots, y_n)dx_1 dx_2...dx_k \omega(y_1)\omega(y_2)\cdots \omega(y_k)$$

Show: $$\int_{\mathbb R} |(g_{k-1} - g_k)(y_1, y_2, ...,y_{k-1},x,y_{k+1},\cdots, y_n)| |x| dx \in L^1 (\mathbb R^{n-1})$$

I think the key is to use the first condition: $\int_{\mathbb R^n} \mid x \mid \mid g(x) \mid dx< \infty$, but this is for $n$ dimensions and I want 1 dimension, so it's hard to apply the condition here.

$\endgroup$
  • $\begingroup$ Did you mean $g_0 = g$? $\endgroup$ – user263732 Apr 23 '17 at 1:21
  • $\begingroup$ @JuanDiegoRojas Yes, sorry. Just corrected. $\endgroup$ – Que Er Apr 23 '17 at 1:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.