Suppose that $\int f=0$, $f \in L^2$ and $f$ has a compact support. Let $\phi$ be radial, and such that $\mathrm{supp}(\phi) \in B(0,1)$. Plus, assume that $\int_{R^+} |\hat{\phi}(t\xi)|^2t^{-1}dt$=1 for every $\xi \in R^d-\{0\}$. Define $\phi_{t}(x)=t^{-d}\phi(t^{-1}x)$. Show that for any $x \in R^d$ the following integral is finite. $$\int_{0}^{\infty}\int_{R^d}|\phi_{t}(x-y)||(\phi_t*f)(y)|\frac{dt}{t}dy<\infty$$
I split the interval $(0,\infty)$ into $(0,1)$ and $[1,\infty)$. Consider the second one. By the Cauchy-Schwarz inequality it is dominated by
$$\int_{1}^\infty \left(\int |\frac{1}{t^d}\phi(\frac{x-y}{t})|^2dy\right)^{1/2}\left(\int|\phi_t*f|^2(y)dy\right)^{1/2}\frac{dt}{t}$$
Apply Cauchy-Schwarz inequality again. Then, $$\left(\int_{1}^{\infty}\int_{R^d}t^{-2d-1}|\phi(\frac{x-y}{t})|^2dydt\right)^{1/2}\left(\int_{1}^\infty\int_{R^d} |\phi_t*f|^2(y) dy\frac{dt}{t}\right)^{1/2}$$
By the Plancherel theorem, we get the desired result.
We cannot apply the same method to the case $(0,1)$ because of singularity.
Thanks in advance.
p.s. I got the desired result. By applying Cauchy-Schwarz inequality twice, we get
$$(\int_{R^+}\int_{R^d}|\phi_{t}(x-y)|^2 \frac{dt}{t}dy)^{1/2}(\int_{R^+}\int_{R^d}|(\phi_t *f)(y)|^2\frac{dt}{t}dy)^{1/2}$$
Now, apply Plancherel's theorem in $y$ variable. Then
$$(\int_{R^+}\int_{R^d}|\hat{\phi}(t\xi)|^2\frac{dt}{t}d\xi)^{1/2}(\int_{R^+}\int_{R^d}|\hat{\phi}(t\xi)\hat{f}(\xi)|^2\frac{dt}{t}d\xi)^{1/2}$$
The second term has no problem. For the first term, use $\hat{\phi}(0)=0$, mean value theorem and $|\hat{\phi}(t\xi)| \leq C/(1+|t\xi|^2)^N$.