How to prove $ (1+|x|^2)^{s/2} (f\ast f)(x), (x\in \mathbb R, s<-1)$ square integrable?

Question

Let $M> 100$ and $f(y)= M^{-s} \chi_{[-10, 10]} (|y|-M), y \in \mathbb R.$ Define $F(x)= (1+|x|^2)^{s/2} (f\ast f)(x), x\in \mathbb R, s<-1.$

Can we say that $\|F\|_{L^2} \leq C$ for some constant $C$

My thoughts: We know that $L^{1}\ast L^p \subset L^p$ by young's inequality. Here we have weight as well. I'm interested to get rid of $M$, that is, can we expect C indecent of $M$?

Answer 1

Yes $\|F\|_2 <\infty$ so for some $C>0$ we have $\|F\|_2 <C$. Why is this true well F is actually a bounded function with compact support so It will be in every $L^p$ space. Why does $F$ have compact support? Well $f$ does and we know $supp(f* g) \subset \overline{supp(f)+supp(g)}$ So $F$ is a continuous function on a bounded interval so it must be bounded and hence $$\|F\|_p=\big(\int|F|^pd\mu\big)^\frac{1}{p}\leq\big(\|F\|_{\infty}^p\int_{supp(F)} d\mu\big)^\frac{1}{p}=\|F\|_{\infty} \mu(supp(F))^\frac{1}{p}< \infty$$ This is true $\forall p \geq1$

Answer 2

No, this function is not bounded in $L^2$ with a bound independent of $M$ with the scaling you propose. To find the good scaling, consider $f(x) = M^{a}\,\chi_{[-10,10]}(|y|-M)$ (your function $f$ but replacing $s$ by $-a$ with $a>1$), and $F(x) = (1+|x|^2)^{-b/2} (f*f)(x)$ (your function $F$ where I replace $s$ by $-b$ with $b>1$)

Now remark that $f*f$ can actually be computed explicitly since $$ (f*f)(x) = M^{2a} \int_0^\infty \chi_{[M-10,M+10]}(|y|)\, \chi_{[M-10,M+10]}(|x-y|)\,\mathrm{d}y. $$ By drawing the function inside the integral in the plane $(x,y)$, we see that the resulting function is made of two "hat" functions. More precisely, since $M>100$, for $x\geq0$, $$ (f*f)(x) = M^{2a}\left\{ \begin{array}{lll} x-(M-30) & \text{ if } x\in[M-30,M-10] \\ 20 & \text{ if } x\in[M-10,M+10] \\ (M+30)-x & \text{ if } x\in[M-10,M+30] \end{array}\right. $$ and $(f*f)(-x)=(f*f)(x)$. We could get an exact value for the weighted $L^2$ norm, but since we just need to estimate it uniformly in $M$, we can use the fact that from the above formula we deduce for $x\geq0$ $$ 20\,M^{2a}\,\chi_{[M-10,M+10]} ≤ (f*f)(x) ≤ 20\,M^{2a}\,\chi_{[M-30,M+30]}. $$ Therefore $$ \begin{align*} \|F\|_{L^2}^2 &\leq 2\cdot40\, M^{4a} \int_{M-30}^{M+30} (1+|x|^2)^{-b}\,\mathrm{d}x \\ &≤ 80\,M^{4a} (1+(M-30)^2)^{-b} \int_{M-30}^{M+30} 1 ≤ C\, M^{4a-2b} \\ \|F\|_{L^2}^2 &\geq 80\, M^{4a} \int_{M-10}^{M+10} (1+|x|^2)^{-b}\,\mathrm{d}x \\ &\geq 80\,M^{4a} (1+(M+10)^2)^{-b} \int_{M-10}^{M+10} 1 \geq c\, M^{4a-2b} \end{align*} $$ for two constants $0<c<C<\infty$ which do not depend on $M$.

So if $a=b=-s$, as in your case, then the norm is of order $\|F\|_{L^2} ≃ M^{2a-b} ≃ M^{-s}$.

If you want a bound independent of $M$, you need $2a≤ b$. This works in particular with $$ \begin{align*} f(x) &= M^{-s}\,\chi_{[-10,10]}(|y|-M) \\ F(x) &= (1+|x|^2)^{s} (f*f)(x). \end{align*} $$