Estimating $\lVert f \rVert_1$ in terms of $\int_{\mathbb{R}}\frac{f(x)}{(1+x^2)^n} $?

Question

Let $f \colon \mathbb{R} \to \mathbb{C}$ be a Schwartz function, for safety.

Is there a bound for $\int_\mathbb{R} \lvert f(x)\ \mathrm dx$ if we know the value of \begin{equation} \int\limits_{\mathbb{R}}\frac{f(x)}{(1+x^2)^n} \ \mathrm dx \end{equation} for some $n \in \mathbb{N}$?

This looks similar to the negative Sobolev norm of index $-n$, but I cannot find a relevant reference for the estimate of $\lVert f \rVert_1$ in terms of the above quantity.

Could anyone help me?

Answer 1

No, this is impossible: for $f\equiv 1$ the integral $\int_{\mathbb R} \frac{|f|}{(1+x^2)^n}$ is finite, while $f$ is not integrable on $\mathbb R$.

Now take $f_k\in C_c^\infty(\mathbb R)$, such that $|f_k|\le 1$ on $\mathbb R$ and $f_k = 1$ on $(-k,k)$. Then $\int |f_k|\to \infty$, while $( \int_{\mathbb R} \frac{|f_k|}{(1+x^2)^n})$ is a bounded sequence.

That is, there is no $C>0$ such that $$ \int_{\mathbb R} |f| \le C \int_{\mathbb R} \frac{|f|}{(1+x^2)^n} $$ for all $f\in C_c^\infty(\mathbb R)$.

Answer 2

If $g(x)>0$ is unbounded and $f(x)$ is absolutely integrable there is no constant $c$ such that $$\int\limits_{-\infty}^\infty |f(x)|\,g(x)\,dx\le c \int\limits_{-\infty}^\infty |f(x)|\,dx$$ as $(L^1)^*=L^\infty.$ Equivalently there is no constant $c$ such that $$\int\limits_{-\infty}^\infty |f(x)|\,dx\le c \int\limits_{-\infty}^\infty |f(x)|\,g(x)^{-1}\,dx$$ The space $C_c(\mathbb{R})$ is dense in $L^1$ so smoothness does not help.

In particular we can apply that to $g(x)=x^2+1.$ The above remains valid for spaces $L^1(X,\mu)$ with $\sigma$-finite measure space.