Continuity of a functional on space of compactly supported smooth function

Assume $n\geq 3$. Let $C_c^\infty(\mathbb{R}^n)$ be the space of compactly supported real valued smooth functions on $\mathbb{R}^n$ endowed with $L^\infty$ norm. For $f\in C_c^\infty(\mathbb{R}^n)$, let $\hat{f}$ donote it's fourier transform. Define the functional $T: C_c^\infty(\mathbb{R}^n)\to\mathbb{R}$ by $$T(f) = \int_{\mathbb{R}^n}\frac{|\hat{f}(x)|^2}{||x||^2}\,dx$$

Show that T is well-defined. Is T continuous?

I was able to show that T is well-defined i.e the integral is finite but could not prove the continuity part. Any help is appreciated!

Update: I am starting to think that T may not be continuous in max norm. So I have changed the question.

• Which bound did you get for the integral to show that it is finite? – Sebastian Bechtel Dec 22 '17 at 8:06
• I used the local integrability of $\frac{1}{||x||^2}$ to bound the integral on ball and the polynomial decay of $\hat{f}$ outside the ball to bound the other part of the integral. – Rob Dec 22 '17 at 8:19
• Because $f \mapsto |\hat{f}|^2$ is continuous $C^\infty_c \to L^1 \cap L^\infty$ and $\frac{1}{\|x\|^2} \in L^1(\|x\| < 1) , L^\infty(\|x\| \ge 1)$ then yes $f \mapsto T(f)$ is continuous for the $C^\infty_c$ topology. – reuns Dec 22 '17 at 10:08

More generally, consider the bilinear form $$C(f,g)=\int_{\mathbb{R}^n}\frac{\overline{\widehat{f}(x)}\widehat{g}(x)}{||x||^2}\ dx\ .$$ It is easy to show it is continuous since it satisfies a bound $$|C(f,g)|\le \rho(f)\rho(g)$$ for some continuous seminorm $\rho$. Then compose with the diagonal map $f\mapsto(f,f)$ to get that $T(f)=C(f,f)$ is continuous.