If $f$ and $g$ have the same $L^{2}$ norm, does it imply $hf$ and $hg$ have the smae $L^{2}$ norm? Let $f,g \in L^{2}(\mathbb R)$ with $\|f\|_{L^{2}}=\|g\|_{L^{2}}$( that is, $f$ and $g$ have the same $L^{2}-$ norm). We choose $h\in \mathcal{S}(\mathbb R)$(= Schwartz space) so that $hf, hg \in L^{2}(\mathbb R).$

My Question: Can we expect $\|hf\|_{L^{2}} =\|hg\|_{L^{2}},$ that is, $\int_{\mathbb R} |h(x)f(x)|^{2} dx = \int_{\mathbb R} |h(x)g(x)|^{2} dx$ ? (In other words, after multiplying $f$ and $g$ with $h$, can we expect $hf$ and $hg$ have the same $L^{2}-$ norm)


Let $f=I_{[0,1]}$ be the indicator function on the interval $[0,1]$.  In other words, $f$ is $0$ except on the interval $[0,1]$, where it is $1$.  Let $g=I_{[2,3]}$.  Then, the $L^2$ norms of these are the same (they are both 1).
But, $\|hf\|_{L^2}$ would be the $L^2$ norm of $h$ over the interval $[0,1]$ and $\|hg\|_{L^2}$ would be the $L^2$ norm of $h$ over the interval $[2,3]$.  For most Schwartz functions, the norm over these intervals is not the same.
