It is known that we may choose smooth $f:\mathbb R \to [0,1]$ such that $f(x)=1$ if $x\geq \frac{3}{4} $ and $f(x)=0$ if $x\leq -\frac{3}{4}+1.$

Define $h(x)= \sin (\frac{\pi}{2} f(x+1))$ if $x\leq 0$ and $h(x)= \cos (\frac{\pi}{2} f(x))$ if $x\geq 0.$

We note that support of $h$ is $[-3/4, 3/4]$ and $h^2(x)+ h^2(x-1)=1$ for all $x\in [0,1],$ $\|h\|_{L^2}=1,$ and this $h:\mathbb R \to \mathbb C$ is smooth.

My Question: Can we expect to choose $h:\mathbb R \to \mathbb C$ such that support of $h$ is $[-1/3, 1/3]$ and $h^2(x)+ h^2(x-1)=1$ for all $x\in [0,1],$ and $$\|h\|_{L^2}^2=\int_{-3/4}^{3/4} |h(x)|^2 dx =3/2?$$