Mfrhtyj

We denote the set of functions $q : mathbb Rtomathbb C$ for which
$$
|q|_{L^2_u} := left(sup_{ninmathbb N}int_n^{n+1}|q(x)|^2,dxright)^{1/2} < infty
$$
by $L^2_u(mathbb R)$ (uniformly locally $L^2$). It is well-known that $L^2_u(mathbb R)$ is exactly the class of functions $q$ for which $qfin L^2(mathbb R)$ for all $fin H^1(mathbb R)$. Each such function induces a bounded operator from $H^1(mathbb R)$ to $L^2(mathbb R)$. Hence, there exists $C = C_q> 0$ such that
$$
|qf|_2,le,C_q|f|_{H^1}.
$$
The constant $C_q$ can be bounded by $|q|_{L^2_u}$, that is, we have
$$
|qf|_2,le,C|q|_{L^2_u}|f|_{H^1}
$$
for all $qin L^2_u(mathbb R)$ and all $fin H^1(mathbb R)$.

I would like to know the best constant $C$ for this. I have derived one, but it is very bad. I got it in the following way. First, for $qin L^2$ one has
$$
|qf|_2 = |hat q * hat f|_2,le,|q|_2|hat f|_1,le,sqrtpi|q|_2|f|_{H^1}.
$$
Now, I let $qin L^2_u$ and pick a smooth partition of unity $(phi_n)$ with $operatorname{supp}phi_nsubset [n-1,n+1]$. Then
$$
|qf|_2^2 = sum_n|q(sqrt{phi_n}f)|_2^2,dx,le,pisum_n|q|_{L^2(n-1,n+1)}^2|sqrt{phi_n}f|_{H^1}^2,le,2pi|q|_{L^2_u}^2sum_n|sqrt{phi_n}f|_{H^1}^2.
$$
Now, it remains to estimate the last sum:
begin{align*}
sum_n|sqrt{phi_n}f|_{H^1}^2
&= sum_nintphi_n|f|^2 + Big|frac{phi_n'f}{sqrt{phi_n}} + sqrt{phi_n}f'Big|^2,dx\
&le 2|f|_2^2 + 2sum_nintBig|frac{phi_n'f}{sqrt{phi_n}}Big|^2 + phi_n|f'|^2,dx\
&le 2|f|_2^2 + 4|f'|_2^2 + 2sum_nint_{n-1}^{n+1}frac{|phi_n'|^2}{phi_n}|f|^2,dx\
&le (2+4S)|f|_2^2 + 4|f'|_2^2,
end{align*}
where $S = sup(|phi_0'|^2/phi_0)$, because the $phi_n$ are just translated versions of each other. Now, my $S$ is about $144$ or so. It would be much nicer to have it close to one.

So, I am either looking for a partition of unity $(phi_n)$ for which $|phi_n'|^2/phi_n$ is small or for just another way of treating the whole thing. Maybe someone here has a reference?

I have been able to reduce the constant to a minimum (at least with respect to the above technique). I define
$$
phi_0(x) :=
begin{cases}
2(1+x)^2 &text{if }xin [-1,-tfrac 1 2],\
1-2x^2 &text{if }xin[-tfrac 12,tfrac 12],\
2(1-x)^2 &text{if }xin [tfrac 12,1]
end{cases}.
$$
Then $phi_0in C^1(mathbb R)$, its support is $[-1,1]$ and $|phi_0'|^2/phi_0le 8$. Moreover, the functions $phi_n(x) = phi_0(x-n)$ are a partition of unity as above. Since also $|phi_0'|le 2$, we get
begin{align*}
sum_{ninmathbb Z}big|sqrt{phi_n}fbig|_{H^1}^2
&= sum_{ninmathbb Z}int_{n-1}^{n+1}left(phi_n|f|^2 + Big|frac{phi_n'}{2sqrt{phi_n}}f + sqrt{phi_n}f'Big|^2right),dx\
&= sum_{ninmathbb Z}int_{n-1}^{n+1}left(phi_n|f|^2 + frac{|phi_n'|^2}{4phi_n}|f|^2 + phi_n|f'|^2 + phi_n'operatorname{Re}(overline{f}f')right),dx\
&le sum_{ninmathbb Z}int_{n-1}^{n+1}left(3|f|^2 + |f'|^2 + 2|f'f|right),dx,le,8|f|_{H^1}^2.
end{align*}
Thus,
$$
|qf|_2,le,4sqrtpi|q|_{L^2_u}|f|_{H^1}.
$$