Theorem: Strictly negative regularized control response

For every finite $r_{\mathrm{c}}>0$, the function

is strictly concave on $z>0$. Consequently

for the three separations used here. On every compact interval of finite positive regularization lengths, its negative magnitude has a strictly positive minimum.

Proof. Let $I(d)={{\int }_0}^d{r}^2{w}_{r_{\mathrm{c}}}(r)\mathrm{d}r$. The shell theorem gives

Writing $\psi (z)={\Gamma }_{r_{\mathrm{c}}}(\sqrt{z})$, differentiation gives

Strict decrease of $w_{r_{\mathrm{c}}}$ implies $3I(d)>d^3{w}_{r_{\mathrm{c}}}(d)$, so ${\psi }^{\prime \prime }<0$.

For $z_i={d_i}^2$, strict concavity gives

Since $z_2-z_1=3R^2/16$ and $z_3-z_2=5R^2/16$, their slope difference equals $16{\mathcal{B}}_{r_{\mathrm{c}}}/(15R^2)$. This proves the sign.

Continuity in $r_{\mathrm{c}}$ follows by dominated convergence in the spectral integral on compact positive intervals. A continuous strictly positive $-{\mathcal{B}}_{r_{\mathrm{c}}}$ has a positive minimum on such an interval.