Sparse Gaussian scalar certificate and extremal profile

Put $w=z^2$. The hyperbolic series give $w\sinh w-q(\cosh w-1)={\sum }_{k\ge 1}\frac{(2k-q)w^{2k}}{(2k)!}\mathrm{.}$ Every coefficient is positive because $q<2$. Replacing $q$ by $2$ makes only the first coefficient zero and leaves every later coefficient positive. The positive denominators in $a,b$ therefore prove $a(z),b(z)>0$.

For fixed $z>0$ and $v>0$, put $f(v)=(\cosh (zv)-1-a(z)v^2)\mathrm{/}{v}^q$. Differentiation and the absolutely convergent series show that $\frac{f^{\mathrm{\prime }}(v)}{v^{1-q}}={\sum }_{k\ge 1}\frac{(2k-q)z^{2k}{v}^{2k-2}}{(2k)!}-(2-q)a(z)\mathrm{.}$ This expression is strictly increasing in $v$: its terms with $k\ge 2$ have strictly positive coefficients and positive powers. It is zero at $v=z$ by the definition of $a$. Consequently $f$ decreases to $z$ and increases after $z$. Direct substitution gives $f(z)=-b(z)$, so $\cosh (zv)-1\ge a(z)v^2-b(z)v^q$. Evenness proves the assertion for negative $v$, and direct substitution proves equality at $0,z,-z$.

The quotient representation of $\varphi$ in the Statement is continuous. To prove strict increase, write its numerator series before the factor $z^{2-q}$ as $P(w)={\sum }_{k\ge 1}{p}_k{w}^{2k}$ and its denominator as $Q(w)={\sum }_{k\ge 1}{q}_k{w}^{2k}$, where $p_k=(2k-2)\mathrm{/}(2k)!$ and $q_k=(2k-q)\mathrm{/}(2k)!$. The ratios $p_k\mathrm{/}{q}_k=(2k-2)\mathrm{/}(2k-q)$ strictly increase with $k$. Pairing terms with indices $k>\mathrm{\ell }$ in $P^{\mathrm{\prime }}Q-P{Q}^{\mathrm{\prime }}$ gives positive multiples of $(k-\mathrm{\ell })(p_k{q}_{\mathrm{\ell }}-p_{\mathrm{\ell }}{q}_k)w^{2k+2\mathrm{\ell }-1}$, so this derivative numerator is positive for $w>0$. The products and paired sums are justified first for finite partial sums and then by absolute convergence on compact intervals. Thus $P\mathrm{/}Q$ is strictly increasing; multiplying it by the positive strictly increasing function $z^{2-q}$ proves that $\varphi$ is strictly increasing.

The first nonzero numerator and denominator terms are $w^4\mathrm{/}12$ and $(2-q)w^2\mathrm{/}2$, respectively. Hence $\varphi (z)\sim \frac{z^{6-q}}{6(2-q)}(z\downarrow 0)\mathrm{.}$ The exponential representations instead give $\varphi (z)\sim z^{2-q}$ as $z\to \mathrm{\infty }$, since the ratio of its two bracketed expressions tends to one. Thus its endpoint limits are zero and infinity. Strict increase and [F3] give exactly one inverse value at each positive target. The inverse is continuous: values on either side of its value have images on either side of the target, which is the epsilon-delta criterion for the inverse at that target. The inverse power asymptotics follow by trapping $\varphi (z)$ between $(1\pm \eta )$ times its stated positive leading power and then taking the corresponding positive roots.

Expanding $a$ at zero gives $a(z)\sim z^2\mathrm{/}2$, so $\psi (z)\sim z^{-2}$. At infinity, $a(z)\sim e^{z^2}\mathrm{/}[2(2-q)]$, so $\psi (z)\sim (2-q)e^{-z^2}$. Combining these with the inverse asymptotics in step 3.1 proves all endpoint formulas for $z(x)$ and $d(x)$. In particular $z(x)$ is comparable to $x^{\kappa }$ at infinity. The functions $\psi (z)z^2$ and $\psi (z)(e^{z^2}-1)$ have finite limits at zero and infinity; continuity bounds them also on the intervening compact interval. Similarly $e^{z^2}\psi (z{)}^2(e^{z^2}-1)$ is bounded for $z\ge 1$, proving the three global bounds.

The definitions give $d(x)a(z(x))=x^{2\beta }\mathrm{/}2$ and $d(x)b(z(x))=x^{qs}\mathrm{/}2$, since $\varphi (z(x))=x^L$ and $L+2\beta =qs$. Also $a(z)z^2-b(z)z^q=\cosh (z^2)-1$. Thus $d(x{)}^2(\cosh (z(x{)}^2)-1)=\frac{1}{2}(d(x)z(x{)}^2{x}^{2\beta }-d(x)z(x{)}^q{x}^{qs})\mathrm{.}$ Near zero the integrands for $c_1,c_2,c_0$ are asymptotic to positive constants times $x^{4\beta }$, $x^{4\beta +4L\mathrm{/}(6-q)}$, and $x^{4\beta }$, respectively; those for $M_1,M_2$ have respective exponents $6\beta +2L\mathrm{/}(6-q)$ and $6\beta +10L\mathrm{/}(6-q)$. All these exponents exceed $-1$. At infinity each integrand is bounded by $C{x}^m{e}^{-c{x}^{2\kappa }}$ for suitable positive $C,c$ and some nonnegative $m$. This envelope is integrable: choosing an integer $k$ with $2\kappa k>m+2$ and using $e^{c{x}^{2\kappa }}\ge (c{x}^{2\kappa }{)}^k\mathrm{/}k!$ bounds it by a constant times $x^{-2}$ for large $x$. The integral of $x^{-2}$ on $[b,R]$ is $1\mathrm{/}b-1\mathrm{/}R$ by the compact fundamental theorem in [F4], and is bounded as $R\to \mathrm{\infty }$. The small-$x$ power bounds likewise integrate to $a^{p+1}\mathrm{/}(p+1)$ when $p>-1$. Compact intervals are covered by [F4]. The five improper integrals therefore exist, are finite, and are positive. Integrating the pointwise identity proves $c_0=(c_1-c_2)\mathrm{/}2>0$.

For fixed $n,h_0>0$, the large-$x$ bound $d(x)\le C{x}^{2\beta }{e}^{-c{x}^{2\kappa }}$ makes ${\sum }_i{h}_{0}d(i\mathrm{/}n)$ finite by comparison with a convergent power tail. There are only finitely many remaining indices, including those close to zero. Multiplying the pointwise inequality of step 2.1 by $h_i$ and using $h_{i}a(z_i)=h_0(i\mathrm{/}n{)}^{2\beta }\mathrm{/}2$ and $h_{i}b(z_i)=h_0(i\mathrm{/}n{)}^{qs}\mathrm{/}2$ proves the finite-vector scalar certificate in the Statement. This argument places no upper bound on the score coefficient $h_i$.

Let $f$ be any of the five integral integrands in step 5.1. On a fixed $[a,b]\subset (0,\mathrm{\infty })$, its mesh-$1\mathrm{/}n$ sums converge to its integral by trapping the sums between compact-interval upper and lower sums from [F4]; the two endpoint pieces have lengths at most $1\mathrm{/}n$ and bounded values. For small $x$, $f(x)\le C{x}^p$ with one of the positive exponents $p$ in step 5.1, so $\frac{1}{n}{\sum }_{i\mathrm{/}n<a}f(i\mathrm{/}n)\le C(a^{p+1}+\frac{a^p}{n})\mathrm{.}$ At infinity use the envelope $g(x)=C{x}^m{e}^{-c{x}^{2\kappa }}$, increasing $m,C$ if necessary. Its derivative has the sign of $m-2c\kappa x^{2\kappa }$, so it is eventually decreasing. Consequently its discrete tail is bounded by its integral tail plus one mesh term. First let $n\to \mathrm{\infty }$, then $a\downarrow 0$ and $b\to \mathrm{\infty }$. These two tail bounds prove the five full Riemann limits. Substituting $h_i=h_{0}d(i\mathrm{/}n)$ gives all five scaled sum equivalents, irrespective of how the positive factor $h_0$ varies with $n$.

The two scaling equations have the unique positive solution $n^L=\frac{c_1}{c_2}\frac{{\epsilon }^{2-q}}{r^2},h_0=\frac{r^2{\epsilon }^{-2}}{c_1{n}^{2\beta +1}}\mathrm{.}$ Let $V={\sum }_i{h}_i^2(\cosh (z_i^2)-1)$. Step 6.2 gives $V\sim U^2=c_0{h}_0^{2}n$ whenever $n\to \mathrm{\infty }$. For any feasible relaxed pair $(g_i,w_i)$ of finite cost $W$, expansion of the positive kernel gives $\cosh (zw)-1={\sum }_{k\ge 1}\frac{z^{2k}{w}^{2k}}{(2k)!}\mathrm{.}$ Apply [F5] to the arrays with entries $h_i{z}_i^{2k}\mathrm{/}\sqrt{(2k)!}$ and $g_i{w}_i^{2k}\mathrm{/}\sqrt{(2k)!}$ for finitely many $i,k$, and increase the truncations. It follows that ${\sum }_i{h}_i{g}_i(\cosh (z_i{w}_i)-1)\le \sqrt{VW}\mathrm{.}$ This is a deterministic kernel inequality, with no Gaussian probability space.

Multiply the scalar certificate in step 2.1 at $v=w_i$ by $h_i{g}_i$ and sum up to an index $m$. Step 7.1 bounds the left side by $\sqrt{VW}$, and the partial budget is at most ${\epsilon }^{-q}$. The partial energy is therefore bounded above independently of $m$ by the resulting inequality, so its full sum is finite. Increasing $m$ now yields $\sqrt{VW}\ge \frac{h_0}{2}(n^{-2\beta }{r}^2{\epsilon }^{-2}-n^{-qs}{\epsilon }^{-q})=\frac{h_0^{2}n}{2}(c_1-c_2)=U^2\mathrm{.}$ Thus every finite-cost feasible pair has $W\ge U^4\mathrm{/}V$, and infinite-cost pairs do not lower the infimum. Hence $J_{\epsilon }(r)\ge U^4\mathrm{/}V\sim U^2$.

Fix numbers $0<a<b<\mathrm{\infty }$ and $0<\eta <1$. Write $c_{j,[a,b]}$ for the integral defining $c_j$ restricted to $[a,b]$, for $j=0,1,2$. These constants are positive. Choose $n^{\mathrm{\prime }},h_0^{\mathrm{\prime }}>0$ from $c_{1,[a,b]}{h}_0^{\mathrm{\prime }}(n^{\mathrm{\prime }}{)}^{2\beta +1}=(1+\eta )r^2{\epsilon }^{-2},c_{2,[a,b]}{h}_0^{\mathrm{\prime }}(n^{\mathrm{\prime }}{)}^{qs+1}=(1-\eta ){\epsilon }^{-q}\mathrm{.}$ Division gives $\frac{n^{\mathrm{\prime }}}{n}={\left(\frac{c_{1,[a,b]}\mathrm{/}{c}_{2,[a,b]}}{c_1\mathrm{/}{c}_2}\frac{1-\eta }{1+\eta }\right)}^{1\mathrm{/}L},\frac{h_0^{\mathrm{\prime }}}{h_0}=(1+\eta )\frac{c_1}{c_{1,[a,b]}}{\left(\frac{n}{n^{\mathrm{\prime }}}\right)}^{2\beta +1}\mathrm{.}$ For this fixed window and margin these ratios are fixed positive constants. Hence $n^{\mathrm{\prime }}\to \mathrm{\infty }$ and $h_0^{\mathrm{\prime }}\to 0$ under the claimed hypotheses.

Put $g_i=h_0^{\mathrm{\prime }}d(i\mathrm{/}{n}^{\mathrm{\prime }})$ and $w_i=z(i\mathrm{/}{n}^{\mathrm{\prime }})$ when $a\le i\mathrm{/}{n}^{\mathrm{\prime }}\le b$, and set both to zero otherwise. There are finitely many nonzero coordinates. The compact-window maximum of $d$ is finite, so eventually $0\le g_i\le 1$. Compact Riemann convergence gives actual energy $(1+\eta +o(1))r^2{\epsilon }^{-2}$ and actual budget $(1-\eta +o(1)){\epsilon }^{-q}$, hence eventual feasibility. Its cost divided by $U^2$ tends to $\frac{c_{0,[a,b]}}{c_0}{\left(\frac{h_0^{\mathrm{\prime }}}{h_0}\right)}^2\frac{n^{\mathrm{\prime }}}{n}\mathrm{.}$ As $a\downarrow 0$, $b\to \mathrm{\infty }$, and $\eta \downarrow 0$, this expression tends to one by step 5.1 and the explicit ratios in step 8.2. For any fixed positive tolerance one may first fix a sufficiently wide window and small margin, and then take sufficiently small $\epsilon$. This proves $\mathrm{lim\hspace{0.17em}sup}{J}_{\epsilon }(r)\mathrm{/}{U}^2\le 1$ without choosing an infinite family of windows. With step 8.1 it proves $J_{\epsilon }(r)\sim U^2$. In particular, no untruncated claim that $h_{0}d(i\mathrm{/}n)\le 1$ has been used.

Substituting the value of $n$ from step 7.1 into $U=\sqrt{c_0}{h}_0\sqrt{n}$ gives $U=\frac{\sqrt{c_0}}{c_1}{\left(\frac{c_1}{c_2}\right)}^{-(2\beta +1\mathrm{/}2)\mathrm{/}L}{\epsilon }^{-[2q\alpha +1-q\mathrm{/}2]\mathrm{/}L}{r}^{(2qs+1)\mathrm{/}L}\mathrm{.}$ Since $D=L\mathrm{/}q$, this is precisely the stated power with exponents $A\mathrm{/}D$ and $B\mathrm{/}D$. Similarly $h_0^L=C_2{r}^{2(qs+1)}{\epsilon }^{-(2q\alpha +2-q)}$ for a parameter-only $C_2>0$, so $h_0\to 0$ is equivalent to $r=o({\epsilon }^{a_c})$. Finally $a_c-\frac{2-q}{2}=\frac{qL}{2(qs+1)}>0.$ Thus this condition also forces $n\to \mathrm{\infty }$ by the formula for $n^L$, while $n\to \mathrm{\infty }$ and $h_0\to 0$ plainly include $h_0\to 0$. This proves the asserted equivalence, the restricted power asymptotic, and all remaining claims.