Proof of A Linear Perturbation Producing a Sequentially Strict Maximum under a Coercive Bound

Let $\gamma\in\mathbb{R}$ be positive and fix $z\in A$, which is possible because $A$ is nonempty.

Preliminaries. The map $x\mapsto|x|$ is continuous on $H$, since $\bigl||x|-|x'|\bigr|\le|x-x'|$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §reverse-triangle; hence, by Continuity Between Metric Spaces is Equivalent to Sequential Continuity and claims 2 and 3 of Arithmetic of Limits of Real Sequences, for every $c\in\mathbb{R}$ the map $x\mapsto c\,|x|^{2}$ is continuous on $H$.

Put $B=C-\Phi(z)+\tfrac{\kappa}{2}$. From the hypothesis, $\kappa|z|^{2}\le C-\Phi(z)$, and $0\le\kappa|z|^{2}$ by claim 5 of Elementary Arithmetic in an Ordered Field, so $0\le C-\Phi(z)$ and hence $\tfrac{\kappa}{2}\le B$ by claim 3 of Elementary Arithmetic in an Ordered Field; since $0<\tfrac{\kappa}{2}$ by claim 8 of Elementary Order Arithmetic in an Ordered Field, claim 2 of that lemma gives $0<B$. Let $\beta$ be the nonnegative real number with $\beta^{2}=\tfrac{2B}{\kappa}$, given by Existence and Uniqueness of the Nonnegative Square Root. Then $\beta+4$ is positive, and we may choose a positive $\mu\in\mathbb{R}$ with

for instance the least of $\tfrac{\kappa}{2}$ and $\tfrac{\gamma}{2(\beta+4)}$, which is one of them and hence positive, by claim 9 of Elementary Order Arithmetic in an Ordered Field.

The auxiliary function. Let $\Phi_{\mu}:A\to\mathbb{R}$ be given by $\Phi_{\mu}(x)=\Phi(x)+\mu|x|^{2}$. By the preliminaries the map $g:H\to\mathbb{R}$, $g(x)=-\mu|x|^{2}$, is continuous, and $\Phi_{\mu}(x)=\Phi(x)-g(x)$, so $\Phi_{\mu}$ has closed superlevel sets in $H$ by claim 3 of Functions with Closed Superlevel Sets: Sequential Characterisation, Semicontinuity, Perturbation and Limits. Moreover, for $x\in A$,

since $0\le\tfrac{\kappa}{2}\le\kappa-\mu$ and $0\le|x|^{2}$, using claim 5 of Elementary Arithmetic in an Ordered Field. So $\Phi_{\mu}$ is bounded above and $\sup_{x\in A}\Phi_{\mu}(x)$ exists.

A near-maximiser of bounded norm. By claim 3 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ there is $x_{0}\in A$ with

Since $\Phi_{\mu}(z)\le\sup_{x\in A}\Phi_{\mu}(x)$ and $\Phi(z)\le\Phi_{\mu}(z)$, we get $\Phi(z)-\mu\le\Phi_{\mu}(x_{0})$. Combining with the displayed bound $\Phi_{\mu}(x_{0})\le C-(\kappa-\mu)|x_{0}|^{2}\le C-\tfrac{\kappa}{2}|x_{0}|^{2}$ gives

hence $|x_{0}|^{2}\le\tfrac{2B}{\kappa}=\beta^{2}$ and therefore $|x_{0}|\le\beta$, both numbers being nonnegative, by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field and trichotomy.

Applying the perturbed maximum principle. Apply A Perturbed Maximum Principle of Borwein-Preiss Type in a Real Hilbert Space to $H$, $A$, $\Phi_{\mu}$, the positive numbers $\mu$ and $\lambda=1$, and the point $x_{0}$, whose defining inequality is the displayed one above since $\mu\lambda^{2}=\mu$. It yields $\bar{y}\in H$ and $\bar{x}\in A$ such that $|\bar{y}-x_{0}|\le4$ and such that the function $x\mapsto\Phi_{\mu}(x)-\mu|x-\bar{y}|^{2}$ attains a sequentially strict maximum on $A$ at $\bar{x}$.

By The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle, $|\bar{y}|\le|\bar{y}-x_{0}|+|x_{0}|\le\beta+4$. Set $p=-2\mu\,\bar{y}$. Then, by Elementary Identities in a Real Inner Product Space §homogeneity,

Identifying the perturbation. By Elementary Identities in a Real Inner Product Space §expansion, $|x-\bar{y}|^{2}=|x|^{2}-2\langle x,\bar{y}\rangle+|\bar{y}|^{2}$, so for $x\in A$

where the last step uses $\langle p,x\rangle=\langle-2\mu\bar{y},x\rangle=-2\mu\langle\bar{y},x\rangle=-2\mu\langle x,\bar{y}\rangle$, by conditions (a) and (c) of Real Inner Product Space §inner-product.

Thus the function $x\mapsto\Phi(x)-\langle p,x\rangle$ differs from $x\mapsto\Phi_{\mu}(x)-\mu|x-\bar{y}|^{2}$ by the constant $\mu|\bar{y}|^{2}$, and therefore attains a sequentially strict maximum on $A$ at $\bar{x}$ by claim 3 of Elementary Properties of Sequentially Strict Extrema. Since $|p|\le\gamma$, this proves the corollary.