16.28.31 Variation of continuous weights485B15BBE554

Proof. Nonemptiness follows from theorem 16.28.27. Since the test functions are continuous, $\mathcal J_ a(K)$ is closed and convex, and the equality defining active pairs is a closed affine condition. For an active pair, (16.28.17), applied with the weight $q$, gives

The pairs in $\mathcal J_ a(K)$ with bounded constants form a compact set, as in the proof of theorem 16.28.29; hence the set of active pairs is compact.

With $q$ fixed, lemma 16.28.18 and (16.28.17), applied with the weight $q$, bound $|F_{\eta ,q}(a)-F_{\delta ,q}(a)|$ by $|\eta -\delta |(V_ K^*(a)+\operatorname {osc}_ Kq/\epsilon )$. Comparison of obstacles and translation by constants bound $|F_{\eta ,p}(a)-F_{\eta ,q}(a)|$ by $\| p-q\| _\infty $. Adding these bounds gives (16.28.34).

For the sequence in the statement, the constants $c_ j$ have a common bound, by the estimate above, because the weights $q_ j$ are uniformly bounded and the parameters $\delta _ j$ are bounded below by a positive constant. Weak compactness of probability measures gives a convergent subsequence of pairs, and its limit belongs to $\mathcal J_ a(K)$, which is closed. The estimate (16.28.34) gives $F_{\delta _ j,q_ j}(a)\to F_{\delta ,q}(a)$, while uniform convergence of the $q_ j$ and weak convergence of the $\mu _ j$ give $\int _ Kq_ j\, \mathrm{d}\mu _ j\to \int _ Kq\, \mathrm{d}\mu $. Passing to the limit in the equality defining active pairs proves the last assertion. $\square$

Proof. By compactness of the set of active pairs, the minimum on the right is attained. For a fixed pair $(\mu ,c)$ active at $(\delta ,q)$, the representation (16.28.28) at the perturbed parameters gives

Choosing an active pair that attains the minimum on the right, we see that the upper limit of the difference quotients is at most this minimum. Conversely, choose a pair $(\mu _ h,c_ h)$ active at $(\delta +hs,q+hr)$. The representation at $(\delta ,q)$ gives

The estimate (16.28.34) bounds the difference quotients. Along a sequence of values of $h$ realizing their lower limit, lemma 16.28.33 gives a subsequence of pairs converging to a pair active at $(\delta ,q)$. For this pair, $\int _ Kr\, \mathrm{d}\mu +sc$ is at least the minimum on the right. This proves (16.28.36).

For convergent directions, the same two inequalities hold. The fixed active pair attaining the minimum gives an upper bound with the right limit. The active pairs at the perturbed parameters still have uniformly bounded constants, and their subsequential limits are active at $(\delta ,q)$. Uniform convergence of the $r_ j$ controls the integrals $\int _ Kr_ j\, \mathrm{d}\mu _{h_ j}$, and $s_ j\to s$ controls the terms $s_ jc_{h_ j}$. The lower bound therefore has the same limit. Note that all perturbed growth allowances $\delta +h_ js_ j$ are positive for small $h_ j$. $\square$

Proof. Apply (16.28.36) with $r=0$ and $s=\pm 1$. The direction $s=-1$ gives minus the left derivative, which proves (16.28.38). Concavity gives

Hence the open intervals $(D^+_{\delta ,q}(a),D^-_{\delta ,q}(a))$, which are nonempty exactly at the parameters where differentiability fails, are pairwise disjoint. Choosing a rational number in each of them proves countability. $\square$

Proof. With $s=0$, (16.28.36) shows that the right derivative in the direction $r$ is the minimum of $\int _ Kr\, \mathrm{d}\mu $ over active pairs, and the direction $-r$ shows that the left derivative is the maximum. If all active pairs have the same probability measure, these values agree and define a continuous linear derivative. Conversely, if the map is Gâteaux differentiable, the minimum and the maximum agree for every continuous $r$. Continuous functions separate Borel probability measures on $K$, so all active pairs have the same probability measure; their constant is then determined by (16.28.32), since $\delta >0$. $\square$

Proof. For $t\geq 0$, each competitor is at most $-t$ on the unit circle, so the maximum principle for subharmonic functions bounds it by $-t$ on $D$. This bound persists after regularization on $D$, and the constant $-t$ is a competitor. Hence $F_{\delta ,q_ t}(0)=-t$.

For $t=-b$ with $0\leq b\leq \delta /2$, define

This function is continuous and subharmonic: its Laplacian has density $4b$ inside the disk, density zero outside, and a nonnegative contribution on the circle coming from the jump $\delta -2b\geq 0$ of the radial derivative. It belongs to $\mathcal L_\delta $ and equals $q_{-b}$ on $K$. Hence $F_{\delta ,q_{-b}}(0)\geq u_ b(0)=0$. On $D$ the raw envelope is at most the continuous function $b|z|^2$, so its regularization at zero is at most zero. This proves (16.28.41). In particular, the right and left derivatives in $t$ at $t=0$ are $-1$ and $0$, and corollary 16.28.39, applied in the direction $r=-|z|^2$, gives the final assertion. $\square$