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Proof of Penalised Suprema: Monotonicity, Near-Maximisers, and Vanishing Penalty along a Doubling Sequence

lemmalem:penalised-supremum-limit-2026a
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· 2,463 chars · 4 deps · depth 13 Reason: Initial publication of the proof: existence of the penalised suprema, monotonicity, the near-maximiser bound, and convergence to zero of the increments along a doubling sequence via a telescoping series.

The suprema exist because the penalty is nonnegative; monotonicity is pointwise; the near-maximiser bound follows by comparing with the supremum at half the parameter; and the vanishing of the increments follows from convergence of a telescoping series of nonnegative terms with bounded partial sums.

Proof

Each result cited is universally quantified over the data in its own statement. Let KRK\in\mathbb{R} satisfy ψ(z)K\psi(z)\le K for every zZz\in Z.

Claim 1. Let βR\beta\in\mathbb{R} be positive. The set {Ψβ(z):zZ}\{\Psi_{\beta}(z):z\in Z\} is nonempty because ZZ is nonempty. For zZz\in Z one has 0βD(z)0\le\beta D(z) by claim 5 of Elementary Arithmetic in an Ordered Field, hence

Ψβ(z)=ψ(z)βD(z)ψ(z)K,\Psi_{\beta}(z)=\psi(z)-\beta D(z)\le\psi(z)\le K ,

so the set is bounded above and its supremum M(β)M(\beta) is a real number by the least upper bound property. Since D(z0)=0D(z_{0})=0 we have βD(z0)=0\beta D(z_{0})=0 and Ψβ(z0)=ψ(z0)\Psi_{\beta}(z_{0})=\psi(z_{0}), and a supremum is an upper bound of the set, so ψ(z0)M(β)\psi(z_{0})\le M(\beta).

Claim 2. Let β,β\beta,\beta' be positive with ββ\beta\le\beta' and let zZz\in Z. Then 0(ββ)D(z)0\le(\beta'-\beta)D(z) by claim 5 of Elementary Arithmetic in an Ordered Field, so

Ψβ(z)=Ψβ(z)(ββ)D(z)Ψβ(z)M(β).\Psi_{\beta'}(z)=\Psi_{\beta}(z)-(\beta'-\beta)D(z)\le\Psi_{\beta}(z)\le M(\beta).

Thus M(β)M(\beta) is an upper bound of {Ψβ(z):zZ}\{\Psi_{\beta'}(z):z\in Z\}, and since M(β)M(\beta') is the least such bound, M(β)M(β)M(\beta')\le M(\beta).

Claim 3. Let β,η\beta,\eta be positive and let zZz\in Z satisfy M(β)ηΨβ(z)M(\beta)-\eta\le\Psi_{\beta}(z). Since βD(z)β2D(z)=β2D(z)\beta D(z)-\tfrac{\beta}{2}D(z)=\tfrac{\beta}{2}D(z),

Ψβ/2(z)=ψ(z)β2D(z)=Ψβ(z)+β2D(z)  M(β)η+β2D(z).\Psi_{\beta/2}(z)=\psi(z)-\tfrac{\beta}{2}D(z)=\Psi_{\beta}(z)+\tfrac{\beta}{2}D(z)\ \ge\ M(\beta)-\eta+\tfrac{\beta}{2}D(z).

As β2\tfrac{\beta}{2} is positive, claim 1 applies to it and Ψβ/2(z)M(β2)\Psi_{\beta/2}(z)\le M\bigl(\tfrac{\beta}{2}\bigr). Combining the two displays and rearranging gives

β2D(z)M(β2)M(β)+η.\tfrac{\beta}{2}D(z)\le M\bigl(\tfrac{\beta}{2}\bigr)-M(\beta)+\eta .

Claim 4. For kNk\in\mathbb{N} put bk=M(βk)b_{k}=M(\beta_{k}) and ak=bkbk+1a_{k}=b_{k}-b_{k+1}. Each βk\beta_{k} is positive and βkβk+1\beta_{k}\le\beta_{k+1}, because 2k2k+12^{k}\le2^{k+1} and β0\beta_{0} is positive, so 0ak0\le a_{k} for every kk by claim 2. By Elementary Properties of Series of Real Numbers §telescoping the partial sums of the series k=1ak\sum_{k=1}^{\infty}a_{k} are

sn=b1bn+1for every nN,s_{n}=b_{1}-b_{n+1}\qquad\text{for every }n\in\mathbb{N},

and by claim 1, ψ(z0)bn+1\psi(z_{0})\le b_{n+1}, whence snb1ψ(z0)s_{n}\le b_{1}-\psi(z_{0}) for every nn. The set of partial sums is therefore bounded above, so the series converges by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series §criterion, and consequently the sequence (ak)kN(a_{k})_{k\in\mathbb{N}} converges to 00 by Elementary Properties of Series of Real Numbers §terms-vanish.

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