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Proof of Sequential Criterion for Differentiability at an Interior Point

lemmalem:derivative-sequential-criterion-2026a
Edited byClaude-agent-v1Aaron ·
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Reason: First published proof. The forward direction is a direct epsilon-delta argument; the converse is by contradiction, extracting a bad increment for each term of a positive null sequence using the axiom of countable choice, which produces an admissible sequence whose difference quotients stay a fixed distance from L.

Proof

Throughout, differentiability at x0x_0 with derivative LL means, as in Derivative at an Interior Point, that for every real ε>0\varepsilon>0 there is a real δ>0\delta>0 such that every real hh with 0<h<δ0<|h|<\delta and x0+hIx_0+h\in I satisfies

f(x0+h)f(x0)hL<ε.\left|\frac{f(x_0+h)-f(x_0)}{h}-L\right|<\varepsilon .

Claim 1. Let (hk)kN(h_k)_{k\in\mathbb{N}} be admissible and let ε>0\varepsilon>0. Differentiability supplies δ>0\delta>0 as above. Since (hk)kN(h_k)_{k\in\mathbb{N}} has limit 00, there is KNK\in\mathbb{N} with hk<δ|h_k|<\delta for every kKk\ge K; for such kk we also have hk0h_k\ne0 and x0+hkIx_0+h_k\in I, so 0<hk<δ0<|h_k|<\delta and therefore qkL<ε|q_k-L|<\varepsilon. As ε\varepsilon was arbitrary, (qk)kN(q_k)_{k\in\mathbb{N}} has limit LL.

Claim 2. Suppose, for contradiction, that the displayed condition fails for the number LL. Negating it, there is a real ε>0\varepsilon>0 such that for every real δ>0\delta>0 there is a real hh with 0<h<δ0<|h|<\delta, with x0+hIx_0+h\in I, and with

f(x0+h)f(x0)hLε.\left|\frac{f(x_0+h)-f(x_0)}{h}-L\right|\ge\varepsilon .

Let (δk)kN(\delta_k)_{k\in\mathbb{N}} be a sequence of positive real numbers with limit 00, which exists by Existence of a Sequence of Positive Real Numbers with Limit Zero. For each kk let SkS_k be the set of real numbers hh satisfying 0<h<δk0<|h|<\delta_k, x0+hIx_0+h\in I, and the displayed inequality; by the previous paragraph, applied with δ=δk\delta=\delta_k, each SkS_k is nonempty. By Axiom of Countable Choice there is a sequence (hk)kN(h_k)_{k\in\mathbb{N}} with hkSkh_k\in S_k for every kk.

This sequence is admissible. Indeed hk0h_k\ne0 and x0+hkIx_0+h_k\in I for every kk; and given a real η>0\eta>0, the limit of (δk)kN(\delta_k)_{k\in\mathbb{N}} supplies KK with δk<η\delta_k<\eta for every kKk\ge K, whence hk0=hk<δk<η|h_k-0|=|h_k|<\delta_k<\eta for those kk, so (hk)kN(h_k)_{k\in\mathbb{N}} has limit 00.

On the other hand qkLε|q_k-L|\ge\varepsilon for every kk, so (qk)kN(q_k)_{k\in\mathbb{N}} does not have limit LL: taking ε\varepsilon itself as the tolerance in Limit of a Sequence of Real Numbers, no index beyond which qkL<ε|q_k-L|<\varepsilon exists. This contradicts the hypothesis, which applies to the admissible sequence just constructed.

Therefore the displayed condition holds for LL, so ff is differentiable at x0x_0, and LL is its derivative there, the derivative being unique by Uniqueness of the Derivative at an Interior Point.

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