TheoremBase

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+h∈Ix_0+h\in I satisfies

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

Claim 1. Let (hk)k∈N(h_k)_{k\in\mathbb{N}} be admissible and let ε>0\varepsilon>0. Differentiability supplies δ>0\delta>0 as above. Since (hk)k∈N(h_k)_{k\in\mathbb{N}} has limit 00, there is K∈NK\in\mathbb{N} with ∣hk∣<δ|h_k|<\delta for every k≥Kk\ge K; for such kk we also have hk≠0h_k\ne0 and x0+hk∈Ix_0+h_k\in I, so 0<∣hk∣<δ0<|h_k|<\delta and therefore ∣qk−L∣<ε|q_k-L|<\varepsilon. As ε\varepsilon was arbitrary, (qk)k∈N(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+h∈Ix_0+h\in I, and with

∣f(x0+h)−f(x0)h−L∣≥ε.\left|\frac{f(x_0+h)-f(x_0)}{h}-L\right|\ge\varepsilon .

Let (δk)k∈N(\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+h∈Ix_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)k∈N(h_k)_{k\in\mathbb{N}} with hk∈Skh_k\in S_k for every kk.

This sequence is admissible. Indeed hk≠0h_k\ne0 and x0+hk∈Ix_0+h_k\in I for every kk; and given a real η>0\eta>0, the limit of (δk)k∈N(\delta_k)_{k\in\mathbb{N}} supplies KK with δk<η\delta_k<\eta for every k≥Kk\ge K, whence ∣hk−0∣=∣hk∣<δk<η|h_k-0|=|h_k|<\delta_k<\eta for those kk, so (hk)k∈N(h_k)_{k\in\mathbb{N}} has limit 00.

On the other hand ∣qk−L∣≥ε|q_k-L|\ge\varepsilon for every kk, so (qk)k∈N(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 ∣qk−L∣<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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