Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A bounded function of two real variables whose every coordinate slice is real analytic is continuous

Statement

False claim: every bounded function f:R2R whose slices xf(x,b) and yf(a,y) are real analytic for every fixed real a,b is continuous on R2.

The point of the witness is exactly that the several-variable holomorphic theorem "locally bounded and separately holomorphic implies holomorphic" has no parallel with "real analytic" in place of "holomorphic".

Facts & Assumptions

Given: The witness

f(x,y)={xyx2+y2,(x,y)(0,0),0,(x,y)=(0,0).

[L1]

A locally bounded separately holomorphic function is holomorphic (Locally bounded and separately holomorphic implies holomorphic), and separate holomorphy means one-variable holomorphy on every coordinate slice (Separately holomorphic functions).

[L4]

The Cauchy inequalities bound one-variable holomorphic derivatives by a sup norm on a circle (Cauchy's inequalities bound every derivative by a boundary bound on a compactly contained circle).

Refutation

technique · direct
1.1

The false claim is the real-variable analogue of [L1], so refuting it shows that the several-variable holomorphic theorem is genuinely complex-analytic and not a formal separate-regularity statement.

L1
1.2

The displayed formula defines a real-valued function on all of R2.

given
2.1

The function is bounded: for (x,y)(0,0), the inequality 2xyx2+y2 gives f(x,y)1/2, and at the origin f(0,0)=0, so f(x,y)1/2 everywhere.

step 1.2algebra
2.2

The function is not continuous at (0,0): if it were, then by [L3] applied with ε=1/4 there would be δ>0 such that (x,y)<δ implies f(x,y)<1/4; but for any real t with 0<t<δ/2 one has (t,t)=2t<δ and f(t,t)=1/2, a contradiction.

step 1.2L3
3.1

Fix b0. Then the slice xf(x,b)=bx/(x2+b2) is a quotient of polynomials, and the denominator never vanishes on R, so [L2] makes this slice real analytic on all of R. The same argument applies to the slice yf(a,y) when a0.

step 2.1L2
4.1

If b=0, then f(x,0)=0 for every real x, so the slice xf(x,0) is identically zero and therefore real analytic; by the symmetry f(x,y)=f(y,x), the same is true of the slice yf(0,y). Thus every coordinate slice of f is real analytic.

step 1.2step 3.1L2
5.1

So this bounded function has real-analytic slices in each variable but is not jointly continuous, and the false claim fails. The published B-page item named cex-partial-derivatives-without-continuity uses the same witness for a different failure, but it is homed on an examples page and is therefore named here only in prose, without a wikilink or a dependency edge. The complex case is different precisely because [L4] turns a one-variable bound on a holomorphic slice into a uniform derivative bound, and no real-analytic hypothesis supplies such a bound.

step 4.1step 2.2L4

Depends on

Used by

Dependency tree · two levels

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Sources