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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 whose slices and are real analytic for every fixed real is continuous on .
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
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).
Sums, products, quotients by a nowhere-zero function, and compositions preserve real analyticity on open real domains (Real-analytic functions are closed under sums, products and compositions, and under quotients where the denominator is nonzero, A real-analytic function on an open subset of is locally represented by a convergent real power series).
Continuity at a point is the - condition in metric spaces (Continuity of a map between metric spaces, at a point and globally, in the - form, Vector-valued functions , their limits and continuity, with the dictionary to the metric notions).
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
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.
The displayed formula defines a real-valued function on all of .
The function is bounded: for , the inequality gives , and at the origin , so everywhere.
The function is not continuous at : if it were, then by [L3] applied with there would be such that implies ; but for any real with one has and , a contradiction.
Fix . Then the slice is a quotient of polynomials, and the denominator never vanishes on , so [L2] makes this slice real analytic on all of . The same argument applies to the slice when .
If , then for every real , so the slice is identically zero and therefore real analytic; by the symmetry , the same is true of the slice . Thus every coordinate slice of is real analytic.
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.
Depends on
- A real-analytic function on an open subset of $\mathbb{R}$ is locally represented by a convergent real power series
- Real-analytic functions are closed under sums, products and compositions, and under quotients where the denominator is nonzero
- Locally bounded and separately holomorphic implies holomorphic
- Separately holomorphic functions
- Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$-$\delta$ form
- Vector-valued functions $f : A \to \mathbb{R}^m$, their limits and continuity, with the dictionary to the metric notions
- Cauchy's inequalities bound every derivative by a boundary bound on a compactly contained circle
Used by
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Sources
- J. Lebl, Tasty Bits of Several Complex Variables, v4.4, Ex. 1.1.5 (standard reference, not scraped)