ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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$x y \sin (1 / (x^{2} + y^{2}))$ is differentiable at the origin with unbounded partial derivatives nearby

Example

Define $f : R^{2} \to R$ by

$f (0, 0) = 0, f (x, y) = x y \sin ⁣ (\frac{1}{x^{2} + y^{2}}) when (x, y) \neq (0, 0) .$

Then $f$ is totally differentiable at the origin with $D f (0, 0) = 0$ . Both partial derivatives exist at every point, but each is unbounded on every neighbourhood of the origin.

Facts & Assumptions

Given: The function $f$ in the Example and $s = x^{2} + y^{2}$ .

[L1]

For every real $t$ , $∣ \sin t ∣ \leq 1$ and $∣ \cos t ∣ \leq 1$ (Parity and the Pythagorean identity for sine and cosine).

[L2]

The functions $\sin$ and $\cos$ are differentiable on $R$ , with $(\sin t)^{'} = \cos t$ and $(\cos t)^{'} = - \sin t$ ; also $\sin 0 = 0$ and $\cos 0 = 1$ (The derivatives of sine and cosine are cosine and minus sine).

[L3]

If real functions $g$ and $h$ are differentiable at the relevant points, then $(g \circ h)^{'} = (g^{'} \circ h) h^{'}$ (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[L4]

Products, sums, scalar multiples, and quotients with nonzero denominator obey their usual derivative rules (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L5]

A map is totally differentiable at the origin with derivative zero when $∣ f (h) ∣ / ∥ h ∥_{2} \to 0$ as $h \to 0$ through nonzero vectors (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder).

[L6]

The coordinate partial derivatives are the derivatives of the two coordinate-line restrictions (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ).

[L7]

For $(x, y) \in R^{2}$ , $∥ (x, y) ∥_{2} = \sqrt{x^{2} + y^{2}}$ (The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ ).

[L8]

Sine and cosine have period $2 π$ (The zero sets of sine and cosine and the least positive common period 2 pi).

[L9]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L10]

Every nonnegative real $a$ has a unique nonnegative square root $a^{1 / 2}$ , positive when $a > 0$ (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ).

[L11]

The number $π$ is positive (Pi as twice the smallest positive zero of cosine).

Verification

technique · direct

1.1algebra

For all real $x, y$ , $2 ∣ x y ∣ \leq x^{2} + y^{2} = s$ , since $(∣ x ∣ - ∣ y ∣)^{2} \geq 0$ .

1.2givenL6

Both coordinate-line restrictions through the origin are identically zero, so $\partial_{x} f (0, 0) = \partial_{y} f (0, 0) = 0$ .

1.3givenL2L3L4algebra

At every point with $s > 0$ , the derivative rules give $\partial_{x} f = y \sin (1 / s) - 2 x^{2} y \cos (1 / s) / s^{2}$ and $\partial_{y} f = x \sin (1 / s) - 2 x y^{2} \cos (1 / s) / s^{2}$ .

1.4L9L10L11algebra

For $k \geq 1$ put $a_{k} = 1 / \sqrt{4 π k}$ . Then $a_{k} > 0$ and $a_{k} \to 0$ : for $ε > 0$ , choose $N \geq 1$ with $1 / N < 4 π ε^{2}$ by [L9]; if $k \geq N$ , then $0 < a_{k}^{2} = 1 / (4 π k) < ε^{2}$ , hence $a_{k} < ε$ .

2.1givenstep 1.1L1L7

If $(x, y) \neq (0, 0)$ , then $∣ f (x, y) ∣ \leq ∣ x y ∣ \leq s / 2 = ∥ (x, y) ∥_{2}^{2} / 2$ .

2.2step 1.3L2L8algebra

Since $1 / (2 a_{k}^{2}) = 2 π k$ , periodicity and the values in [L2] give $\sin (1 / (2 a_{k}^{2})) = 0$ and $\cos (1 / (2 a_{k}^{2})) = 1$ ; step 1.3 therefore gives $\partial_{x} f (a_{k}, a_{k}) = \partial_{y} f (a_{k}, a_{k}) = - 1 / (2 a_{k})$ .

3.1step 2.1L5algebra

Dividing step 2.1 by $∥ (x, y) ∥_{2} > 0$ gives $∣ f (x, y) ∣ / ∥ (x, y) ∥_{2} \leq ∥ (x, y) ∥_{2} / 2 \to 0$ , so $f$ is totally differentiable at the origin with derivative zero.

4.1step 1.4step 2.2L7choose∎

Given any neighbourhood radius $δ > 0$ and any bound $B > 0$ , step 1.4 permits a $k$ with $a_{k} < \min {δ / \sqrt{2}, 1 / (2 B)}$ ; then $∥ (a_{k}, a_{k}) ∥_{2} < δ$ while step 2.2 gives $∣ \partial_{x} f (a_{k}, a_{k}) ∣ = ∣ \partial_{y} f (a_{k}, a_{k}) ∣ > B$ . Thus both partial derivatives are unbounded on every neighbourhood of the origin.

Remarks

Differentiability at one point controls the size of the function's increment there. It does not impose a bound on derivatives at nearby points, and the rapidly oscillating reciprocal phase makes that distinction explicit.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

70 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis II, sections 8.3 and 11.4 (standard reference, not scraped)
University of Toronto MAT237, section 2.1 Differentiation of real-valued functions (standard reference, not scraped)

xysin⁡(1/(x2+y2))xysin(1/(x2+y2)) is differentiable at the origin with unbounded partial derivatives nearby