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Part CCSIR NET December 2023

Partial derivatives, differentiability, chain rule

For real numbers a, b, c, d, e, f, consider the function F:R2→R2F : \mathbb{R}^{2} \to \mathbb{R}^{2} given by F(x, y) = (ax + by + c, dx + ey + f). Which of the following statements are true?

  1. A.F is continuous.✓ Correct
  2. B.F is uniformly continuous.✓ Correct
  3. C.F is differentiable.✓ Correct
  4. D.F has partial derivatives of all orders.✓ Correct

The idea, in plain words

This is just a straight-line map: it stretches, rotates and shifts, and does nothing else. Every smoothness property you could ask for holds automatically, which is why all four statements are true at once.

Solution

An affine map is Lipschitz (hence uniformly continuous), differentiable with constant derivative, and all higher partials are zero.

Why each option is right or wrong

  • A.Correct

    True.

    Each component is a polynomial of degree one in (x, y), and polynomials are continuous.

  • B.Correct

    True.

    An affine map is Lipschitz: |F(p) − F(q)| ≤ ‖A‖·|p − q| where A is the matrix of linear coefficients.

    A Lipschitz map is uniformly continuous, with δ=ε/\delta = \varepsilon/‖A‖ working everywhere at once.

    Note this holds on all of R2\mathbb{R}^{2} — no compactness needed

    which is why the usual continuous-but-not-uniformly-continuous worry does not arise.

  • C.Correct

    True.

    F is differentiable with DF(x, y) = A, the constant matrix [[a, b], [d, e]].

    The remainder in the definition of the derivative is identically zero

    since F is already affine.

  • D.Correct

    True.

    The first partials are the constants a, b, d, e, and every partial of order two or higher is zero.

    Constants are infinitely differentiable

    so F is C∞C^\infty.

    All four options are correct — worth noticing

    since a Part C question with no false statement is unusual and can make you doubt correct work.

The answer key is official; the reasoning above could not be confirmed by symbolic computation, so it has had no second check.

Related counterexample: If all partial derivatives exist at a point then f is continuous there

More on this topic

The chapter behind this: Differentiability in several variables — free to read

From Functions of Several Variables › Partial derivatives, differentiability, chain rule

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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