Pointwise vs uniform

The mode of convergence — or of continuity — was the wrong one.

Would you have fallen for these?

Each is a real option from a previous-year paper. Read it cold and decide before opening it.

• Option D

  g is uniformly continuous on $[0,\infty )$.

  Why this is wrong, and why it looks right

  False, and this is the point of the question. Uniform continuity is not preserved under composition, because $x\mapsto x^2$ is not uniformly continuous on $[0,\infty )$. Concretely, take $x_n=\sqrt{n+1/2}$ and $y_n=\sqrt{n}$: the gap $x_n-y_n\to 0$, yet $g(x_n)-g(y_n)$ stays at 1, because squaring compresses each successive tent into a shorter interval. SymPy makes the mechanism explicit: $(t+1/t{)}^2-t^2=2+1/t^2$, which stays bounded away from zero even as the input gap 1/t → 0.

  From this question

  Let $f:[0,\infty )\to \mathbb{R}$ be the periodic function of period 1 given by f(x) = 1 − |2x − 1| for x ∈ [0, 1], and define $g:[0,\infty )\to \mathbb{R}$ by $g(x)=f(x^2)$. Which of the following statements are true?

  See the full question

  Part C · december 2023 · free

• Option B

  $(f_n)$ converges uniformly on compact subsets of [0, 1) but not on [0, 1).

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  $(f_n)$ converges uniformly on [0, 1) but not on [0, 1].

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

• Option C

  J[y] ≥ J[0] for all $y\in B_0(0,\epsilon )$

  The analysis of this one comes with the PYQ pack.

  Part C · december 2023

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