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 B

  $f_{n} \to 0$ uniformly on [0,1]

  Why this is wrong, and why it looks right

  False: $f'_{n}(x)=n(1-x)^{n}$⁻$^{1}(1-(n+1)x)$ vanishes at x=1/(n+1), where $f_{n}$ equals $[n/(n+1)]^{n}$⁺$^{1}=1/(1+1/n)^{n}$⁺$^{1}\to1/e\ne0$. The peak height doesn't shrink even though every fixed point does.

  From this question

  Let $f_{n}(x) =$ nx$(1 - x)^{n}$ on [0, 1]. Which of the following are true?

  Part C · free

• Option B

  {$f_{n}$} converges uniformly on [0, 1].

  Why this is wrong, and why it looks right

  $sup_{x\in[0,1]}$ |$x^{n} - f(x)$| = 1 for every n, approached as x→1⁻, so the sup never shrinks to 0 no matter how large n is. Convergence here is only pointwise, not uniform.

  From this question

  For n ≥ 1 let $f_{n} : [0, 1] \to \mathbb{R}$ be $f_{n}(x) = x^{n}$. Which of the following are true?

  Part C · free

• Option D

  The pointwise limit is continuous on [0, 1].

  Why this is wrong, and why it looks right

  The pointwise limit jumps from 0 on [0,1) to 1 at x=1, so it is discontinuous at x=1. This is the standard illustration that a pointwise limit of continuous functions need not be continuous — that guarantee needs uniform convergence.

  From this question

  For n ≥ 1 let $f_{n} : [0, 1] \to \mathbb{R}$ be $f_{n}(x) = x^{n}$. Which of the following are true?

  Part C · free

• Option B

  $\int{}f_{n} \to 0$

  Why this is wrong, and why it looks right

  $\int_{0}^{1}f_{n} = n \cdot ($length of (0,1/n)) = n · (1/n) = 1 for every n, so $\int{}f_{n} \to 1$, not 0. The spike gets narrower but taller in exactly the way that conserves its total mass.

  From this question

  Let $f_{n}(x) = n\cdot1_{(0, 1/n)}(x)$ on [0,1]. Which of the following are true?

  Part C · free

• 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 is uniformly continuous on $\mathbb{R}$.

  Why this is wrong, and why it looks right

  Take $x_{n}=n, y_{n}=n+1/n$: |$x_{n}-y_{n}$|=1/n→0, but |$f(x_{n})-f(y_{n})$| = |$n^{2}-(n+1/n)^{2}$| $= 2+1/n^{2} \to 2$. No single $\delta$ works for all of $\mathbb{R}$ at $\varepsilon=1$ — continuity at each point doesn't give a domain-wide modulus.

  From this question

  Let $f(x) = x^{2}$ on $\mathbb{R}$ and $g(x) = \sqrt{x}$ on $[0, \infty)$. Which of the following are true?

  Part C · free

• Option A

  $f_{n} \to 0$ pointwise but not uniformly on $\mathbb{R}$.

  Why this is wrong, and why it looks right

  This is the guess you'd make by analogy with sequences like nx$(1 - x)^{n}$, which really do converge pointwise but not uniformly on [0, 1]. Here, though, the supremum can be computed directly: $f_{n}'(x) = (1 -$ nx$^{2})/(1 +$ nx$^{2})^{2}$, which vanishes at $x = \pm1/\sqrt{n}$, where |$f_{n}(1/\sqrt{n})$| $= (1/\sqrt{n})/(1 + 1) = 1/(2\sqrt{n})$. Since this is the global maximum of |$f_{n}$| on $\mathbb{R} (f_{n} \to 0$ at x = 0 and as $x \to \pm\infty$ too), sup_x |$f_{n}(x) - 0$| $= 1/(2\sqrt{n}) \to 0$ — which is exactly uniform convergence. The trap is reasoning from a family resemblance to a different sequence instead of computing this one's actual sup.

  From this question

  Let $f_{n}(x) = x/(1 +$ nx$^{2})$ on $\mathbb{R}$. Which statement is correct?

  Part B · free

• Option B

  J[y] ≤ J[0] for every $y \in B_{1}(0, \varepsilon)$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  J[y] ≥ J[0] for every $y \in B_{0}(0, \varepsilon)$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  $f_{n}(x) = \sin($nx) is equicontinuous on [0, 1].

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option A

  $(g_{n})_{n}\ge_{1}$ converges uniformly on any compact subset of $\mathbb{R}$ but not on $\mathbb{R}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  There exists a subsequence of $(h_{n})_{n}\ge_{1}$ that converges uniformly on $\mathbb{R}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  For every compact subset $K \subseteq \mathbb{R}$, there exists a subsequence of $(h_{n})_{n}\ge_{1}$ that converges uniformly on K.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  The series $\sum(n\ge1) f_{n}$ converges pointwise at all points of 𝔻 but not uniformly on some compact subsets of 𝔻.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option A

  $(f_{n})_{n}\ge_{1}$ converges uniformly to 0 on $\mathbb{R}$, and $(f'_{n})_{n}\ge_{1}$ converges uniformly to 0 on the interval (−M, M) for some positive real number M.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option A

  $(f_{n})_{n}\ge_{1}$ does not converge pointwise on [0, 1].

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  $(f_{n})_{n}\ge_{1}$ converges pointwise to a discontinuous function on [0, 1].

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option A

  The series is neither absolutely convergent nor uniformly convergent.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option B

  The series is both absolutely convergent and uniformly convergent.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  The series is uniformly convergent but not absolutely convergent.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  $\sum \sin($nx)/n on $\mathbb{R}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option A

  uniformly to 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  nowhere

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  The sequence $(p_{n}(x))$ converges to f(x) uniformly on [0, 1)

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option B

  $\sum(n\ge1) f_{n}$ is uniformly convergent on the interval [−1, 1], but $\sum(n\ge1) g_{n}$ is not.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  $\sum(n\ge1) g_{n}$ is uniformly convergent on the interval [−1, 1], but $\sum(n\ge1) f_{n}$ is not.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  Neither $\sum(n\ge1) f_{n}$ nor $\sum(n\ge1) g_{n}$ is uniformly convergent on the interval [−1, 1].

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  J[y] ≥ J[0] for all $y \in B_{0}(0, \varepsilon)$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option B

  $(f_{n})$ converges uniformly on compact subsets of [0, 1) but not on [0, 1).

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option C

  $(f_{n})$ converges uniformly on [0, 1) but not on [0, 1].

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option B

  If each $f_{n}$ is bounded and $(f_{n})_{n}\ge_{1}$ converges to f pointwise, then f is bounded.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option A

  $(f_{n})_{n}\ge_{1}$ converges uniformly on [0, 1] to a continuous function f.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option D

  $(f_{n})_{n}\ge_{1}$ does not converge pointwise on [0, 1].

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option D

  F is not uniformly continuous on $\mathbb{R}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

The counterexamples that settle these

One example is enough to kill a false statement. These are the standard ones for the claims above.

• Continuous on a bounded interval ⇒ bounded
• Uniformly continuous ⇒ Lipschitz
• Pointwise limit of continuous functions is continuous
• Bounded and continuous on ℝ ⇒ uniformly continuous
• Product of uniformly continuous functions is uniformly continuous
• fₙ → f uniformly ⇒ fₙ′ → f′

Other mistakes that cost marks