KindiakMath

Strengthening Point-wise Convergence

Problem 1. Let \{f_n\} be a sequence of differentiable functions on [0, 1] with the following properties:

  • f_n \to f pointwise on [0, 1],
  • f is continuous on [0, 1],
  • there exists M > 0 such that \displaystyle \sup \{\|f_n'\|_\infty : n \in \mathbb N\} \leq M.

Prove that f_n \to f uniformly.

(Click for Solution)

Solution. The key insight here is that f being continuous on [0, 1] strengthens it to being uniformly continuous, so that we can gain some global control over our estimates.

Fix \epsilon > 0. Since f is (uniformly) continuous on [a, b], for any k_1 > 0, there exists \delta > 0 such that for any u,v \in [0, 1],

|u-v| < \delta \quad \Rightarrow \quad |f(u) - f(v)| < k_1 \cdot \epsilon.

By the Archimedean property, find N_1 \in \mathbb N so that 1/N_1 < \delta. Partition [0, 1] into \{x_0, x_1,\dots,x_{N_1}\}, x_i = i/N_1. For each i, point-wise convergence yields that for any k_2 > 0, there exists N_2 \in \mathbb N such that whenever n > N_2,

n > N_2 \quad \Rightarrow \quad |f_n(x_i) - f(x_i)| < k_2 \cdot \epsilon,\quad i=0,1,2,\dots,N_1.

Finally, fix x \in [0, 1] and suppose x\in [x_0, x_1] without loss of generality. Using the uniform bound on the derivatives, the mean value theorem yields

\displaystyle |f_n(x) - f_n(x_0)| \leq M \cdot |x - x_0| \leq M \cdot \frac 1{N_1}.

Then for any n > N_2, combining the estimates yields

\displaystyle \begin{aligned} |f_n(x) - f(x)| &\leq |f_n(x)-f_n(x_0)| + |f_n(x_0) - f(x_0)| + |f(x_0) - f(x)|\\ &< M \cdot \frac 1{N_1}+ k_2 \cdot \epsilon + k_1 \cdot \epsilon \\ &=M \cdot \frac 1{N_1} + (k_1 + k_2) \cdot \epsilon. \end{aligned}

Taking care of the dependencies, set k_1 = k_2 = 1/3 and further stipulate 1/N_1 < \epsilon/(3M) to yield the desired result.

—Joel Kindiak, 31 Jan 25, 1221H

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