Does Pointwise Convergence to a Continuous Function Imply Uniform Convergence?

Recently in my advanced calculus class we discussed how the uniform limit of a sequence of continuous functions is itself continuous.  One of my students turned the question around: If the limit of a pointwise sequence of continuous functions is continuous, does that mean that the convergence is uniform?

The answer is “No,” although I couldn’t think of a counterexample off the top of my head.  I found one (not surprisingly) on Math.SE.  Although there are some good discussion and multiple good answers to this question given there,  I’m going to discuss Ilmari Karonen’s answer.

Let \displaystyle f_n: [0,1] \mapsto \mathbb{R} be given by

\displaystyle f_n(x) = \begin{cases} nx, & 0 \leq x \leq 1/n; \\ 2 - nx, & 1/n < x \leq 2/n; \\ 0, & \text{ otherwise}. \end{cases}

Each f_n is continuous (piecewise, but that’s fine).  For any x \in (0,1], there exists N such that x > 2/n for all n \geq N.  Since f_n(0) = 0 for all n, this means that f_n(x) \to 0 for all x \in [0,1].  The zero function is clearly continuous.

However, for each n \in \mathbb{N} we have that f_n(1/n) = 1.  This means that when \epsilon < 1 there cannot be an N \in \mathbb{N} such that |f_n(x)| < \epsilon for all x \in [0,1] and n \geq N.  Thus the convergence cannot be uniform.

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2 Responses to Does Pointwise Convergence to a Continuous Function Imply Uniform Convergence?

  1. Adhvaitha says:

    An even more simpler example is \(f_n: (0,1) \mapsto [0,1)\), where \(f_n(x)=x^n\). \(f_n(x)\) is clearly continuous and converges to \(0\) on \((0,1)\) (again continuous point-wise), whereas the convergence is not uniform.

    • mzspivey says:

      I think in class we had restricted ourselves to closed and bounded domains for \{f_n\}. But you’re right; for the question I stated in the post that is definitely a simpler example.

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