Suppose $(f_n)$ is a sequence of functions and $f_n:[0,1) \to [0,1)$ such that $f_n(x) = x^n$. We need to prove that it converges pointwise but not uniformly.

$f_n$ converges to $f$ pointwise, if there exists an $N(\epsilon , x)$, dependent on both $\epsilon > 0$ and $x \in E$, such that $\mid f_n(x) - f(x)\mid < \epsilon$ when $n \ge N(\epsilon , x)$.

So, for all $n \ge N(\epsilon , x)$ $$\mid x^n \mid = x^n \le x^N < \epsilon$$ Take $N > \dfrac{\ln \epsilon}{\ln x}$. This proves that the sequence converges pointwise. Now, my question is since we are able to find $N$ dependent on both $\epsilon$ and $x$, wouldn't this imply that the sequence does not converge uniformly (because in the statement of uniform convergence $N$ is independent of $x$)?


The $N$ that you found is dependent on $x$, yes, but you should prove that you can't find an $N$ that is independent of $x$. Consider that $x^N$ is strictly decreasing in $N$ for $x\in (0, 1)$, so if $x^N = \epsilon$ for $N = \log_x(\epsilon)$, then we have $x^N < \epsilon$ iff $N > \log_x(\epsilon)$. Then, we note that $\sup_{x\in [0, 1)} \log_x(\epsilon) = \infty$, as $\lim_{x\to 1^-} \log_x(\epsilon)$ diverges to infinity for all $\epsilon > 0$. Therefore, there is no finite $N$ that satisfies the uniform convergence condition on $[0, 1)$ for any $\epsilon > 0$. In contrast, for any $0 < \alpha < 1$, we can find an $N$ that satisfies uniform convergence on $[0, \alpha]$ by letting $N > \log_{\alpha}(\epsilon)$.

  • $\begingroup$ Is it enough to note that $$\sup_{x \in [0,1)}\mid x^n \mid$$ is not equal to $0$ as $n \to \infty$? $\endgroup$
    – Parth
    Apr 11 '17 at 19:18
  • 1
    $\begingroup$ Yes, that would be an equivalent formulation. As $$\lim_{n\to \infty} \sup_{x\in [0, 1)} \lvert x^n\rvert\neq 0$$ we do not have convergence of $x^n$ to $0$ in the uniform topology. $\endgroup$
    – Michael L.
    Apr 11 '17 at 19:20

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