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Let $X$ be a compact subset of $\mathbb{R}$. Let $f_n:X\to (0,\infty)$ be a sequence of function converging uniformly to a function $f:X\to(0,\infty)$. Show that the sequence of functions $\left(\frac{1}{n}\log(f_n)\right)$ converges uniformly to the identically zero function. Hint: Use the concavity of the logarithm function.

I wrote down the assumptions and conclusions using definition of uniform convergence, but I'm really stuck on them. Also I don't know where to use the concavity of the log.

The thing I know is that $\left(\frac{1}{n}\log(f_n)\right)$ converges pointwise to $0$, because $\frac{1}{n}\to0$ and $\log(f_n(x))\to\log(f(x))$ is bounded. Maybe I should use Dini's theorem, but I must show that each sequence $\left(\frac{1}{n}\log(f_n(x))\right)$ is monotone. Any hint will be appreciated.

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  • $\begingroup$ Continuity of log suffices $\endgroup$ – Bananach May 3 at 5:47
  • $\begingroup$ @Bananach can you give more detail? The only result I've seen about continuity and uniform convergence is that the uniform limit of a sequence of continuous function is continuous. But the converse does not hold ($f_x(x) = x^n(1-x^n)\to0$ over $[0,1]$ but this convergence is not uniform). $\endgroup$ – AnalyticHarmony May 3 at 6:08
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    $\begingroup$ See the answer by Kavi. In words: f is strictly bounded away from 0 and from infinity. Because of uniform convergence $f_n$ is too for large enough $n$. Afterwards, the log term is bounded and you still have $1/n$ (as you see, you don't even need continuity, it's even enough that $\log$ is bounded on any closed interval in $(0,\infty)$) $\endgroup$ – Bananach May 3 at 6:58
  • $\begingroup$ By the way: Your example is false; the convergence is uniform. It's still true that pointwise convergence of continuous functions doesn't imply uniform convergence even on compact sets, imagine a hump sliding to the boundary of a compact interval $\endgroup$ – Bananach May 3 at 7:02
  • $\begingroup$ The convergence is not uniform. For any $n$ the point $x = (1/2)^{1/n}$ is such that $f_n(x) = 1/4$. $\endgroup$ – AnalyticHarmony May 3 at 7:10
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Without any continuity assumptions this is false. Let $X=[0,1]$ and $f_n(x)=f(x)$ and $f(x)=\frac 1 x$ for $x>0, f(0)=1$. Then $\frac 1 n \log(f_n(x))$ does not tend to $0$ uniformly.

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  • $\begingroup$ I'm sorry but I didn't understand why we should have $f_{n_k}(x_k)\to f(x)$. Are you using that if $x_k$ has a subsequence converging to some $x$ and $f_n\to f$ uniformly then any (numerical) subsequence of $(f_n(x_k))$ converges to $f(x)$? $\endgroup$ – AnalyticHarmony May 3 at 7:14
  • $\begingroup$ @AnalyticHarmony I have a new answer. $\endgroup$ – Kavi Rama Murthy May 3 at 7:32

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