# If $f_n\to f$ uniformly, then $\frac{1}{n}\log(f_n)\to 0$ uniformly

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.

• Continuity of log suffices – Bananach May 3 at 5:47
• @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). – AnalyticHarmony May 3 at 6:08
• 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 $(0,\infty)$) – Bananach May 3 at 6:58
• 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 – Bananach May 3 at 7:02
• The convergence is not uniform. For any $n$ the point $x = (1/2)^{1/n}$ is such that $f_n(x) = 1/4$. – AnalyticHarmony May 3 at 7:10

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.
• 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 $f(x)$? – AnalyticHarmony May 3 at 7:14