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Consider the sequence of functions $\{f_n \}$ on $\Bbb R$ defined by $$f_n(x) = n \log \left (1 + \frac {x^2} {n} \right ),\ \ x \in \Bbb R$$ for all $n \geq 1.$ Show that the sequence of functions $\{f_n \}$ is not uniformly convergent on $\Bbb R.$

My attempt $:$ First I observe that the sequence of functions $\{f_n \}$ converges pointwise to the everywhere continuous function $f$ defined by $$f(x) = x^2,\ \ x \in \Bbb R.$$

If the sequence of functions $\{f_n \}$ converges uniformly to the continuous limit function $f$ on whole of $\Bbb R$ then in particular the sequence of functions $\{f_n \}$ converges uniformly to the continuous limit function $f$ on $[0,1].$

But then we have $$\lim\limits_{n \rightarrow \infty} \int_{0}^{1} f_n(x)\ dx = \int_{0}^{1} f(x)\ dx.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

Now what I found is that $$\int_{0}^{1} f_n(x)\ dx = n \log \left (1 + \frac 1 n \right ) - 2n + 2n^{\frac 3 2} \arctan \left (\frac {1} {\sqrt n} \right ).$$ So we have $$\lim\limits_{n \rightarrow \infty} \int_{0}^{1} f_n(x)\ dx = - \infty.$$ where as $$\int_{0}^{1} f(x)\ dx = \frac 1 3.$$ which is a contradiction to $(1).$ Hence our assumption is false. So the sequence of functions $\{f_n \}$ is not uniformly convergent on $\Bbb R,$ as required.

EDIT $:$ I found the wrong limit. By L'Hospital the limit $$\lim\limits_{n \rightarrow \infty} n \left ( \log \left (1 + \frac 1 n \right ) - 2 \right ) = 1.$$

Now what will be $$\lim\limits_{n \rightarrow \infty} 2n^{\frac 3 2} \arctan \left (\frac {1} {\sqrt n} \right )\ ?$$

By L'Hospital I got infinity. Isn't it the case? But then the overall limit becomes $+\infty.$

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  • $\begingroup$ Oh! God. Online calculator shows that the limit is $\frac 1 3.$ $\endgroup$ Jan 16, 2020 at 15:12
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    $\begingroup$ Clearly, your calculations must be wrong because the integrand is positive everywhere, so the integral couldn't be negative. $\endgroup$ Jan 16, 2020 at 15:21
  • $\begingroup$ @Fimpellizieri I think the limit is $+\infty.$ Am I still doing any mistake? $\endgroup$ Jan 16, 2020 at 15:28
  • $\begingroup$ As it turns out, $f_n$ does converge uniformly to its limit on $[0,1]$. So, your line of reasoning is bound to fail. $\endgroup$ Jan 16, 2020 at 15:34

2 Answers 2

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Here's an approach: note that $$ f_{n}(x) - f(x) = n \log \left (1 + \frac {x^2} {n} \right ) - \log(e^{x^2}) = \log \left( \frac{(1 + x^2/n)^n}{e^{x^2}} \right). $$ Now, we note that for any fixed $n$, we have $$ \begin{align} \lim_{x \to \infty} \frac{(1 + x^2/n)^n}{e^{x^2}} &= \left(\lim_{x \to \infty} \frac{1 + x^2/n}{e^{x^2/n}}\right)^n = 0 \end{align} $$ (since the numerator is a polynomial) from which we may conclude that (again, for any fixed $n$) the set $\{|f_n(x) - f(x)|: x \in \Bbb R\}$ is unbounded. Thus, if we fix (for instance) $\epsilon = 1$, we can see that for any $n$ there exists an $x$ such that $|f_n(x) - f(x)| \geq \epsilon$, which is to say that $\|f_n - f\|_{\sup} \geq \epsilon$.

So, the sequence is not uniformly convergent.


Regarding your limit: with the substitution $m = \sqrt{1/n}$, we have $$ \begin{align} \lim_{m \to 0^+} 2m^{-3} \arctan \left (m \right ) &= 2\lim_{m \to 0^+}\frac{\arctan(m)}{m^3} = 2\lim_{m \to 0^+}\frac{\frac{1}{m^2 + 1}}{3m^2} = \frac 23 \lim_{m \to 0^+} \frac{1}{m^2(m^2 + 1)} = +\infty \end{align} $$

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  • $\begingroup$ Very nice! +1 ${}$ $\endgroup$ Jan 16, 2020 at 15:38
  • $\begingroup$ @F Thanks! And good catch. $\endgroup$ Jan 16, 2020 at 15:38
  • $\begingroup$ Can you help me in the computation of the above limit? If the given sequence of functions was uniformly convergent on $[0,1]$ then the limit should be equal to $\frac 1 3.$ $\endgroup$ Jan 16, 2020 at 15:45
  • $\begingroup$ See my latest edit. We could do the second limit, for instance, with two iterations of L'Hospital $\endgroup$ Jan 16, 2020 at 15:50
  • $\begingroup$ I misunderstood, I thought you meant the limit from my question. I'm not particularly interested in ironing that limit of integrals out. Regardless of uniform convergence, the limit has to be $\frac 13$ by the monotone convergence theorem. $\endgroup$ Jan 16, 2020 at 15:51
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@mathmaniac.: (too big for a comment) using Taylor series, you rewrite $$n\log\Big(1+\frac{1}{n}\Big)-2n+2n\sqrt n\arctan\Big(\frac{1}{\sqrt{n}}\Big)$$ as $$n\Big(\frac{1}{n}+o\Big(\frac{1}{n}\Big)\Big)-2n+2n\sqrt{n}\Big(\frac{1}{\sqrt n}-\frac{1}{3(\sqrt n)^3}+o\Big(\frac{1}{(\sqrt n)^3}\Big)\Big)=1-\frac{2}{3}+o(1)\underset{n\to\infty}{\longrightarrow}\frac{1}{3},$$

so perhaps your mistake was to only use degree $1$ for the development of $\arctan$.

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    $\begingroup$ Great answer. Thanks @Ballon. +1 $\endgroup$ Jan 16, 2020 at 16:45
  • $\begingroup$ @mathmaniac.: You're welcome! $\endgroup$
    – Balloon
    Jan 16, 2020 at 16:46

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