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I want to check whether the following function is uniformly converges: $f_n(x)=n\cos^nx\sin x$ for $x \in \left[0,\frac{\pi}{2} \right]$.

I proved that the $\lim \limits_{n \to \infty}f_{n}(x)=0$ for every $x$. I'd love your help with the uniformly continues convergence. I always get confused with it. I already showed that $|f_n(x) - 0|< \epsilon$. What else should I show or how should I refute the claim?

Thanks a lot.

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    @D.Thomine: The maximum is at $x=\frac{1}{\sqrt{n+1}}$, and I use that in my counterexample below.2011-12-13

3 Answers 3

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We use the following claim:

Let $a,b$ two real numbers and $\{f_n\}$ a sequence of continuous functions on $\left[a,b\right]$ which converges uniformly to $f$ on $[a,b]$. Then $\lim_{n\to\infty}\int_a^bf_n(t)dt=\int_a^bf(t)dt.$

Indeed, we have $\left|\int_a^bf_n(t)dt-\int_a^bf(t)dt\right|\leq (b-a)\sup_{a\leq x\leq b}|f_n(x)-f(x)|,$ which converges to $0$ thanks to the uniform convergence on $[a,b]$.

In our case, we have $\int_0^{\frac{\pi}2}f_n(x)dx=n\left[-\frac{(\cos x)^{n+1}}{n+1}\right]_0^{\frac \pi 2}=\frac n{n+1}\to 1,$ whereas $f_n$ converges pointwise to $0$. This shows that the convergence cannot be uniform.

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    You mean, the fact that a uniformly convergent sequence of continuous functions has a continuous limit? Well, fix \delta>0, and $n_0$ such that $||f-f_{n_0}||\leq\delta$. Then for a fixed $x_0$, $|f(x)-f(x_0)|\leq |f(x)-f_{n_0}(x)|+|f_{n_0}(x)-f_{n_0}(x_0)|+|f_{n_0}(x_0)-f(x_0)|\leq 2\delta+|f_{n_0}(x)-f_{n_0}(x_0)|$ and you conclude using the continuity of $f_{n_0}$ at $x_0$.2011-12-13
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Let $x_n=\sin^{-1}\left(\frac{1}{\sqrt{n+1}}\right)$. Then $\sin(x_n)=\frac{1}{\sqrt{n+1}}$ and $\cos(x_n)=\frac{1}{\sqrt{1+1/n}}$.

Thus, $ \begin{align} \lim_{n\to\infty}\frac{f_n(x_n)}{\sqrt{n}} &=\lim_{n\to\infty}\frac{n\;\cos^n(x_n)\sin(x_n)}{\sqrt{n}}\\ &=\lim_{n\to\infty}\frac{1}{(1+1/n)^{n/2}}\sqrt{\frac{n}{n+1}}\\ &=e^{-1/2}\tag{1} \end{align} $ Therefore, $ f_n(x_n)\sim e^{-1/2}\sqrt{n}\tag{2} $ The asymptotic growth in $(2)$ says that, although $\lim\limits_{n\to\infty}f_n(x)=0$ pointwise, $f_n(x)$ does not converge uniformly to $0$ on $[0,\frac{\pi}{2}]$ since there is always an $x$ so that $f_n(x)>1$.

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You might like to simplify first (otherwise the argument is that of robjohn):

Put $t=\cos(x)$ ($0\le t\le1$), then we look at $g(t)= n t^n\sqrt{1-t^2}$ or, even better, we might consider $h(t) = g(t)^2= n^2 t^{2n}(1-t^2)$. Then h'(t)= 2n^3t^{2n-1}(1-t^2)-2n^2t^{2n+1}=2n^2t^{2n-1}(n-(n+1)t^2). Note from this that $t_n=\sqrt{\frac{n}{n+1}}$ is the maximum of $h$, and that $h(t_n)=n^2\left(\frac{n}{n+1}\right)^n \left(1-\frac{n}{n+1}\right)= n\cdot\frac{1}{\left(\frac{n}{n+1}\right)^n}\cdot \frac{n}{n+1}\sim n\cdot\frac{1}{e}\cdot 1 \qquad \text{as $n\to\infty$}$ Which should have been $0$ if the limit was uniform.