# Does $f_{n}(x)=n\cos^n x \sin x$ uniformly converge for $x \in [0,\frac{\pi}{2}]$?

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.

-
The form of $f_n$ can incite us to integrate. If the convergence where uniform on the whole interval, we would have $\int_0^{\frac{\pi}2}f_n(x)dx=0$. Is it the case? –  Davide Giraudo Dec 13 '11 at 10:46
Sorry, I didn't understand you. can you please explain your claim? –  Jozef Dec 13 '11 at 11:02
See the answer below. I give the details. –  Davide Giraudo Dec 13 '11 at 11:03
Alternatively, you could evaluate $f_n$ at $1/n$, and see what happens as $n$ goes to $+ \infty$ (edit : or even better, at $1/\sqrt{n}$). –  D. Thomine Dec 13 '11 at 11:07
@D.Thomine: The maximum is at $x=\frac{1}{\sqrt{n+1}}$, and I use that in my counterexample below. –  robjohn Dec 13 '11 at 11:19

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.

-
Just a nitpick, but in your claim, you're assuming $f$ is integrable (which, of course, it is under the hypotheses of the claim; but doesn't this requires a separate argument?). –  David Mitra Dec 13 '11 at 11:11
In fact we don't need $f$ integrable. I will modify the claim in order to work with continuous functions (but it also requires the argument that an uniform limit of continuous functions is continuous). –  Davide Giraudo Dec 13 '11 at 11:14
Great explanation, Thanks! –  Jozef Dec 13 '11 at 14:27
@DavideGiraudo: How do you prove this claim? –  Jozef Dec 13 '11 at 21:07
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 $x_0$. –  Davide Giraudo Dec 13 '11 at 21:11

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 $f_n(x)>1$.

-

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.

-