uniform convergence check for $f_n(x)=\cos ^nx (1-\cos ^nx)$ for $x\in[o,\frac{\pi}{2}]$

Question

Question is :

Uniform convergence check for $f_n(x)=\cos ^nx (1-\cos ^nx)$ for $x\in[o,\frac{\pi}{2}]$

my idea is to check if $d_n\rightarrow 0 $ where $d_n =\sup \{ |f_n(x)-f(x)| :x\in[o,\frac{\pi}{2}] \}$

Now, I fix $x =0$ then I would have $\cos x=1$ so $f_n(x)=0$ i.e., $f_n\rightarrow 0$

Now, I fix $x\neq 0$ then I wuld have $|\cos x| <1$ so, $\cos ^n x\rightarrow 0$ i.e., $f_n\rightarrow 0$

So, I could see that this $f_n$ converges pointwise to $0$.

Now, to find supremum I would like to take help of derivative test.

$f_n(x)=\cos ^nx -\cos ^{2n}x)\Rightarrow f_n'(x)=-n\cos^{n-1}x\sin x+2n\cos^{2n-1}x\sin x=0$

i.e., $n(\cos^{n-1}x)(\sin x)(-1+2\cos^nx)=0$

If I choose $\sin x=0$ then $\cos x=1$ then I would end up with $f_n(x)=0$

If I choose $\cos x=0$ then I would end up with $f_n(x)=0$

So, I would not get any supremum in this case.

So, I would wish to take $-1+2\cos^nx=0$ i.e., $\cos^nx=\frac{1}{2}$

So, $f_n(x)=\frac{1}{2}(1-\frac{1}{2})=\frac{1}{4}$

So, supremum would then be $\frac{1}{4}$ i.e.,$d_n=\frac{1}{4}$ for all $n\geq 2$

Which would imply $d_n\rightarrow \frac{1}{4} \neq 0$

SO, I would like to say this is not uniform convergence.

The problem is :

I am sure i was not fair in saying "So, I would wish to take $-1+2\cos^nx=0$".

I am not sure If i can do this.

Please help me to see if there is any thing missing in the argument.

Thank you :)

Answer 1

I am sure i was not fair in saying "So, I would wish to take $−1+2\cos^n x = 0$".

There's one small piece of the puzzle missing, you must argue that there is an $x \in [0, \pi/2]$ such that $\cos^n x = \frac12$. If you have done that, everything is fine. To argue the existence of such an $x$, use the intermediate value theorem for the function $x \mapsto \cos^n x$, which attains the value $1$ for $x = 0$ and the value $0$ for $x = \pi/2$ (if $n > 0$).

Answer 2

Put $g(t) = t(1-t)$, $0\le t\le 1$. This function has a maximum when $t = 1/2$. Hence $g(\cos^n(t))$ has a maximum when $\cos^n(t) = 1/2$. At that value, $g$ will attain the value of $1/4$. These functions cannot converge uniformly to zero.