Let $\{f_n\}_{nā„0}$ be a sequence of function defined by:
$f_n : [0,1] \rightarrow \mathbb{R}, t \mapsto \Bigg(\sin\Big(\dfrac{t \pi}{2}\Big)\Bigg)^n\Bigg(\cos\Big(\dfrac{\pi t}{2}\Big)\Bigg)$
I am asked to find $ \|f_n\|_p$ with $p \in \{1, 2, \infty\}$.
I didn't know how to go about and I need help.
Hint: You should probably apply different techniques for $p \in \{1,2\}$ and $p=\infty$.
$p=\infty$ asks you to find the maximum of a derivable function on $[0,1]$, so the standard method (derivate, find annulation point, ...) works.
$p=1$ and $p=2$ is an integral computation. An idea could be linearizing using trigonometric formulas.
For $p=2$, $\cos^2+\sin^2=1$ should be interesting. See also Wallis' integrals.
For $p=1,2$ we have: \begin{eqnarray} \|f_n\|_1&=&\int_0^1|f_n(t)|\,dt=\int_0^1\left|\sin^n\left(\dfrac{\pi t}{2}\right)\cos\left(\dfrac{\pi t}{2}\right)\right|\,dt=\dfrac2\pi\int_0^{\pi/2}|\sin^nx\cos x|\,dx\\ &=&\dfrac2\pi\int_0^{\pi/2}\sin^nx\cos x\,dx=\dfrac{2}{(n+1)\pi}\sin^{n+1}x\Big|_0^{\pi/2}=\dfrac{2}{(n+1)\pi}.\\ \|f_n\|_2^2&=&\int_0^1|f_n(t)|^2\,dt=\int_0^1\left|\sin^n\left(\dfrac{\pi t}{2}\right)\cos\left(\dfrac{\pi t}{2}\right)\right|^2\,dt=\dfrac2\pi\int_0^{\pi/2}|\sin^nx\cos x|^2\,dx\\ &=&\dfrac2\pi\int_0^{\pi/2}\sin^{2n}x\cos^2 x\,dx=\dfrac{2}{\pi}\int_0^{\pi/2}(\sin^{2n}x-\sin^{2n+2}x)=\dfrac{2}{\pi}(A_{2n}-A_{2n+2}) \end{eqnarray} with \begin{eqnarray} A_n&=&\int_0^{\pi/2}\sin^nx\,dx=-\int_0^{\pi/2}\sin^{n-1}x(\cos x)'\,dx=(n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x\,dx\\ &=&(n-1)(A_{n-2}-A_n) \end{eqnarray} Hence, $$ A_n=\dfrac{n-1}{n}A_{n-2} $$ and it follows that $$ A_{2n}=\dfrac{(2n-1)(2n-3)\ldots1}{(2n)\ldots(2\cdot1)}A_0=\dfrac{(2n)!\pi}{2^{2n+1}(n!)^2} $$ We deduce that $$ \|f_n\|_2^2=\dfrac{(2n)!}{2^{2n}(n!)^2}-\dfrac{(2n+2)!}{2^{2n+2}[(n+1)!]^2}=\dfrac{(2n)!}{2^{2n}(n!)^2}\left[1-\dfrac{(2n+2)(2n+1)}{2^2(n+1)^2}\right]=\dfrac{(2n)!}{2^{2n}(n!)^2(2n+2)} $$ Notice $$ f_n(t)=\sin^n\left(\dfrac{\pi t}{2}\right)\cos\left(\dfrac{\pi t}{2}\right)=g_n(x)=x^n\sqrt{1-x^2}, $$ with $$ x=\sin\left(\dfrac{\pi t}{2}\right) \in [0,1] $$ Obviously $g_n(0)=g_n(1)=0$, and $g_n(x)\ge 0$, therefore $g_n$ achieves its maximum at some $x_n \in (0,1)$. Since $g_n$ is differentiable for $x\in (0,1)$ and $$ g_n'(x)=nx^{n-1}\sqrt{1-x^2}-\dfrac{x^{n+1}}{\sqrt{1-x^2}}=\dfrac{[n(1-x^2)-x^2]x^{n-1}}{\sqrt{1-x^2}}=\dfrac{[n-(n+1)x^2]x^{n-1}}{\sqrt{1-x^2}} $$ we deduce that for $n\ge 2$ the equation $g_n'(x)=0$ admits $x=0$ and $x_n=\sqrt{\dfrac{n}{n+1}}$ as solutions. Since $g_n'(x)<0$ for $x\in (0,x_n)$,a dn $g_n'(x)>0$ for $x\in (x_n,1)$, it is clear that $g_n$ has a maximum at $x_n$. Thus $$ \|f_n\|_\infty=\max_{x\in [0,1]}g_n(x)=g_n(x_n)=\dfrac{1}{\sqrt{n+1}}\left(\dfrac{n}{n+1}\right)^{n/2} $$
Hence $$ \|f_n\|_p= \begin{cases} \dfrac{2}{(n+1)\pi}& \text{ if } p=1\\ \dfrac{1}{2^nn!}\sqrt{\dfrac{(2n)!}{2n+2}} & \text{ if } p=2\\ \dfrac{1}{\sqrt{n+1}}\left(\dfrac{n}{n+1}\right)^{n/2} & \text{ if } p=\infty \end{cases} $$