How to prove that $\sup|\int_{(0,1)}\ \cos(tx)dt| < \frac{1}{1+n} $?

Question

Given the function: $$ f(x)=\frac{\sin(x)}x,\qquad\qquad x\in\mathbb R^*$$ and $f(0)=1$.

I have shown that$$f(x) = \int_{(0,1)}\ \cos(tx)dt $$ and that $$f∈ C^{\infty}(\mathbb R).$$

The next question is to prove that $$\sup|f^{(n)}(x)| \leq \frac{1}{1+n}.$$ Here is my try: let $g(t,x) = \cos(tx)$, we know that since $g∈ C^{\infty}([0,1]\times \mathbb R)$, $f∈ C^{\infty}(R)$ and $$f^{(n)}(x) = \int_{(0,1)}\ \frac{\partial^n g(t,x)}{\partial x^n }dt,$$ I found that: for n odd $$f^{(n)}(x)= -t^n\sin(tx) ,$$ so $|t^n\sin(tx)|<t^n<1$ since $0<t<1$. And for n even $$f^{(n)}= t^n\cos(tx).$$ But I have no idea on how to continue from here. Any help please?

Answer 1

For all $t\in\mathbb R$, you have $\displaystyle\left|\frac{\partial^n}{\partial x^n}[\cos(tx)]\right|\leqq|t|^n$ since \begin{align} \left|\frac{\partial^n}{\partial x^n}[\cos(tx)]\right| &= \begin{cases} |t|^n|\sin(tx)|&\mathrm{if}\quad n : \mathrm{odd}\\ |t|^n|\cos(tx)|&\mathrm{if}\quad n : \mathrm{even} \end{cases} &\leqq|t|^n. \end{align}

Thus, for all $x\in\mathbb R$, you have \begin{align} |f^{(n)}(x)| &=\left|\int_0^1\frac{\partial^n}{\partial x^n}[\cos(tx)]\ dt\right|\\ &\leqq\int_0^1\left|\frac{\partial^n}{\partial x^n}[\cos(tx)]\right|dt\\ &\leqq\int_0^1 t^n dt \\ &=\frac{1}{1+n}. \end{align} Therefore $$\sup_{x\in\mathbb R}|f^{(n)}(x)|\leqq\frac{1}{1+n}.$$

Answer 2

The expressions you found for the derivatives are missing the integrals. What you have is $$ f^{(n)}(x)=\int_0^1t^ng_n(tx)\,dt, $$ where $$g_n(t)=\begin{cases}\pm \sin t,&\ n\ \text{even,}\\ \pm\cos t,&\ n\ \text{odd}\end{cases} $$ So $$ |f^{(n)}(x)|≤\int_0^1t^{n}\,dt=\frac1{n+1}. $$