A bound on the Fourier coefficients of an $\alpha$-Lipschitz function

Question

I am asked to show that if $0 < \alpha < 1$, and if $f \in \Lambda^\alpha(\mathbb{T})$, then we have for $k\neq 0$, $$|\widehat{f}(k)| \leq \pi^\alpha \frac{\|f\|_{\Lambda^1}}{k^\alpha}$$

I applied some properties of inequalities and integrals, but must have gotten a bit carried away because my final bound ended up being far too big as you can see below.

Update: I am getting closer. I just don't know where the factor of $k^\alpha$ is coming from.

It has been advised that this theorem might be useful:

Theorem (Fejér): If $f\in L_p(\mathbb T^d)$, then $\|\sigma_n (f) - f\|_p \to 0$. (Here, $\sigma_n(f) = \frac1{n}\sum\limits_{j=0}^{n-1} D_j$).

Attempt #3:

$$ \begin{align*} |\widehat{f}(k)| &= \left|\frac1{2\pi}\int_{-\pi}^\pi f(t)e^{ikt}\mathrm dt\right|\\ &\leq \frac1{2\pi}\int_{-\pi}^\pi |f(t)|\cdot|e^{ikt}|\mathrm dt\\ &= \frac1{2\pi}\int_{-\pi}^\pi |f(t)|\mathrm dt\\ &= \frac1{2\pi}\int_{-\pi}^\pi |f(t+\pi) + f(t) - f(t+\pi)|\mathrm dt\\ &\leq \frac1{2\pi}\int_{-\pi}^\pi |f(t + \pi)| + |f(t +\pi) - f(t)|\mathrm dt\\ &= \frac{\pi^\alpha}{2\pi}\int_{-\pi}^\pi \frac{|f(t + \pi)|}{\pi^\alpha} + \frac{|f(t +\pi) - f(t)|}{\pi^\alpha}\mathrm dt\\ &\leq \frac{\pi^\alpha}{2\pi}\int_{-\pi}^\pi |f(t + \pi)| + \frac{|f(t +\pi) - f(t)|}{\pi^\alpha}\mathrm dt\\ &\leq \frac{\pi^\alpha}{2\pi}\int_{-\pi}^\pi \|f\|_{\Lambda^{\alpha}}\mathrm dt\\ &= \pi^\alpha \|f\|_{\Lambda^\alpha} \end{align*} $$

Answer 1

For integer $k\ne0$ $$ c_k=\frac{1}{2\pi}\int_{0}^{2\pi}f(t)e^{-ikt}dt= -\frac{1}{2\pi}\int_{0}^{2\pi}f(t)e^{ik(t-\pi/k)}dt= -\frac{1}{2\pi}\int_{0}^{2\pi}f\left(t+\frac\pi k\right)e^{ikt}dt. $$ Taking one half of the sum of these expressions for $c_k$ we have $$ |c_k|=\frac{1}{4\pi}\left|\int_{0}^{2\pi}\left(f(t)-f\left(t+\frac\pi k\right)\right)e^{-ikt}dt\right|\leq\frac{1}{4\pi}\int_{0}^{2\pi}\left|f(t)-f\left(t+\frac\pi k\right)\right|dt\leq $$ $$ \frac{||f||_{\Lambda^{\alpha}}}{4\pi}\int_0^{2 \pi } \left|\frac{\pi }{k}\right|^{\alpha } \, dt= \frac{\pi^\alpha}{2|k|^\alpha}||f||_{\Lambda^{\alpha}}. $$

Answer 2

As I mentioned in my comment, we want to use the smoothness of $f$ to get a decay in $\hat{f}$. To make use of the smoothness of $f$ (which is given as a bound on $\Delta_s f$), consider the Fourier Transform of $\Delta_s f(t)=f(t+s)-f(t)$: $$ \hat{f}(k)(e^{2\pi iks}-1)=\int_{\mathbb{T}}(f(t+s)-f(t))\;e^{-2\pi ikt}\;\mathrm{d}t\tag{1} $$ Therefore, we have $$ \begin{align} |\hat{f}(k)||e^{2\pi iks}-1| &\le\int_{\mathbb{T}}|f(t+s)-f(t)|\;\mathrm{d}t\\ &\le |s|^\alpha\|f\|_{\Lambda(\alpha)}\tag{2} \end{align} $$ Since $|e^{2\pi iks}-1|=|2\sin(\pi ks)|\ge|4ks|$ for $|ks|\le\frac{1}{2}$, we get $$ |\hat{f}(k)|\le \frac{|s|^{\alpha-1}}{|4k|}\|f\|_{\Lambda(\alpha)}\tag{3} $$ If we choose $s=\frac{1}{2k}$ (so that $|ks|\le\frac{1}{2}$), $(3)$ becomes $$ |\hat{f}(k)|\le \frac{1}{2^{\alpha+1}|k|^\alpha}\|f\|_{\Lambda(\alpha)}\tag{4} $$ I think the difference in constant is due to our use of different normalizations of the Fourier Transform. Since this is homework, I will let you convert.

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More Motivation: In a comment, I mentioned that one dervative of smoothness in $f$ yields one power of $k$ in the decay of $\hat{f}$. This is usually accomplished using integration by parts to show that $$ \int_\mathbb{T}f^{\;\prime}(t)\;e^{-2\pi ikt}\;\mathrm{d}t=2\pi ik\int_\mathbb{T}f(t)\;e^{-2\pi ikt}\;\mathrm{d}t\tag{5} $$ Unfortunately, we can't take derivatives, but as we see above $$ \int_\mathbb{T}\Delta_sf(t)\;e^{-2\pi ikt}\;\mathrm{d}t=(e^{2\pi iks}-1)\int_\mathbb{T}f(t)\;e^{-2\pi ikt}\;\mathrm{d}t\tag{6} $$ which can be used the same way.