A bound on the Fourier coefficients of an $\alpha$-Lipschitz function 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:
$$
\begin{align*}
|\widehat{f}(k)| &= \left|\frac1{2\pi}\int_{-\pi}^\pi f(t)e^{ikt}\mathrm dt\right|\\
&\leq \frac1{2\pi}\int_{-\pi}^\pi |f(t)|\cdot|e^{ikt}|\mathrm dt\\
&= \frac1{2\pi}\int_{-\pi}^\pi |f(t)|\mathrm dt\\
&= \frac1{2\pi}\int_{-\pi}^\pi |f(t+\pi) + f(t) - f(t+\pi)|\mathrm dt\\
&\leq \frac1{2\pi}\int_{-\pi}^\pi |f(t + \pi)| + |f(t +\pi) - f(t)|\mathrm dt\\
&= \frac{\pi^\alpha}{2\pi}\int_{-\pi}^\pi \frac{|f(t + \pi)|}{\pi^\alpha} + \frac{|f(t +\pi) - f(t)|}{\pi^\alpha}\mathrm dt\\
&\leq \frac{\pi^\alpha}{2\pi}\int_{-\pi}^\pi |f(t + \pi)| + \frac{|f(t +\pi) - f(t)|}{\pi^\alpha}\mathrm dt\\
&\leq \frac{\pi^\alpha}{2\pi}\int_{-\pi}^\pi \|f\|_{\Lambda^{\alpha}}\mathrm dt\\
&= \pi^\alpha \|f\|_{\Lambda^\alpha}
\end{align*}
$$
 A: 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.

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.
A: 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
$$
|c_k|=\frac{1}{4\pi}\left|\int_{0}^{2\pi}\left(f(t)-f\left(t+\frac\pi k\right)\right)e^{-ikt}dt\right|\leq\frac{1}{4\pi}\int_{0}^{2\pi}\left|f(t)-f\left(t+\frac\pi k\right)\right|dt\leq
$$
$$
\frac{||f||_{\Lambda^{\alpha}}}{4\pi}\int_0^{2 \pi } \left|\frac{\pi }{k}\right|^{\alpha } \, dt= \frac{\pi^\alpha}{2|k|^\alpha}||f||_{\Lambda^{\alpha}}.
$$
