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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:

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

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for $h>1$, the Lipschitz quotient is less than twice the $L^\infty$ bound. –  robjohn Oct 1 '11 at 2:34
    
This is certainly true but I don't think it helps. But I think by the mean value theorem any high value of the Lipschitz quotient should also occur for small $h$ values (or as a limit as $h$ becomes small). This would be helpful because it would allow me to replace $||f||_{\infty}$ with $||f||_{\Lambda^{1}}$ in an inequality chain. –  Kyle Schlitt Oct 1 '11 at 4:23
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I'm afraid but what you've done is quite a stretch from your goal. Note that what you want to prove implies that $|\hat{f}(k)| \sim \frac{1}{k^\alpha} \to 0$ as $k \to \infty$, while with what you've done you only get that it is bounded. I guess that you have Fejér's theorem at hand (I know of no other method) and that you have seen how to do similar estimates for $C^k$-functions, for example, but I wanted to make sure before starting an answer involving facts you don't know. –  t.b. Oct 4 '11 at 5:07
    
We have one theorem available in my notes which is credited to Fejer which I just edited into the question. As for performing estimates on $C^k$ functions, I'm not familiar with it. –  Kyle Schlitt Oct 4 '11 at 10:19
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Attempt #3 does not show any decay. Each derivative of smoothness of a function usually translates to a power of $\xi$ decay in the fourier transform (and vice versa). You are trying to show that this works for the fractional Lipschitz smoothness as well. –  robjohn Oct 4 '11 at 15:28

2 Answers 2

up vote 8 down vote accepted
+50

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

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Wow. I would never have thought to do that. Is this a common technique? I'm trying to see what should have suggested this. I suppose that the $(\frac{\pi}{k})^{\alpha}$ should have tipped me off that I would need to somehow obtain $|f(x) - f(y)|$ in my expression, where $|x - y| = \frac{\pi}{k}$. I will be honest in my homework and make it clear that someone showed me this trick. Thank you. –  Kyle Schlitt Oct 4 '11 at 14:14
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@Kyle Yes, it's a common technique in the theory of trigonometric series for estimating moduli of continuity. See, for example, in Zygmund, Trigonometric Series, vol.1, ch.2, $\S4$, ebookee.org/Trigonometric-Series-Vol-1_212310.html –  Andrew Oct 4 '11 at 14:49

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.

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I confess I don't follow what you are doing in the first half of your answer. I wouldn't worry too much though, as the question has been answered already. As for the second part, yes we definitely proved this in class. –  Kyle Schlitt Oct 4 '11 at 17:31
    
@Kyle: From the sound of your comments to Andrew, you are still somewhat uncertain about what is going on. I would like to help make you more certain. Equation $(1)$ is simply equation $(6)$. That transfers the smoothness of $f$ to the decay of $\hat{f}$. Equation $(2)$ simply quantifies the Lipschitz condition. As you mentioned, we are interested in the Lipschitz condition for small shifts, and $|e^{2\pi iks}-1|\sim 2\pi ks$ for small $ks$. Therefore, we divide the left of $(2)$ by $|e^{2\pi iks}-1|$ and the right by $4ks$ to get $(3)$. –  robjohn Oct 4 '11 at 18:51
    
@Kyle: Equation $(3)$ is true for any $s\le\frac{1}{2k}$, but the smaller we make $s$, the worse the estimate is ($\alpha-1<0$), so we choose as large an $s$ as we can, i.e. $\frac{1}{2k}$. That gives us $(4)$. –  robjohn Oct 4 '11 at 18:52
    
To be totally honest, I do understand the steps in Andrew's answer. I just don't understand the approach behind choosing them yet. I will have to spend some time thinking about it still. I may have to read your explanation more carefully before I follow the steps precisely. Thanks for your help. –  Kyle Schlitt Oct 4 '11 at 20:41
    
@ robjohn: I dont know why I didn't understand this earlier. I just read it now and it is very clear. Thanks for this alternative approach. (Spending Thanksgiving weekend with family is clearly a good brain refresher. :) Hope yours is going well. Cheers!) –  Kyle Schlitt Oct 9 '11 at 17:59

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