Lower bounds on (truncated) Fourier transform of functions of constant modulus and bounded derivative Let $f(x)=e^{i\phi(x)}$ define a function from $[0,1]$ to the complex unit circle through the real smooth function $\phi(x)$. Also, this function is periodic: $\phi(0)=\phi(1)=0\text{ mod }2\pi$ and has bounded derivatives for $x\in[0,1]$: $$\vert d\phi(x)/dx\vert\le \omega.$$ Consider the (truncated) Fourier transform of $f(x)$: $$\hat{f}(k)=\int_0^1 f(x) e^{ikx}dx.$$
What is a lower bound on $E(K) := \Vert f\Vert_{L^2,[0,K]}$? More explicitly what is a lower bound for $$E(K)= \int_0^K \vert \hat{f}(k)\vert^2 dk$$
A similar question was asked and answered here where there are no  restrictions on $f$ resulting on a lower bound of zero for the value of $\vert \hat{f}(k)\vert$. The case where $\phi(x)$ is a piecewise constant function is answered here where the problem is equivalent to bounding polynomials (even with non-integer exponents) and is related to the following physics paper. From physical considerations, I expect the lower bound to depend on $\omega$.
I suspect that: 


*

*the lower bound scales proportional to $1/(\omega)^2$ for large enough $\omega$ but I could be way off. 

*a lower bound might best be described by a different quantity other than the derivative that I have provided: another sort of complexity will have to go into $\phi(x)$.

*This could be related to some uncertainty principle as we want to minimize the support of $\hat{f}$. Lemma 3, in Terry Tao's blog entry on Hardy's Uncertainty Principle seems to be related except that it works only for even order derivatives and the connection is vague anyways.

*I am missing a trivial point. Maybe constant modulus, periodic, etc is unnecessary and something can be said directly using $\sup\vert f(x)\vert$ and  $\sup\vert f'(x)\vert$? That would somewhat annul my next question though.

*My physics background automated me to discretize the problem (i.e. piece-wise linear $\phi(x)$) but I was not successful although the approach seemed promising. See 1 also.


How could I then find the functions $\phi(x)$ that saturate the lower bound? 
 A: It is an interesting problem which is related to some recent work of mine. The reason for why I started to work on similar problems  is because  connections to a problem of Ramachandra on Dirichlet polynomials, connections to the nordic school of Hardy classes of Dirichlet series (Hedenmalm, Saksman, Seip, Olsen, Olofsson, Lindqvist and others), as well as universality questions for zeta-functions and their properties on the line Re(s)=1.
While my papers are not quite finished, I have put two early preprints on my homepage,  On a problem of Ramachandra and approximation of functions by Dirichlet polynomials with bounded coefficients and  On generalized Hardy classes of Dirichlet series. I have talked about some of these problems at analytic number theory conferences in India. Like in your paper I have considered Dirichlet series (it should be possible to obtain something like Theorem 2.1 in your paper by my method also, although I have not stated a direct analogue in my paper).
Now your problem in the question is rather easy for small $\omega$ so we will from now on assume that $\omega>1/2$.  In fact if $\omega<1/2$, then $|f(0)|>1/2$ and $\int_0^K |\hat f(t)^2|dt \geq \min(1/10,K/10)$ (constants not chosen in an optimal way)
In my papers on Dirichlet series I have used a somewhat different method than you use in your paper, namely the Jensen inequality on the logarithmic integral in a half-plane. This method is applicable for the problem at hand.  Lemma 7 in my paper ``On generalized Hardy classes of Dirichlet series'' can be used with  $\sigma=0$ and $L(it)=\hat f(-t)$   and we obtain
$$\frac D \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {D^2+t^2} dt \leq     \frac D \pi \int_{-\infty}^\infty \frac {\log^+  |\hat f (t)|} {D^2+t^2} dt -         \log  |\hat f(iD)|.
$$ 
For similar results see also Koosis - The logarithmic integral. (Remark Feb 16: The above inequality is an equality if the function is non-zero on a half plane. The inequality follows from Jensen's formula on a disc by mapping the half plane on the disc by the standard holomorphic bijection where $iD$ goes to $0$)
The reason why we can do this is that with the definition of the fourier-transform in your question it means that $ \hat f(z)$ will be a bounded analytic function in the half plane Im$(z) \geq 0$. 
Now in this case we also have that $\log^+  |\hat f (t)|=0$ since $ |\hat f (t)| \leq 1$. Thus the inequality simplifies to
$$\frac D \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {D^2+t^2} dt \leq   -         \log  |\hat f(iD)|.$$ 
It is not too difficult to see that for $\omega>1/2$
$$
 |\hat f(i\omega)|= \left|\int_0^1 e^{i \phi(x)-\omega x} dx \right|>\frac {1} {10 \omega}.
$$ 
(The constant $10$ not chosen optimally). Thus we can choose $D=\omega$ and it is clear that
$$
 \int_0^K \log^- |\hat f(t)| dt < \frac \pi {\omega} \left({\omega^2+K^2} \right) \frac {\omega} \pi \int_{-\infty}^\infty \frac {\log^- |\hat f(t)|} {\omega^2+t^2} dt
$$
From these estimates we see that
$$
  \frac 1 K \int_0^K  \log^- |\hat f(t)| dt< \frac {\pi(\omega^2+K^2)}{\omega K} \log (10 \omega).
$$
Now we can use the Jensen inequality
$$
  \exp\left(\frac 1 K \int_0^K  \log |\hat f(t)| dt\right)< \sqrt{\frac 1 K \int_0^K  |\hat f(t)|^2 dt}
 $$
We get the lower bound
$$
 K \left(\frac 1 {10 \omega} \right)^{2\pi (\omega^2+K^2)/(K \omega)} \leq  \int_0^K  |\hat f(t)|^2 dt
$$
for $\omega>1/2$.  If $c>2 \pi$ and $\omega/K$  is sufficiently  large this gives a lower bound
$$\omega^{-c \omega/K} \leq \int_0^K |\hat f(t)|^2 dt$$
which is  weaker than your expected $e^{-c \omega/K}$. At least we have an explicit lower bound. 
Updated Feb 16: In the case where both $\omega$ and $K$ are large but still $\omega>K$ this can be improved by the following trick. Let $g$ be the convolution of $\hat f$
 with a non negative test-function $\Phi(t/K)$, such that $\hat \Phi(0)>0$ where  $\Phi$ has support on $[0,1/2]$ . Then use Jensen's inequalities on the function $g$ instead of $\hat f$ as above. The advantage with this is that it then follows that $|\hat g(iw)| \gg K/\omega$ and thus we can get the lower bound (by using Jensen's inequality w.r.t the L^1-norm instead of the L^2-norm.)
$$(\omega/K)^{-c \omega/K} \leq \frac 1 K \int_0^{K/2} |g(t)| dt$$
for some constant $c>0$. Since
$$ g(t)=\int_0^t \Phi((t-x)/K) \hat f(x) dx$$ it is clear by the triangle inequality that
$$\frac 1 K \int_0^{K/2} |g(t)| dt = \frac 1 K \int_0^{K/2} \left|\int_0^t \Phi((t-x)/K)\hat f(x) \right| dx \leq $$ 
 $$\leq \frac 1 K \int_0^{K/2} |f(x)| dx  \int_0^{K/2}  |\Phi(x/K)| dx  \leq c \int_0^{K/2} |\hat f(x)| dx $$
The inequality
$$K^{-1} (\omega/K)^{-c \omega/K} \leq \int_0^{K/2} |\hat f(t)|^2 dt$$
follows by the Cauchy-Schwarz inequality for some constant $c>0$.
This formula just use involves dimensionless quantity $\omega/K$ as expected. Since the function $E(K)$ is increasing in $K$ it gives the lower bound $E(K) > C_0 K^{-1}>0$ for $1 \leq \omega \leq K$ for some absolute constant $C_0$.
