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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:

1. the lower bound scales proportional to $1/(\omega)^2$ for large enough $\omega$ but I could be way off.
2. 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 $\phi(x)$.
3. 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.
4. 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.
5. 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?

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:

1. the lower bound scales proportional to $1/(\omega)^2$ for large enough $\omega$ but I could be way off.
2. 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 $\phi(x)$.
3. 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.
4. 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.
5. 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?

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 and is related to the following physics paper. From physical considerations, I expect the lower bound to depend on $\omega$.

I suspect that:

1. the lower bound scales proportional to $1/(\omega)^2$ for large enough $\omega$ but I could be way off.
2. 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 $\phi(x)$.
3. 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.
4. 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.

How could I then find the functions $\phi(x)$ that saturate the lower bound?

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:

1. the lower bound scales proportional to $1/(\omega)^2$ for large enough $\omega$ but I could be way off.
2. 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 $\phi(x)$.
3. 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.
4. 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.
5. 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?

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 without restrictions on $f$ the lower bound would be zero on the value of $\vert \hat{f}(k)\vert$. I suspect that:

1. the lower bound scales proportional to $1/(\omega)^2$ for large enough $\omega$ but I could be way off.
2. 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 $\phi(x)$.
3. 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.
4. 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.

How could I then find the functions $\phi(x)$ that saturate the lower bound?

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 and is related to the following physics paper. From physical considerations, I expect the lower bound to depend on $\omega$.

I suspect that:

1. the lower bound scales proportional to $1/(\omega)^2$ for large enough $\omega$ but I could be way off.
2. 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 $\phi(x)$.
3. 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.
4. 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.

How could I then find the functions $\phi(x)$ that saturate the lower bound?