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edited May 24 2011 at 18:31 |
Let's create a proof a la Koosis. All the techniques used below can be found in his book "The Logarithmic Integral".
Take $a>1$ and put $f(z)=a\prod_j(1+z/z_j)^{p_j}$. That is a nice analytic function and, while its absolute value is somewhat hard to understand, its argument is very simple: it is an $n$-piece piecewise linear function on the circle with slope $\frac 12 n$ (with respect to the usual circle length) and jumps at $z_j$. Let $I$ be the image of one of the arcs between two adjacent points $z_j$ and $z_{j+1}$ under the mapping $z\mapsto \operatorname{arg}f(z)$. We can transplant all functions defined on the circle arc $[z_j,z_{j+1}]$ to $I$ using this mapping. Note that the integral of any function over the arc with respect to the circle length is just $2/n$ times the integral of its transplant over $I$ with respect to the line length.
Let $\Phi$ be the transplant of $|f|$. Assume that $\Phi<2$ on $I$. The transplant of $f$ is then just $F(t)=\Phi(t)e^{it}$. The key observation is the following:
$$
\int_I \log|2-F(t)|dt\ge \log 2(|I|-\pi).
$$
Assuming that it is true, we conclude that the full integral of $\log|2-f|$ over the unit circle is at least $2/n$ times the sum of the right hand sides over the intervals corresponding to all arcs, which is $0$. On the other hand, if $a>1$, then $\log|2-af(0)|=\log|2-a|\log|2-f(0)|=\log|2-a|<0$, so $2-af$ 2-f$ must have a root inside the disk and the maximum principle finishes the story.
Now let us prove the observation claim. The only thing we really know about $\Phi$ is that it is log-concave and, thereby, unimodal. Fortunately, that's all we need. So, in what follows, $\Phi$ will be just any unimodal function on $I$ with values in $[0,2]$. Since we can always extend $\Phi$ by $0$ outside $I$, we can switch to any larger interval we want without making the inequality easier. So, WLOG, $I=[-2\pi n-\frac\pi 2,2\pi n+\frac\pi 2]$
Now, let us observe that for every fixed $t$, the integrand is minimized for $\Phi(t)=2\max(0,\cos t)$ (that is just the nearest point on the line) and that the farther we go away from this optimal value, the larger the integrand is. Therefore, to minimize the left hand side, we need to stay as close to the black regime on the picture (the graph of $2\cos_+ t$) as we can.
Suppose that the actual $\Phi$ is given by the blue line. Then, replacing $\Phi$ by the red line $\Psi$, we come closer to the optimum at every point. But the red line consists of several full periods (horizontal pieces) and several pieces that together constitute one full positive arc of $\cos t$. Now, each full period means running over some circle around the origin, so the average value of $\log|2-\Psi(t)e^{it}|$ over each full period is exactly $\log 2$. At last, the $2\cos t$ part gives $\int_0^\pi\log (2|\sin t|)dt=0$, which is exactly the loss of $\pi \log 2$ compared to $\log 2$ times its length $\pi$.
That's it. Feel free to comment and/or ask questions.
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edited May 24 2011 at 17:01 |
Let's try to do things create a proof a la Koosis(we'll fail but we'll learn a couple of tricks on . All the way)techniques used below can be found in his book "The Logarithmic Integral". Put Take $f(z)=-\prod_j(1-z/z_j)^{p_j}$. a>1$ and put $f(z)=a\prod_j(1+z/z_j)^{p_j}$. That is a nice analytic function and, while its absolute value is somewhat hard to understand, its argument is very simple: it is an $n$-piece piecewise linear function on $[0,2\pi]$ the circle with slope $\frac 12 n$ (with respect to the usual circle length) and jumps at arguments of $z_j$. Now, consider Let $g(z)=1+\frac 12f(z)$. If there are no zeroes inside I$ be the circle (otherwise we are done by image of one of the maximum principle), we must have arcs between two adjacent points $\int_0^{2\pi}\log|g(e^{it}|dt=-2\pi\log 2$, z_j$ and $z_{j+1}$ under the same as for mapping $f(z)=z-1$. Now observe that if we know z\mapsto \operatorname{arg}f(z)$. We can transplant all functions defined on the argument circle arc $\sigma$ of [z_j,z_{j+1}]$ to $f$ alone, we already know I$ using this mapping. Note that $\log|g|\ge \log|\frac{e^{2i\sigma}-1}2|$, which is always the equality for $z-1$. Also, integral of any function over the smaller arc with respect to the argument circle length is ($\mod 2\pi$), the less just $2/n$ times the bound is. So, it will be nice integral of its transplant over $I$ with respect to compare the distribution functionsline length. Let $z-1$ gives just \Phi$ be the uniform distribution transplant of argument $|f|$. Assume that $\Phi<2$ on $[-\pi/2,\pi/2]$ with density I$. The transplant of $2$. If the run for each piece f$ is at most then just $2\pi$, F(t)=\Phi(t)e^{it}$. The key observation is the following:\int_I \log|2-F(t)|dt\ge \log 2(|I|-\pi).Assuming that it is true, we can say conclude that the preimage of the interval full integral of length $L$ centered at $0$ \log|2-f|$ over the unit circle is at most least $2L$ ($n$ 2/n$ times the sum of the right hand sides over the intervals corresponding to all arcs, which is $2L/n$ on each due to 0$. On the slope). Soother hand, if the angles between adjacent points are all less than $4\pi/n$, we are in good shape. This takes care of a>1$, then $n\le 5$ because \log|2-af(0)|=\log|2-a|<0$, so $(2\sin\frac\pi 5)^5>2$. Otherwise we still can say that 2-af$ must have a root inside the preimage is of length at most $4L$ doubling disk and the possible value of maximum principle finishes the integral (which story. Now let us prove the observation claim. The only thing we really know about $\Phi$ is crude that it is log-concave andbad, but youthereby, probablyunimodal. Fortunately, want that's all or nothingwe need. So, so I'll not try in what follows, $\Phi$ will be just any unimodal function on $I$ with values in $[0,2]$. Since we can always extend $\Phi$ by $0$ outside $I$, we can switch to squeeze epsilons)any larger interval we want without making the inequality easier. Note that this estimate So, WLOG, $-4\pi\log 2$ works I=[-2\pi n-\frac\pi 2,2\pi n+\frac\pi 2]$ Now, let us observe that for every fixed $1+af$ just as well as t$, the integrand is minimized for $f$ \Phi(t)=2\max(0,\cos t)$ (that is just the nearest point on the line) and that the farther we used go away from this optimal value, the argumentof $f$ only)larger the integrand is. ThusTherefore, if $a=\frac 34$to minimize the left hand side, we get a contradiction, so need to stay as close to the black regime on the picture (the graph of $4/3$ 2\cos_+ t$) as we can. Suppose that the actual $\Phi$ is always yoursgiven by the blue line. At lastThen, if all jumps are replacing $\pi$ (\Phi$ by the integer case; I'm aware red line $\Psi$, we come closer to the optimum at every point. But the red line consists of Vieta, several full periods (horizontal pieces) and several pieces that together constitute one full positive arc of course, but still)$\cos t$. Now, we have a perfect linear run if we care just about lines instead of rays when talking about argumenteach full period means running over some circle around the origin, so the distribution on lines average value of $\log|2-\Psi(t)e^{it}|$ over each full period is just perfect but some rays may be wrong. Sorry for the brevityexactly $\log 2$. I'll have to run now butAt last, since you said you couldn't even do the $n=3$, I concluded that 2\cos t$ part gives $\int_0^\pi\log (2|\sin t|)dt=0$, which is exactly the above technique may be unknown to you and I didn't want loss of $\pi \log 2$ compared to keep you waiting without a good reason :). Uphhh... The hard part done. Now $\log 2$ times its length $\pi$. That's itjust remains . Feel free to rewrite the proofcomment and/or ask questions.Give me a couple of hours :).
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edited May 24 2011 at 16:18 |
Let's try to do things a la Koosis (we'll fail but we'll learn a couple of tricks on the way).
Put $f(z)=-\prod_j(1-z/z_j)^{p_j}$. That is a nice analytic function and, while its absolute value is somewhat hard to understand, its argument is very simple: it is an $n$-piece piecewise linear function on $[0,2\pi]$ with slope $\frac 12 n$ and jumps at arguments of $z_j$. Now, consider $g(z)=1+\frac 12f(z)$. If there are no zeroes inside the circle (otherwise we are done by the maximum principle), we must have $\int_0^{2\pi}\log|g(e^{it}|dt=-2\pi\log 2$, the same as for $f(z)=z-1$. Now observe that if we know the argument $\sigma$ of $f$ alone, we already know that $\log|g|\ge \log|\frac{e^{2i\sigma}-1}2|$, which is always the equality for $z-1$. Also, the smaller the argument is ($\mod 2\pi$), the less the bound is. So, it will be nice to compare the distribution functions. $z-1$ gives just the uniform distribution of argument on $[-\pi/2,\pi/2]$ with density $2$. If the run for each piece is at most $2\pi$, we can say that the preimage of the interval of length $L$ centered at $0$ is at most $2L$ ($n$ intervals, $2L/n$ on each due to the slope). So, if the angles between adjacent points are all less than $4\pi/n$, we are in good shape. This takes care of $n\le 5$ because $(2\sin\frac\pi 5)^5>2$. Otherwise we still can say that the preimage is of length at most $4L$ doubling the possible value of the integral (which is crude and bad, but you, probably, want all or nothing, so I'll not try to squeeze epsilons). Note that this estimate $-4\pi\log 2$ works for $1+af$ just as well as for $f$ (we used the argumentof $f$ only). Thus, if $a=\frac 34$, we get a contradiction, so $4/3$ is always yours. At last, if all jumps are $\pi$ (the integer case; I'm aware of Vieta, of course, but still), we have a perfect linear run if we care just about lines instead of rays when talking about argument, so the distribution on lines is just perfect but some rays may be wrong.
Sorry for the brevity. I'll have to run now but, since you said you couldn't even do $n=3$, I concluded that the above technique may be unknown to you and I didn't want to keep you waiting without a good reason :).
Uphhh... The hard part done. Now it just remains to rewrite the proof. Give me a couple of hours :).
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edited May 24 2011 at 16:09 |
Let's try to do things a la Koosis (we'll fail but we'll learn a couple of tricks on the way).
Put $f(z)=-\prod_j(1-z/z_j)^{p_j}$. That is a nice analytic function and, while its absolute value is somewhat hard to understand, its argument is very simple: it is an $n$-piece piecewise linear function on $[0,2\pi]$ with slope $\frac 12 n$ and jumps at arguments of $z_j$. Now, consider $g(z)=1+\frac 12f(z)$. If there are no zeroes inside the circle (otherwise we are done by the maximum principle), we must have $\int_0^{2\pi}\log|g(e^{it}|dt=-2\pi\log 2$, the same as for $f(z)=z-1$. Now observe that if we know the argument $\sigma$ of $f$ alone, we already know that $\log|g|\ge \log|\frac{e^{2i\sigma}-1}2|$, which is always the equality for $z-1$. Also, the smaller the argument is ($\mod 2\pi$), the less the bound is. So, it will be nice to compare the distribution functions. $z-1$ gives just the uniform distribution of argument on $[-\pi/2,\pi/2]$ with density $2$. If the run for each piece is at most $2\pi$, we can say that the preimage of the interval of length $L$ centered at $0$ is at most $2L$ ($n$ intervals, $2L/n$ on each due to the slope). So, if the angles between adjacent points are all less than $4\pi/n$, we are in good shape. This takes care of $n\le 5$ because $(2\sin\frac\pi 5)^5>2$. Otherwise we still can say that the preimage is of length at most $4L$ doubling the possible value of the integral (which is crude and bad, but you, probably, want all or nothing, so I'll not try to squeeze epsilons). Note that this estimate $-4\pi\log 2$ works for $1+af$ just as well as for $f$ (we used the argumentof $f$ only). Thus, if $a=\frac 34$, we get a contradiction, so $4/3$ is always yours. At last, if all jumps are $\pi$ (the integer case; I'm aware of Vieta, of course, but still), we have a perfect linear run if we care just about lines instead of rays when talking about argument, so the distribution on lines is just perfect but some rays may be wrong.
Sorry for the brevity. I'll have to run now but, since you said you couldn't even do $n=3$, I concluded that the above technique may be unknown to you and I didn't want to keep you waiting without a good reason :).
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edited May 24 2011 at 16:00 |
Let's try to do things a la Koosis (we'll fail but we'll learn a couple of tricks on the way).
Put $f(z)=-\prod_j(1-z/z_j)^{p_j}$. That is a nice analytic function and, while its absolute value is somewhat hard to understand, its argument is very simple: it is an $n$-piece piecewise linear function on $[0,2\pi]$ with slope $\frac 12 n$ and jumps at arguments of $z_j$. Now, consider $g(z)=1+\frac 12f(z)$. If there are no zeroes inside the circle (otherwise we are done by the maximum principle), we must have $\int_0^{2\pi}\log|g(e^{it}|dt=-2\pi\log 2$, the same as for $f(z)=z-1$. Now observe that if we know the argument $\sigma$ of $f$ alone, we already know that $\log|g|\ge \log|\frac{e^{2i\sigma}-1}2|$, which is always the equality for $z-1$. Also, the smaller the argument is ($\mod 2\pi$), the less the bound is. So, it will be nice to compare the distribution functions. $z-1$ gives just the uniform distribution of argument on $[-\pi/2,\pi/2]$ with density $2$. If the run for each piece is at most $2\pi$, we can say that the preimage of the interval of length $L$ centered at $0$ is at most $2L$ ($n$ intervals, $2L/n$ on each due to the slope). So, if the angles between adjacent points are all less than $4\pi/n$, we are in good shape. This takes care of $n\le 5$ because $(2\sin\frac\pi 5)^5>2$. Otherwise we still can say that the preimage is of length at most $4L$ doubling the possible value of the integral (which is crude and bad, but you, probably, want all or nothing, so I'll not try to squeeze epsilons). Note that this estimate $-4\pi\log 2$ works for $1+af$ just as well as for $f$ (we used the argumentof $f$ only). Thus, if $a=\frac 34$, we get a contradiction, so $4/3$ is always yours. At last, if all jumps are $\pi$ (the integer case; I'm aware of Vieta, of course, but still), we have a perfect linear run if we care just about lines instead of rays when talking about argument, so the distribution on lines is just perfect but some rays may be wrong.
Sorry for the brevity. I'll have to run now but, since you said you couldn't even do $n=3$, I concluded that the above technique may be unknown to you and I didn't want to keep you waiting without a good reason :).
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answered May 23 2011 at 15:17 |
Let's try to do things a la Koosis (we'll fail but we'll learn a couple of tricks on the way).
Put $f(z)=-\prod_j(1-z/z_j)^{p_j}$. That is a nice analytic function and, while its absolute value is somewhat hard to understand, its argument is very simple: it is an $n$-piece piecewise linear function on $[0,2\pi]$ with slope $\frac 12 n$ and jumps at arguments of $z_j$. Now, consider $g(z)=1+\frac 12f(z)$. If there are no zeroes inside the circle (otherwise we are done by the maximum principle), we must have $\int_0^{2\pi}\log|g(e^{it}|dt=-2\pi\log 2$, the same as for $f(z)=z-1$. Now observe that if we know the argument $\sigma$ of $f$ alone, we already know that $\log|g|\ge \log|\frac{e^{2i\sigma}-1}2|$, which is always the equality for $z-1$. Also, the smaller the argument is ($\mod 2\pi$), the less the bound is. So, it will be nice to compare the distribution functions. $z-1$ gives just the uniform distribution of argument on $[-\pi/2,\pi/2]$ with density $2$. If the run for each piece is at most $2\pi$, we can say that the preimage of the interval of length $L$ centered at $0$ is at most $2L$ ($n$ intervals, $2L/n$ on each due to the slope). So, if the angles between adjacent points are all less than $4\pi/n$, we are in good shape. This takes care of $n\le 5$ because $(2\sin\frac\pi 5)^5>2$. Otherwise we still can say that the preimage is of length at most $4L$ doubling the possible value of the integral (which is crude and bad, but you, probably, want all or nothing, so I'll not try to squeeze epsilons). Note that this estimate $-4\pi\log 2$ works for $1+af$ just as well as for $f$ (we used the argumentof $f$ only). Thus, if $a=\frac 34$, we get a contradiction, so $4/3$ is always yours. At last, if all jumps are $\pi$ (the integer case; I'm aware of Vieta, of course, but still), we have a perfect linear run if we care just about lines instead of rays when talking about argument, so the distribution on lines is just perfect but some rays may be wrong.
Sorry for the brevity. I'll have to run now but, since you said you couldn't even do $n=3$, I concluded that the above technique may be unknown to you and I didn't want to keep you waiting without a good reason :).
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