# Norm of swapped power series in the unit disk

Suppose $f(z)=a_0+a_1z+\cdots+a_nz^n+\cdots$ is defined in the unit disk and $\|f\|_{\infty}\leq 1.$ Lets form another series $g$ by interchanging $a_1$ and $a_k$ i.e. $g(z)=a_0+a_kz+\cdots+a_1z^k+\cdots$. Is $g$ of norm less than or equal to one?if that is not the case can you provide a counterexample?

• Almost any polynomial, suitably scaled, is either $f$ or $g$ for a counterexample. – Robert Israel Nov 21 '14 at 16:33
• If it were true, you would have it for any transposition of coefficients ((k,l)=(1,k)(1,l)(1,k)), so it would be true for all permutations of the coefficients with indices $\ge 1$. Thus, all $H^\infty$ norms of such permuted series would be exactly the same, which is sort of too hard to believe... – fedja Nov 21 '14 at 16:33
• That's what I was also thinking; though passing to the limit from permutations with compact support, to all permutations, seems delicate to me, as I only see local uniform convergence. – Pietro Majer Nov 21 '14 at 17:07
• A question: what can be said about a function $f\in H^\infty(D)$ whose norm is invariant by exchange $a_1\leftrightarrow a_k$ for any $k\ge 1$? – Pietro Majer Nov 21 '14 at 17:18
• E.g. $f$ with all $a_k$ real and positive are OK, as they reach the norm at $z=1$. – Pietro Majer Nov 21 '14 at 17:27

For simplicity, let us assume that the radius of convergence of $f$ is strictly larger than one. By the maximum principle the absolute value of the analytic function $f:D\to\mathbb C$ obtains its maximum at the boundary $\partial D=S^1$. Therefore we are interested in the function $g:[0,2\pi]\to\mathbb C$, $$h(\phi) = \sum_{n=0}^\infty a_ne^{in\phi}.$$ This is a Fourier series. The norm of this function is given by $$|h(\phi)|^2 = \sum_{n,m}a_na_m^*e^{i(n-m)\phi} .$$ If the coefficients are real, we get $$|h(\phi)|^2 = \sum_{n,m}a_na_m\cos((n-m)\phi) .$$ The question is now whether the maximum of the function $\phi\mapsto|h(\phi)|^2$ changes when we change the coefficients.
Let us take a concrete example: $f(z)=z+z^2-z^4$. Now if $k=4$, the other function is $g(z)=\tilde f(z)=-z+z^2+z^4$. Now $$|h(\phi)|^2=3+2\cos(\phi)-2\cos(2\phi)-2\cos(3\phi)$$ and the corresponding norm for $\tilde f$ is $$|\tilde h(\phi)|^2=3-2\cos(\phi)+2\cos(2\phi)-2\cos(3\phi).$$ If you plot these functions, you see that $$\max_{\bar D}|f|^2=\max_{\phi\in[0,2\pi]}|h(\phi)|^2=5$$ but $$\max_{\bar D}|\tilde f|^2=\max_{\phi\in[0,2\pi]}|\tilde h(\phi)|^2=6.$$ (One can of course prove these by hand as well if needed.) Thus the function $\frac1{\sqrt5}f$ with $k=4$ gives a counterexample.