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I am thinking about the following problem: Let $f(z)=2z^2-1.$Then what is the maximum value of $|f(z)|$ on the unit disc $D=\{z \in \mathbb C:|z|\leq 1\}$ ? I guess i have to use the maximum modulus principle.But also i notice that $|f(z)|=|2z^2-1|\leq 2|z|^2+1\leq 2+1=3.$ So, is $3$ is the maximum value of $|f(z)|?$ Can someone point me in the right direction? Thanks in advance for your time. |
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As you correctly noted, the maximum modulus principle tells us that the maximum of $|f|$ is on the boundary. Thus we only have to check for $z=e^{i\phi}$ with $\phi\in\mathbb{R}$ and find the maximum among those points. We can evaluate $$|f(z)|^2 = |2 e^{2i\phi} -1|^2 =[2 \cos(2\phi) -1]^2 + 4 \sin^2(2\phi).$$ To maximize, we take the derivative with respect to $\phi$ and set it to 0. Some straightforward calculation shows that the function assumes its maximum value for $\phi=\pi/2$. The maximum of $|f(z)|^2$ is 9, thus the maximum of the modulus is 3 as you already expected. |
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Given a polynomial $f(z)=a_0z^n+a_1z^{n-1}+\cdots+a_n$ we certainly have $$\max_{|z|\leq1}|f(z)|\leq |a_0|+|a_1|+\cdots+|a_n|.\tag{1}$$ More generally, analytic functions with $\ell^1$ coefficients are bounded in the unit disc and the supremum is bounded by the $\ell^1$-norm - that is if $f(z)=\sum_na_nz^n$ and $$\|f\|_{\ell^1}=\sum|a_n|<\infty$$ then $f$ is bounded in the unit disc (in fact it belongs to the so called disc algebra) and $$\sup_{|z|<1}|f(z)|\leq\|f\|_{\ell^1}$$ In our case, $f(z)=2z^2-1$ it is rather easy to guess a point $z_0$ (or two) where $|f(z_0)|=3$ and hence we have equality in (1). However, in general we cannot get equality in (1), which is easiest realized by an example: Consider $g(z)=(z+1)(z^2-1)=z^3+z^2-z-1$ then $\|g\|_{\ell^1}=4$ - now if we would have equality in (1) then $$4=\max|g|\leq\max|z+1|\cdot|z^2-1|=2\cdot2$$ This means that the max must be attained at $z=1$ where $|z+1|$ is maximized, but this is not the case (since the other factor is 0 there). |
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