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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 I also notice that $|f(z)|=|2z^2-1|\leq 2|z|^2+1\leq 2+1=3.$

So, is $3$ the maximum value of $|f(z)|?$ Can someone point me in the right direction? Thanks in advance for your time.

AD - Stop Putin -
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2 Answers2

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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.

Fabian
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You are totally right that the maximum modulus principle says $$\max_{|z|\leq1}|f(z)|\leq \max_{|z|=1}|f(z)|$$ and also, by the triangle inequality, $$\max_{|z|\leq1}|f(z)|\leq |a_0|+|a_1|+\cdots+|a_n|.\tag{1}$$ where $$f(z)=a_0z^n+a_1z^{n-1}+\cdots+a_n$$ is a polynomial.

Moreover, $$\sup_{|z|<1}|f(z)|\leq \sum_{n=0}^\infty |a_n| $$ for any analytic function $f(z)=\sum_na_nz^n$.

In our case, as you noticed, (1) tell us that $$\sup_{|z|<1}|f(z)|\leq3. \tag{2}$$ where $f(z)=2z^2-1$. Moreover, it is rather easy to guess a point $z_0$ (or two) where we have equality in (2). Hence, we really do have the equality $$\sup_{|z|<1}|f(z)|=3.$$

However, in general we cannot get equality in (1), which is easiest realized by an example.

Consider the polynomial $g(z)=(z+1)(z^2-1)=z^3+z^2-z-1$. Then the right hand side of (1) imply $$\sup_{|z|\leq1}|g(z)|=4.$$ Next, the maximum modulus principle say $$\sup_{|z|\leq1}|g(z)|\leq\sup_{|z|=1}|z+1|\cdot|z^2-1|$$ which is bounded by $2\cdot2$. However, the maximum of the first factor, $|z+1|$, is (only) attained at $z=1$, but then the second factor, $|z^2-1|$, is zero.

That is $$\sup_{|z|\leq1}|g(z)|<4.$$

AD - Stop Putin -
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  • How can we find the suprema in more sophisticated cases like the one you mention or like $z^2 + z - 1$? – Shaheer ziya Mar 14 '24 at 17:29
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    @Shaheerziya Sorry about the late response. Let $p(z)=z^2+z-1$, and note that $$|p(z)|^2 = p(z)\bar{p}(z)=(z^2+z-1)(\bar{z}^2+\bar{z}-1)=$$ $$|z|^4 +|z|^2+1+z|z|^2-z^2+\bar{z}|z|^2-z-\bar{z}^2-\bar{z}$$ Hence, $$\max_{z\leq1}|p(z)|^2=$$ $$\max_{0\leq t<2\pi}(3+e^{it}-e^{2it}+e^{-it}-e^{it}-e^{-2it}-e^{-it})= $$ $$\max_{0\leq t<2\pi}(3-2\cos2t)=5$$ and we get $\max|p| =\sqrt{5}$. – AD - Stop Putin - May 21 '24 at 10:24
  • Sorry about the formatting, I wrote it on my phone. – AD - Stop Putin - May 21 '24 at 10:32