When does equality hold in $\Bigr\lvert\sum_{k=1}^n a_kb_k\Bigr\rvert^2 \le \left(\sum_{k=1}^n |a_k|^2\right)\left(\sum_{k=1}^n |b_k|^2\right)$?

Question

I'm reading Ahlfors' complex analysis. During the proof of Cauchy's inequality, the author uses the following equation: $$ \sum_{k=1}^n \bigr\lvert a_k - \lambda \overline{b_k}\bigr\rvert^2 = \sum_{k=1}^n |a_k|^2 + |\lambda|^2\sum_{k=1}^n |b_k|^2 - 2 \Re\left(\overline{\lambda}\sum_{k=1}^n a_kb_k\right) \tag{1} $$ where the $a_k$'s, $b_k$'s and $\lambda=\frac{\sum_{j=1}^n a_jb_j}{\sum_{j=1}^n |b_j|^2}$ are all some complex numbers.

After the proof is concluded, the author states the following:

From $(1)$ we conclude further that the sign of equality holds in $\Bigr\lvert\sum_{k=1}^n a_kb_k\Bigr\rvert^2 \le \left(\sum_{k=1}^n |a_k|^2\right)\left(\sum_{k=1}^n |b_k|^2\right)$ if and only if the $a_k$ are proportional to the $\overline{b_k}$.

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I understand that the idea here is that, under the hypothesis that $a_k$ and $\overline{b_k}$ are linearly dependent, then in equation $(1)$ both sides should be equal to $0$ (instead of greater or equal to $0$ as in the normal proof). However, I don't see how this is the case.

If I take $a_k = \gamma_k \overline{b_k}$ for some scalars $\gamma_k$'s then $(1)$ reduces to

\begin{align} \sum_{k=1}^n \bigr\lvert \gamma_k \overline{b_k} - \lambda \overline{b_k}\bigr\rvert &= \sum_{k=1}^n |\gamma_k \overline{b_k}|^2 + |\lambda|^2\sum_{k=1}^n |b_k|^2 - 2 \Re\left(\overline{\lambda}\sum_{k=1}^n \gamma_k \overline{b_k}b_k\right) \notag \\ &= \sum_{k=1}^n |\gamma_k|^2 |b_k|^2 + |\lambda|^2\sum_{k=1}^n |b_k|^2 - 2 \Re\left(\overline{\lambda}\sum_{k=1}^n \gamma_k |b_k|^2\right) \notag \end{align}

but from here, I don't see how I could simplify the result further. Could somebody explain to me what the author meant and why it is that equality holds in this case? Thank you!

Answer 1

If each of $a_1,...,a_n,b_1,..,b_n$ is zero, then the equality holds.

If some members of the set $\{b_1,..,b_n\}$ are zero, we can omit them.

Without loss of generality, we can assume that each of the $b_k$ is non zero for $k=1,2,...,n$. $\sum_{k=1}^{n}|a_k-\lambda \bar b_k|^2=0 \iff |a_k-\lambda b_k|^2=0$ for $k=1,2,...,n$ $\iff |a_k-\lambda \bar b_k|=0$ for $k=1,2,...,n$ $\iff a_k=\lambda \bar b_k$ for $k=1,2,...,n$.

Therefore, equality holds if and only if each of $a_k$ and $\bar b_k$ are proportional for $k=1,2,..,n$.

Answer 2

When they proportional, they mean that there exists a scalar $\gamma$ such that $a_k = \gamma b_k$. If the scalar depended on $k$ then almost every vector could be said to be proportional to another vector.

For example take the vectors $a = [1,1,1]$ and $b = [1,1,2]$. If we could vary the parameter then we could choose $\gamma_1, \gamma_2 = 1$ and $\gamma_3 = \frac{1}{2}$, but the Cauchy Schwarz inequality is not an equality.