# Prob 15, Chap. 1 in Baby Rudin: If this condition also sufficient for equality?

Here's Prob. 15, Chap. 1 in the book Principles of Mathematical Analysis by Wlater Rudin, 3rd edition:

Under what conditions does equality hold in the Schwarz inequality?

Now the Schwarz inequality, which is Theorem 1.35 in Rudin, is as follows:

If $$a_1, \ldots, a_n$$ and $$b_1, \ldots, b_n$$ are complex numbers, then $$\left\vert \sum_{j=1}^n a_j \overline{b_j} \right\vert^2 \leq \sum_{j=1}^n \left\vert a_j \right\vert^2 \sum_{j=1}^n \left\vert b_j \right\vert^2.$$

If there is a complex number $$z$$ such that $$a_j = z b_j$$ for each $$j=1, \ldots, n$$, then we have \begin{align} \left\vert \sum_{j=1}^n a_j \overline{b_j} \right\vert^2 &= \left\vert \sum_{j=1}^n z b_j \overline{b_j} \right\vert^2 \\ &= \left\vert z \ \sum_{j=1}^n \left\vert b_j \right\vert^2 \right\vert^2 \\ &= \vert z \vert^2 \cdot \left\vert \sum_{j=1}^n \left\vert b_j \right\vert^2 \right\vert^2 \\ &= \vert z \vert^2 \cdot \left( \sum_{j=1}^n \left\vert b_j \right\vert^2 \right)^2, \end{align} whereas \begin{align} \sum_{j=1}^n \left\vert a_j \right\vert^2 \sum_{j=1}^n \left\vert b_j \right\vert^2 &= \sum_{j=1}^n \left\vert z b_j \right\vert^2 \sum_{j=1}^n \left\vert b_j \right\vert^2 \\ &= \sum_{j=1}^n \vert z \vert^2 \left\vert b_j \right\vert^2 \sum_{j=1}^n \left\vert b_j \right\vert^2 \\ &= \vert z \vert^2 \cdot \left( \sum_{j=1}^n \left\vert b_j \right\vert^2 \right)^2 \\ &= \left\vert \sum_{j=1}^n a_j \overline{b_j} \right\vert^2. \end{align}

Now is this condition also a necessary condition for the equality to hold in the Schwarz "inequality"?

The Schwarz inequality follows from expanding \begin{aligned} 0 &\le \sum_{i=1}^n \sum_{j=1}^n \left\vert a_i b_j - a_j b_i \right\vert^2 \\ &= \sum_{i=1}^n \sum_{j=1}^n (a_i b_j - a_j b_i)\overline{(a_i b_j - a_j b_i)} \\ &= \quad ... \\ &= 2 \sum_{j=1}^n \left\vert a_j \right\vert^2 \sum_{j=1}^n \left\vert b_j \right\vert^2 - 2 \left\vert \sum_{j=1}^n a_j \overline{b_j} \right\vert^2 \end{aligned} and therefore equality holds if and only if $$\tag{*} a_i b_j - a_j b_i = 0 \quad \text{ for all } i, j = 1, \dots n \, .$$ If at least one $b_i$ is not zero then you can define $z = a_i/b_i$ and conclude from $(*)$ that $a_j = z b_j$ for each $j=1, \ldots, n$.
Another way to express $(*)$ is that one of the vectors $(a_1, \ldots, a_n)$ and $(b_1, \ldots, b_n)$ is a constant multiple of the other, or that they are linearly dependent over $\Bbb C$.
Yes: define $\lVert x\rVert:=\sum_{j=1}^n|x_j|^2$; call $a=(a_1,\dots,a_n)$ and $b=(b_1,\dots,b_n)$ (that we assume both non-zero). Then $\left\lVert \lVert a\rVert b-\lVert b\rVert a\right\rVert^2$ equals $0$ if and only if we have equality in Schwarz inequality.
• Not quite, if $b_1 = \dotsc = b_n = 0$, then such a $z$ doesn't exist if there is a nonzero $a_j$. Jan 23, 2016 at 11:56