Prove an if and only if condition for equality in the Cauchy-Schwarz inequality

Recall the Cauchy-Schwarz inequality,

For arbitrary real numbers $a_1, \ldots, a_n$ and $b_1, \ldots, b_n$ we have

$\left( \sum_{k=1}^n a_k b_k \right)^2 \leq \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right)$

The claim is then that the equality sign holds if and only if there is a real number $x$ such that $a_k x + b = 0$ for each $k =1, \ldots, n$ .

Proof. ( $\Rightarrow$ ) If $a_k = 0$ for all $k$ , then equality clearly holds. Assume then that $a_k \neq 0$ for at least one $k$ .

$\left( \sum_{k=1}^n a_k b_k \right)^2 = \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right).$

Then, considering the equation $\sum_{k=1}^n (a_k x + b_k)^2 \geq 0$ and defining,

$A = \sum_{k=1}^n a_k^2, \qquad B = \sum_{k=1}^n b_k a_k, \qquad C = \sum_{k=1}^n b_k^2.$

We have,

$Ax^2 + 2Bx + C \geq 0 \ \implies \ x = \frac{-2B \pm \sqrt{4B^2 - 4AC}}{2A} = \frac{-B \pm \sqrt{B^2 - AC}}{A}.$

But, since we know $B^2 = AC$ (by assumption), we have $x = -\frac{B}{A}$ which is in $\mathbb{R}$ (since $A \neq 0$ since $a_k \neq 0$ for at least one $k$ and each term $a_k^2$ in nonnegative, so the sum is strictly positive).
( $\Leftarrow$ ) Assume there exists $x \in \mathbb{R}$ such that $a_k x + b_k =0$ for each $k =1, \ldots, n$ . Then, $a_k x + b_k = 0 \implies b_k = (-x)a_k$ . So,

$\begin{align*} \left( \sum_{k=1}^n a_k b_k \right)^2 &= \left( -x \left(\sum_{k=1}^n a_k^2 \right) \right)^2 \\ &= x^2 \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n a_k^2 \right) \\ &= \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n x^2 a_k^2 \right) \\ &= \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n (-xa_k)^2 \right) \\ &= \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n b_k^2 \right). \qquad \blacksquare \end{align*}$

3 comments

September 27, 2018 at 8:24 pm

Anonymous says:

The forward implication is actually false. Consider 0 as the a-sequence and 9 as the b-sequence. Then the equality holds but there is no real x which will make ax + b = b = 9 = 0 true.

- November 5, 2018 at 5:33 pm
  
  nu creation says:
  
  I think it might be a mistake in the book. Because the proof technique in the second implication would still work if there is a y s.t. ak + ybk =0.. this would also imply the Cauchy-Schwartz equality which means there’s something’s wrong with the only if since having such an x isn’t the only way to arrive at the equality. Notice if we change the statement to include the bit about the y, his forward implication is true… If sequence a is all zeros, there exists a y st ak + ybk= 0 namely y=0. So make the iff statement require either there is such an x or there is such a y.
  
  - November 2, 2022 at 9:50 pm
    
    supreeth says:
    
    Yess.. I think the statement should be “The equality sign holds if and only if there are real numbers x and y s.t. xak + ybk = 0 for all k.

Stumbling Robot

Prove an if and only if condition for equality in the Cauchy-Schwarz inequality

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