The Cauchy-Schwarz Inequality

To better understand inner product spaces, several applications of the Cauchy–Schwarz inequality have been worked through. The following is a list of problems.

Problem

1. Bounds for Linear Expressions

Suppose $a, b, c, x, y \in \mathbb{R}$ and $a^2 + b^2 + c^2 + x^2 + y^2 \leq 1$.
Prove that $a + b + c + 4x + 9y \leq 10$.

2. Equality Case of Cauchy–Schwarz

Suppose $u, v \in V$ and $|u| = |v| = 1$ and $\langle u, v \rangle = 1$.
Prove that $u = v$.

Solution

1. Bounds for Linear Expressions

To prove this, we leverage the Cauchy-Schwarz Inequality in the inner product space $\mathbb{R}^5$. The inequality states that for any vectors $u, v \in V$, we have

\[|\langle u, v \rangle| \leq \|u\| \|v\|\]

In this case, we define the inner product as the Euclidean inner product (or standard dot product) on $\mathbb{R}^5$. We represent the linear expression as an inner product of two vectors.

Let $u = (a, b, c, x, y)$ and $v = (1, 1, 1, 4, 9)$.

The inner product $\langle u, v \rangle$ corresponds to the expression we want to bound:

\[\begin{aligned} \langle u, v \rangle &= a(1) + b(1) + c(1) + x(4) + y(9) \\ &= a + b + c + 4x + 9y \end{aligned}\]

Applying the inequality \(|\langle u, v \rangle| \leq \|u\| \|v\|\),

\[|a + b + c + 4x + 9y| \leq \sqrt{a^2 + b^2 + c^2 + x^2 + y^2} \sqrt{1^2 + 1^2 + 1^2 + 4^2 + 9^2}\]

From the problem constraints, the norm of $u$ is bounded,

\[\|u\| = \sqrt{a^2 + b^2 + c^2 + x^2 + y^2} \leq \sqrt{1} = 1\]

Calculating the norm of $v$, we get,

\[\|v\| = \sqrt{1 + 1 + 1 + 16 + 81} = \sqrt{100} = 10\]

Plugging these values back into the inequality, we get

\[\begin{aligned} a + b + c + 4x + 9y &\leq (1)(10) \\ a + b + c + 4x + 9y &\leq 10 \end{aligned}\]

Note: Equality happens when $u$ is a scalar multiple of $v$ (i.e., they are linearly dependent).

2. Equality Case of Cauchy–Schwarz

Recall the Cauchy–Schwarz inequality in an inner product space,

\[|\langle u, v \rangle| \leq \|u\| \|v\|.\]

Substituting the given conditions, \(\langle u, v \rangle = 1\) and \(\|u\| = \|v\| = (1)\), we get

\[\langle u, v \rangle = |u\|\|v\|\]

By the equality condition for the Cauchy–Schwarz inequality, the vectors $u$ and $v$ are linearly dependent. Hence, there exists a scalar $\lambda$ such that \(u = \lambda v.\)

\[\begin{aligned} \langle u, v \rangle &= \langle \lambda v, v \rangle \\ &= \lambda \langle v, v \rangle \\ &= \lambda \|v\|^{2} \\ \end{aligned}\]

Hence, $\lambda = 1$. Therefore, $u = v$.