Mathematical Immaturity

1.13 Exercises

Prove that each of the statements in Exercises 3 through 7 is valid for all elements $x$ and $y$ in a real Euclidean space.

• $(x, y) = 0$ if and only if $\|x + cy\| \geq \|x\|$ for all real $c.$
• Recall from Section 11 that for an element $x$ in a Euclidean space $V$ with inner product $(x, x),$ the nonnegative number $\|x\| = (x, x)^{1/2}$ is called the norm of $x.$
• 
  Proof.$\quad$ To show that $(x, y) = 0$ implies $\|x + cy\| \geq \|x\|$ for all real $c,$ we can equivalently show that $(x, y) = 0$ implies $(x + cy, x + cy) \geq (x, x)$ for all real $c.$ First, we rewrite $(x + cy, x + cy)$ as \begin{align*} (x + cy, x + cy) &= (x, x) + 2c(x, y) + c^2(y, y). \end{align*} Then, if $(x, y) = 0,$ the equation becomes $$(x + cy, x + cy) = (x, x) + c^2(y, y).$$ By positivity of the inner product, we have $(x, x) + c^2(y, y) \geq (x, x)$ for all real $c.$ But this means that for all real $c$ we have $(x + cy, x + cy) \geq (x, x),$ and equivalently, $\|x + cy\| \geq \|x\|.$
  
  To prove that $\|x + cy\| \geq \|x\|$ for all real $c$ implies $(x, y) = 0,$ we first note that if $\|x + cy\| \geq \|x\|$ for all real $c,$ then $\|x + cy\|^2 \geq \|x\|^2$ for all real $c,$ giving us \begin{align*} \\ (x, x) + 2c(x, y) + c^2(y, y) &\geq (x, x) \end{align*} Subtracting $(x, x)$ from both sides and rearranging terms, we have $c^2(y, y) \geq -2c(x, y)$ for all real $c.$ If $c \lt 0,$ then $-2c(x, y) = |2c(x, y)| = 2|c||(x, y)|,$ giving us $$|c|(y, y) \geq 2(x, y)$$= for all $c \lt 0.$ But if this is true for all $c \lt 0,$ then $(x, y)$ must be zero. Thus, we have shown that $(x, y) = 0$ if and only if $\|x + cy\| \geq \|x\|$ for all real $c.$ This completes the proof.