Cauchy Schwarz Inequality

Question

Let $x \in \mathbb{R}^k$. Show that if there exists a $c \geq 0$ such that

$x \cdot y \leq c \left\lVert y \right\rVert$ $\hspace{1cm}$ for all $y \in \mathbb{R}^k$

then $\left\lVert x \right\rVert \leq c$

I'm not sure.

Answer 1

Let $ x \in \mathbb{R}^{k}$ and $c \in \mathbb{R}_{\geq 0}$ such that $$ \forall y \in \mathbb{R}^{k}, x \cdot y \leq c ||y|| $$ Then letting $y = x$, we have that $$ x \cdot x = ||x||^{2} \leq c \cdot ||x|| \implies ||x|| \leq c $$ This is the desired result. (Assuming $x \not \equiv 0$).

Answer 2

Let $y=x$ then $$x \cdot y \leq c \left\lVert y \right\rVert\implies $$

$$x \cdot x \leq c \left\lVert x \right\rVert\implies $$

$$\left\lVert x \right\rVert^2 \leq c \left\lVert x \right\rVert\implies $$

$$\left\lVert x \right\rVert\le c $$