Result of non-negativity condition

Question

Consider the non-negativity condition $f(t) = t^2 {\Vert v \Vert}^2 - 2 t \langle u , v \rangle + {\Vert u \Vert}^2 \geq 0$. Where $u$ and $v$ are vectors, and $t$ being a real scalar.

In the process of proving the Cauchy-Schwarz Inequality, one arrives at $4\langle u, v \rangle^2 - 4\| u \|^2\| v \|^2 \leq 0$.

I cannot figure out how one gets to the last expression. Please, be aware that I am not interested in the proof of the Cauchy-Schwarz Inequality, but about the step regarding the discriminant. How does the discriminant come into this exactly ?

How does the discriminant emerge from the quadratic equation and why should it be less than or equal to zero?

Answer 1

This comes back to basic properties of quadratics. Imagine you have a quadratic function $$f(t) = \color{red}at^2 + \color{blue}bt + \color{lime}c,$$ where $\color{red}a > 0$. What we get is a parabola. Given $\color{red}a > 0$, we must have a "smiley" parabola: one that is concave up, and tends to $\infty$ as $t \to \pm \infty$.

Recall the quadratic formula: $$\color{red}at^2 + \color{blue}bt + \color{lime}c = 0 \iff t = \frac{-\color{blue}b \pm \sqrt{\color{blue}b^2 - 4\color{red}a\color{lime}c}}{2\color{red}a}.$$ The bit under the square root is the discriminant: $$\Delta = \color{blue}b^2 - 4\color{red}a\color{lime}c.$$ As we take the square root of the discriminant, its sign determines whether we get real solutions to $f(t) = 0$. In order for the square root to make sense in the reals, we need $\Delta \ge 0$. If $\Delta = 0$, then there is a $0$ to the right of the $\pm$, and so we get only one solution. If $\Delta > 0$, we get two different solutions. If $\Delta < 0$, we get none.

Let's illustrate these cases. First, here's a case where $\Delta > 0$:

Note: there are two roots. Next, where $\Delta = 0$:

Here, there is a single root. Finally, where $\Delta < 0$:

Here, there are no roots; the parabola lies strictly above the axis.

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Now, let's suppose we have a particular quadratic $f(t) = \color{red}at^2 + \color{blue}bt + \color{lime}c$, one you know very little about, except for two things:

1. $\color{red}a > 0$, and
2. $f(t) \ge 0$ for all $t$.

What can we conclude from this? This says something about the discriminant. We could have the parabola floating over the axis, like when $\Delta < 0$. We could even have the parabola just touching the axis, like when $\Delta = 0$. The only thing that we can't have is the parabola dipping below the axis and coming back up (creating two different roots) as in where $\Delta > 0$. So, from our assumptions, we get the conclusion that $\Delta \not> 0$, or equivalently, $\Delta \le 0$. That is, $$\color{blue}b^2 - 4\color{red}a\color{lime}c \le 0.$$ We deduced this just from knowing that $f(t) \ge 0$ for all $t$, and $\color{red}a > 0$. So, we've turned a family of inequalities $f(t) \ge 0$ for all $t$, into a single inequality just involving $\color{red}a$, $\color{blue}b$ and $\color{lime}c$.

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This is what is happening in the proof. You have been given a quadratic: $$\color{red}{\|u\|^2}t^2 + \color{blue}{(-2\langle u, v \rangle)}t + \color{lime}{\|v\|^2}.$$ You know that $\color{red}{\|u\|^2} > 0$ (though, an assumption that $u \neq 0$ is necessary here) and you know that $f(t) \ge 0$ for all $t$ (because $f(t) = \langle tu - v, tu - v \rangle = \|tu - v\|^2$, which is non-negative by assumption). From this, we conclude that $\Delta \le 0$. Here, $\Delta$ is given by: $$\Delta = \color{blue}{(-2\langle u, v \rangle)}^2 - 4\color{red}{\|u\|^2}\color{lime}{\|v\|^2} = 4\langle u, v \rangle^2 - 4\|u\|^2 \|v\|^2.$$ Thus, $4\langle u, v \rangle^2 - 4\|u\|^2 \|v\|^2 \le 0$, as claimed.

Answer 2

Suppose that $u = 0$. Then $|\langle u,v\rangle| = 0 \leq \|u\|\|v\|$.

Consider otherwise that $u\neq 0$.

According to the properties of the inner product, one concludes that

\begin{align*} \langle tu + v, tu + v\rangle \geq 0 & \Longleftrightarrow t^{2}\langle u,u\rangle + 2t\langle u,v\rangle + \langle v,v\rangle \geq 0\\\\ & \Longleftrightarrow t^{2}\|u\|^{2} + 2t\langle u,v\rangle + \|v\|^{2} \geq 0 \end{align*} for every $t\in\mathbb{R}$. This means the discriminant $\Delta \leq 0$ once $\|u\|^{2} > 0$.

In other words, it may be deduced that: \begin{align*} 4|\langle u,v\rangle|^{2} - 4\|u\|^{2}\|v\|^{2} \leq 0 \Longleftrightarrow |\langle u,v\rangle| \leq \|u\|\|v\| \end{align*}

and we are done.

Hopefully this helps!

EDIT

Consider the quadratic function $f(x) = ax^{2} + bx + c$, where $a\neq 0$. It can be written as: \begin{align*} f(x) & = ax^{2} + bx + c\\\\ & = a\left(x^{2} + \frac{bx}{a}\right) + c\\\\ & = a\left(x^{2} + \frac{bx}{a} + \frac{b^{2}}{4a^{2}}\right) + c - \frac{b^{2}}{4a}\\\\ & = a\left(x + \frac{b}{2a}\right)^{2} - \left(\frac{b^{2} - 4ac}{4a}\right) \end{align*}

If $a > 0$ (which is the case), it may be concluded that \begin{align*} f(x) = a\left(x + \frac{b}{2a}\right)^{2} - \left(\frac{b^{2} - 4ac}{4a}\right) \geq -\left(\frac{b^{2} - 4ac}{4a}\right) = - \frac{\Delta}{4a} \end{align*}

Therefore, if $\Delta \leq 0$ and $a > 0$, we may conclude that $f(x)\geq 0$ for every $x\in\mathbb{R}$.

On the other hand, if $f(x)\geq 0$ for every $x\in\mathbb{R}$ and $a > 0$, we conclude that $\Delta \leq 0$.

Indeed, it suffices to consider $x = -b/2a$ to obtain that: \begin{align*} f\left(-\frac{b}{2a}\right) = -\frac{\Delta}{4a} \geq 0 \Rightarrow \Delta \leq 0 \end{align*}

Hence, if $a > 0$, $f(x)\geq 0$ for every real number $x\in\mathbb{R}$ iff $\Delta \leq 0$.

Here I provide a picture for the sake of completeness.