local minima of function in two variables

Question

We just started to learn multivariable calculus, and we got the next definition:
We say that a point $x_{0} \in \mathbb{R}^{n}$ is a local minimum of a function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ if there exists a $\delta > 0$ for which the following holds:
$\left\| x-x_{0}\right\| _{\infty } \leq \delta \Rightarrow f\left( x\right) \geq f\left( x_{0}\right)$
given the function $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ such that $f\left( x_{1}, x_{2}\right) = 2\left(x_{1}\right)^{2} + \left(x_{2}\right)^{2} + 2x_{1}x_{2} - 2$, find it's local minimum, if such a point exists. If there is a local minimum, prove directly from the above definition that it is a local minimum.
So I started by calculate the partial derivatives and got the critic point $x_{0} = \left(0, 0\right)$.
Now I now it's a minimum of $f$, but I got stuck on proving it from the definition.
My try was:
Let $\delta = 1>0$ and denote $x = \left( x_{1}, x_{2}\right)$. We can notice that:

$\left\| x-x_{0}\right\| _{\infty } \leq \delta \Leftrightarrow \left\| x\right\| _{\infty } \leq 1 \Leftrightarrow \left\| \left( x_{1}, x_{2}\right) \right\| _{\infty } \leq 1 \Leftrightarrow max\left\{ \left| x_{1}\right| ,\left| x_{2}\right| \right\} \leq 1 \Leftrightarrow \left| x_{1}\right| \leq 1\wedge \left| x_{2}\right| \leq 1 \Leftrightarrow -1\leq x_{1}\leq 1 \wedge -1\leq x_{2}\leq 1$

So, if $\left\| x-x_{0}\right\| _{\infty } \leq \delta$, then:
$f\left( x_{1}, x_{2}\right) = 2\left(x_{1}\right)^{2} + \left(x_{2}\right)^{2} + 2x_{1}x_{2} - 2 \geq 2\cdot \left( -1\right) ^{2} + \left( -1\right) ^{2} + 2 \cdot \left( -1\right) \cdot \left( -1\right) - 2 = 3 > -2 = f\left( 0, 0\right)$

But then I noticed that if I put the point $\left( \dfrac{1}{2}, \dfrac{1}{2} \right)$ in $f$ I get $f \left( \dfrac{1}{2}, \dfrac{1}{2} \right) = -\dfrac{3}{4} <3$, and I couldn't find my mistake in my proof.

Answer 1

You wrote

$f\left( x_{1}, x_{2}\right) = 2\left(x_{1}\right)^{2} + \left(x_{2}\right)^{2} + 2x_{1}x_{2} - 2 \geq 2\cdot \left( -1\right) ^{2} + \left( -1\right) ^{2} + 2 \cdot \left( -1\right) \cdot \left( -1\right) - 2 = 3 > -2 = f\left( 0, 0\right).$

In this inequality you used that $2x_1x_2 \ge 2 \cdot (-1) \cdot (-1)=2$ for $|x_1| \le 1 $ and $|x_2| \le 1 .$

But this is false. Take $x_1=x_2= 1/4.$

My way:

we have $f(x_1,x_2)= (x_1+x_2)^2+x_1^2-2 \ge -2$ for all $(x_1,x_2)$ and $f(0,0) =-2.$

Hence $$\min \{f(x_1,x_2): (x_1,x_2) \in \mathbb R^2\}=-2=f(0,0).$$