I was looking at this example while learning about unconstrained optimization and I don't think I completely understand it.
Here is the example in question: Consider the two-dimensional linear function $f(x,y)=x+y$ defined over the unit ball $S=B [0,1]={(x,y)^T: x^2+y^2 \le 1}$. Then by the Cauchy-Schwarz inequality we have for any $(x,y)^T∈S $
$ x+y= (x y) \begin{bmatrix}1\\1\end{bmatrix} \le \sqrt{x^2 + y^2\vphantom{b}} \sqrt{1^2 + 1^2\vphantom{b}} \le \sqrt{2\vphantom{b}} $
Therefore, the maximal value of f over S is upper bounded by $\sqrt{2\vphantom{b}} $. On the other hand, the upper bound $\sqrt{2\vphantom{b}} $ is attained at $(x,y)$= $(\frac{1}{ \sqrt{2\vphantom{b}}}, \frac{1}{ \sqrt{2\vphantom{b}}})$. It is not difficult to see that this is the only point that attains this value, and thus $(\frac{1}{ \sqrt{2\vphantom{b}}}, \frac{1}{ \sqrt{2\vphantom{b}}})$ is the strict global maximum point of $f$ over $S$, and the minimal value is $-\sqrt{2\vphantom{b}} $
i) I am aware of how the Cauchy-Schwarz inequality works and maybe my concept of it is not very clear as I am having trouble seeing how it has been applied here. Can someone possibly break down how the maximal value was obtained?
ii) Then I am still not very clear how this gives the strict global maximum point of $(\frac{1}{ \sqrt{2\vphantom{b}}}, \frac{1}{ \sqrt{2\vphantom{b}}})$
Cauchy-Schwarz Inequality states the following:
If $u,v$ are two vectors, then $|\langle u, v\rangle |\le ||u||||v||$, where equality holds if $u$ or $v$ is a scalar multiple of the other.
In the given problem, $u=(x,y)$ and $v=(1,1)$.
Thus $||u||=\sqrt {x^2+y^2}$ and $||v||=\sqrt{1^2+1^2}=\sqrt 2$.
Hence $|\langle u, v\rangle |=x+y\le \sqrt2 \sqrt{(x^2+y^2)}$, where equality holds if $u$ is a scalar multiple of $v$, i.e. $u=\lambda v$ for some $\lambda$.
Hence, $u=\lambda v\implies (x,y)=\lambda (1,1)\implies x=\lambda, y=\lambda$.
Now, $\lambda+\lambda= \sqrt2 \implies 2\lambda=\sqrt 2\implies \lambda=\frac{1}{\sqrt 2}$.
Thus $x=y=\lambda=\frac{1}{\sqrt 2}$