What is the minimum point of $x\mapsto x^Ty$ for $|x|\le 1$ and a fixed $y\in\mathbb{R}^n$? Let $y\in\mathbb{R}^n$. I want to minimize $$f(x):=x^Ty\;\;\;\text{for }|x|\le 1$$ The minimum point should be $$-\frac{y}{\sqrt{y^Ty}}\tag{1}$$ However, how can we derive $(1)$ analytically? Since $f$ is a linear function (and $\nabla f\equiv y\ne 0$ if $y\ne 0$), there is no local minimum point of $f$. 
 A: Your comment is true that there is no local minimum on all of $\mathbb{R}^n$.   However in this problem you are minimizing over the $n$-ball and with respect to that it has a minimium.  
In fact, that means you can solve this problem by restricting to the boundary: |x| = 1.  Given that, this reduces to a standard Lagrange Multiplier problems where you are attempting to find the critical points of:
$$x^Ty - \lambda(x^Tx - 1)$$
A: By the Cauchy–Schwarz inequality, $$|x^T y|\leq \|x\|\|y\|,$$ where $\|\cdot\|$ is the Euclidean norm and $|\cdot|$ is the absolute value. Also, equality prevails if and only if $x$ and $y$ are collinear. Hence, given the constraint $\|x\|\leq 1$ and the fact that $y$ is fixed, you'll have $$x^T y\geq-\|x\|\|y\|\geq-\|y\|.$$ If you can find such an $x\in\mathbb R^n$ that $\|x\|\leq 1$ and $x^Ty=-\|y\|$, then you will have attained the least possible value. This is achieved by setting $\|x\|=1$ and requiring that $x$ and $y$ be collinear. The solution will be $$x=-\frac{1}{\|y\|}y,$$ and this attains the minimum value $-\|y\|$, because $y^T y=\|y\|^2$. (If $y=0$, then this doesn't make sense but then any $x\in\mathbb R^n$ with $\|x\|\leq 1$ trivially solves the minimization problem.)
A: $f$ is continuous on the compact set $C= \left \{ x:\left | x \right | \leq 1\right \}$ so it has a minimum value, which by your comment, lies on the boundary of $C$ and so will be a unit vector. 
Now observe that $f$ is just the dot product of the vetors $x$ and $y$, and as such has its minumum when $x$ is equal to $-y$. 
Therefore, $x$ must be the unit vector equal to $-y$, which is exactly $\frac{-y}{\sqrt{y^{T}y}}$.
