Recent content by Onezimo Cardoso

Another pretty good exercise it to use the exercise above to solve the following one:Exercise: Let ##C \subset \mathbb{R}^n## be a convex and closed set. Let ##f : \mathbb{R}^n \to C## be a function defined as ##f(x)=\overline{x}##, where ##\overline{x}## is the unique point of ##C## such that...

As I promisse, follow a solution for this question. Let us suppose, by contraction, there is ##y \in C## such that ##\langle x-\overline{x}, y - \overline{x} \rangle > 0##. Let ##z=ty+(1-t)\overline{x}##, where ##t \in (0,1)##. By hypothesis, ##C## is convex. Therefore, ##z \in C##. We...

Finally I done this question! Thanks for all the help. I could do it following the tip tnich gave to me. Soon I'll organize the solution and I'll post here as well.

Homework Statement Let ##C \subset \mathbb{R}^n## a convex set. If ##x \in \mathbb{R}^n## and ##\overline{x} \in C## are points that satisfy ##|x-\overline{x}|=d(x,C)##, proves that ##\langle x-\overline{x},y-\overline{x} \rangle \leq 0## for all ##y \in C##. Homework Equations By definition...

In fact vela, we can write any ##x## in the line determined by the line segment ##[a,b]## as ##x = c + k(b-a)##. But follow your tip I stucked at the following red question mark:

Very well noticed Ray Vickson! Ok I can reformulate the question as follow: Homework Statement Given ##a\neq b## vectors of ##\mathbb{R}^n##. Determine ##c## which lies in the line ##r## determined by in the line segment ##[a,b]=\{a+t(b-a) ; t \in [0,1]\}##, such that ##c \perp r##. Conclude...

Homework Statement Given ##a\neq b## vectors of ##\mathbb{R}^n##. Determine ##c## which lies in the line segment ##[a,b]=\{a+t(b-a) ; t \in [0,1]\}##, such that ##c \perp (b-a)##. Conclude that for all ##x \in [a,b]##, with ##x\neq c## it is true that ##|c|<|x|##. Homework Equations The first...

Great mathwonk! I'm just arranging a complete solution using your argument in order to help others that could reach this question. Let ##x,y \in \mathbb{R}^n## not null vectors. Let us suppose that for all ##z \in \mathbb{R}^n## which is orthogonal to ##x## it is true that ##z## is ortogonal to...

Ok I got your point. But let me repharse the proof above and try to express myself better in order to show that the only problem with the proof above (at least for me) is the fact that ##(*)## and ##x-y \neq 0## implies that ##y-\frac{<x,y>}{|x|^2}x = 0##. Let's begin: Let us suppose that...

Homework Statement Let ##x,y \in \mathbb{R^n}## not null vectors. If for all ##z \in \mathbb{R^n}## that is orthogonal to ##x## we have that ##z## is also orthogonal to ##y##, prove that ##x## and ##y## are multiple of each other. Homework Equations We can use that fact that ##<x ...