Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have to prove that the set $$K=\lbrace u\in C[0,1]\mid u(t)\geq a(t)||u|| \text{ on } [0,1]\rbrace,$$ where $a(t)=\displaystyle\frac{t(2p-t)}{p^2}$ with $\frac12<p<1$, and $||u||=\displaystyle\max_{t\in [0,1]}|u(t)|$, is a cone, that is, $K$ is closed, convex, $K \cap \lbrace -K \rbrace =\lbrace 0\rbrace$, and $\lambda K\subset K$ for any $\lambda\ge 0$.

Please. Thank you.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

You just need to show that if $u \in K$, then $\lambda u \in K$ for any $\lambda >0$. Clearly $u \in K$ iff $u(t) \geq a (t)\|u\|$. If $\lambda >0$, then $\lambda u(t) \geq a (t)\|\lambda u\|$, hence $\lambda u \in K$.

(The definition of cone sometimes includes $\lambda =0$; since $0 \in K$, this definition works too.)

You can show $K$ is closed by showing the defining constraint is continuous. More details:

Since the function $f(u) = \max_{t \in [0,1]} (a(t) \|u\| - u(t))$ is continuous, and $K = f^{-1} (-\infty,0]$, it follows that $K$ is closed.

You can use the following to show $K$ is convex:

If $u,v \in K$, then $a(t)\|u+v\| \leq a(t)\|u\| + a(t)\|v\| \leq u(t)+v(t)$, hence $u+v \in K$. It follows that $K$ is convex.

Showing that $K$ contains no rays follows by noticing that if $u \in K$, then $u(t) \ge 0$ for $t \in [0,1]$.

share|improve this answer
    
Ah , what about ,closed and convex , $K\cap \lbrace -K\rbrace =\lbrace 0 \rbrace$ –  Vrouvrou Apr 27 '13 at 6:22
    
You just asked if it was a cone? –  copper.hat Apr 27 '13 at 6:24
    
Our teacher tell us that a set K is said to be a cone if K is closed, convex and * $(\lambda K)\subset K \forall \lambda \geq 0$ * $K\cap \lbrace-K \rbrace =\lbrace 0 \rbrace $ –  Vrouvrou Apr 27 '13 at 6:27
    
There are various definitions of cones. The least restrictive I have seen is the one I have above. I have added some more detail. –  copper.hat Apr 27 '13 at 6:33
    
thank you ,thank you , please $-K =\lbrace u\in C[0,1], -u(t)\geq a(t)||u||,on [0,1] \rbrace $ ? –  Vrouvrou Apr 27 '13 at 6:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.