# How to prove that is a cone

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 $\lambda\ge 0$.

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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 $t \in [0,1]$.

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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