Prove that set is convex.

Question

How to prove that these two sets are convex for certain $p$? And for what p they will not be convex?
$$A= \{(x,y) \in \Bbb R^2 : |x|^p+|y|^p \le 1, p \in \Bbb R\}$$ $$B = \{(x,y) \in \Bbb R^2 : x>0,y>0, x^p+y^p \le 1, p \in \Bbb R\}$$

Answer 1

$\forall x,y \in A \rightarrow \{z \in A: z=(1-\lambda)x+\lambda y \in A, \lambda \in [0,1] \}$

I do not understand how the definition will look like applied to the condition of the set

Obviously your problem is understanding how the set is defined:

The set does not contain elements $x$ and $y$, but it cointas elements of the form $(x,y)$.

So $(x_1,y_1)$ and $(x_2,y_2)$ are two elements in the set.

You'll have to prove that $(x_3,y_3) = ((1-\lambda)x_1+\lambda x_2,(1-\lambda)y_1+\lambda y_2)$ also is an element of the set if $(x_1,y_1)$ and $(x_2,y_2)$ are elements of the set and $0<\lambda<1$.

$B = \{ ... \}$

This set could also be written as:

$B=\{(x,y)\in\mathbb R:x>0,y>0,(x,y)\in A\}$

(By the way: Is the notation "$\mathbb R$" correct here or should it be "$\mathbb R^2$"?)

In the first step you will already have proven that $(x_3,y_3)\in A$ if $(x_1,y_1)\in A \land (x_2,y_2) \in A$.

This means you only need to prove that $x_3>0$ and $y_3>0$ if $(x_1,y_1)\in B \land (x_2,y_2)\in B$.

EDIT

It is not obvious what shall to do next

Unfortunately I don't know if I have the right answer here, either. I was thinking that your only problem is understanding the definition of the set.

However I would try to do the following:

The points $(0,1)$ and $(1,0)$ are elements of the set $A$. If the set is convex, the point $(\frac{1}{2},\frac{1}{2})$ is also an element of the set.

This means:

$(\frac{1}{2})^p+(\frac{1}{2})^p\leq 1\\ \Rightarrow 2(\frac{1}{2})^p\leq 1\\ \Rightarrow 2\leq 2^p$

This condition is obviously not true for $p<1$ so the set $A$ can only be convex for $p\geq 1$.

Proving this for the set $B$ will be harder because $(0,1)\notin B$...

You may argue with the points $(1-2\epsilon,2\delta)$, $(2\delta,1-2\epsilon)$ and with limit calculation: If $\delta$ and $\epsilon$ are small enough...

To prove that $A$ is convex for all $p\geq 1$ I'd first define a modified variant of the set $B$:

$C=\{(x,y) : x\geq 0, y\geq 0, |x|^p+|y|^p\leq 1\}=\{(x,y) : x\geq 0, y\geq 0, x^p+y^p\leq 1\}$

The following inequations are true for every element in the set:

$x_1^p + y_1^p \leq 1\\ x_2^p + y_2^p \leq 1$

And because $0\leq\lambda\leq 1$:

$\lambda x_1^p + \lambda y_1^p \leq \lambda\\ (1-\lambda)x_2^p + (1-\lambda)y_2^p \leq 1-\lambda$

$\Rightarrow \lambda x_1^p + (1-\lambda)x_2^p + \lambda y_1^p + (1-\lambda)y_2^p \leq 1$

We can now have a look at the following function:

$f(x)=x^p, x\in[0,\infty), p>1\\ f'(x)=px^{p-1}\geq 0\\ f''(x)=p(p-1)x^{p-2}\geq 0$

When using the properties of convex and concave functions we see that:

$\lambda f(x_1)+(1-\lambda) f(x_2)\geq f(\lambda x_1+(1-\lambda)x_2)$

(I don't know if you are allowed to use this in some university homework!)

For $f(x)=x$, which means $p=1$, this is also true.

This means:

$\lambda x_1^p+(1-\lambda) x_2^p\geq (\lambda x_1+(1-\lambda)x_2)^p$

Or:

$x_3^p+y_3^p = (\lambda x_1 + (1-\lambda)x_2)^p + (\lambda y_1 + (1-\lambda)y_2)^p \\ \leq \lambda x_1^p + (1-\lambda)x_2^p + \lambda y_1^p + (1-\lambda)y_2^p \leq 1$

Therefore $(x_3,y_3)\in C$.

Now we can also define another set:

$D=\{(x,y) : y\geq 0, |x|^p+|y|^p\leq 1\}=\{(x,y) : (x,y) \in C \lor (-x,y) \in C\}$

It should be easy to prove that this set is also convex; we have to distinguish three cases:

• $x_1,x_2\geq 0 \Rightarrow (x_3,y_3)\in C \Rightarrow (x_3,y_3)\in D$
• $x_1,x_2\leq 0 \Rightarrow (-x_3,y_3)\in C \Rightarrow (x_3,y_3)\in D$
• $x_1$ and $x_2$ have different signs:
  This case is a bit more difficult; we first have to prove that the set contains an element $(0,y_0)$ "between" the points $(x_1,y_1)$ and $(x_2,y_2)$...

Now the set $A$ can be written as: $A=\{(x,y):(x,y)\in D\lor (x,-y)\in D\}$.

The steps needed to prove that $A$ is convex when knowing that $D$ is convex are the same steps needed to prove that $D$ is convex when knowing that $C$ is convex.

EDIT 2

I just thought about the cases $p\leq 0$:

My proof that $A$ is not convex unfortunately will not work for $p\leq 0$; however you can easily show that $A$ cannot be convex for $p<0$ because $(x,y)\in A \Rightarrow (-x,-y)\in A$ but $(0,0)\notin A$.

However $B$ seems to be convex for $p<0$!