如何证明集合X是凸集？X = {x ∈ ℝ² | x₁² ≤ x₂, x₁≥0, x₂≥0}

Hey there! Let's start by covering the standard approach to proving a set is convex, then apply that method to the set ( X ) you've defined. This is a classic convexity problem, so breaking it down step by step will make it clear.

First, let's recap the core definition of a convex set—this is the foundation of every proof:

A set ( S \subseteq \mathbb{R}^n ) is convex if, for any two points ( x, y \in S ) and any scalar ( \lambda \in [0,1] ), the convex combination ( \lambda x + (1-\lambda)y ) is also in ( S ).

Here's the step-by-step workflow to prove convexity:

• Step 1: Pick arbitrary points. Grab two generic points ( x ) and ( y ) that belong to the set—don't use specific values, keep them general.
• Step 2: Define the convex combination. Let ( z = \lambda x + (1-\lambda)y ) where ( 0 \leq \lambda \leq 1 ). Write out each component of ( z ) explicitly (e.g., for 2D sets, ( z_1 = \lambda x_1 + (1-\lambda)y_1 ), ( z_2 = \lambda x_2 + (1-\lambda)y_2 )).
• Step 3: Verify the set's conditions. Since ( x ) and ( y ) are in the set, they satisfy all the set's defining rules. You need to show ( z ) meets every single one of those rules too.
• Step 4: Conclude. If the convex combination ( z ) satisfies all conditions, the set is convex by definition.

Let's apply the method above to set ( X ). We need to show that for any ( x = (x_1, x_2) \in X ), ( y = (y_1, y_2) \in X ), and ( \lambda \in [0,1] ), the convex combination ( z = \lambda x + (1-\lambda)y ) is also in ( X ). That means checking three things for ( z ):

1. ( z_1 \geq 0 )
2. ( z_2 \geq 0 )
3. ( z_1^2 \leq z_2 )

Let's tackle each condition one by one:

1. ( z_1 \geq 0 ) #

Since ( x \in X ) and ( y \in X ), we know ( x_1 \geq 0 ) and ( y_1 \geq 0 ). The scalars ( \lambda ) and ( (1-\lambda) ) are both non-negative (because ( 0 \leq \lambda \leq 1 )). Adding non-negative terms gives a non-negative result:
( z_1 = \lambda x_1 + (1-\lambda)y_1 \geq 0 + 0 = 0 )
Done.

2. ( z_2 \geq 0 ) #

Same logic applies here! ( x_2 \geq 0 ), ( y_2 \geq 0 ), and our coefficients are non-negative. So:
( z_2 = \lambda x_2 + (1-\lambda)y_2 \geq 0 + 0 = 0 )
That's straightforward.

3. ( z_1^2 \leq z_2 ) #

This is the critical part. Let's start by expanding ( z_1^2 ):
( z_1^2 = (\lambda x_1 + (1-\lambda)y_1)^2 = \lambda^2 x_1^2 + 2\lambda(1-\lambda)x_1 y_1 + (1-\lambda)^2 y_1^2 )

Since ( x \in X ), we have ( x_1^2 \leq x_2 ); similarly, ( y_1^2 \leq y_2 ). Substitute these into the expansion to get an upper bound for ( z_1^2 ):
( z_1^2 \leq \lambda^2 x_2 + 2\lambda(1-\lambda)x_1 y_1 + (1-\lambda)^2 y_2 )

Now we need to show this upper bound is less than or equal to ( z_2 = \lambda x_2 + (1-\lambda)y_2 ). Let's rearrange the inequality we need to prove:
( \lambda^2 x_2 + 2\lambda(1-\lambda)x_1 y_1 + (1-\lambda)^2 y_2 \leq \lambda x_2 + (1-\lambda)y_2 )

Subtract the right-hand side from both sides to simplify:
( \lambda^2 x_2 - \lambda x_2 + (1-\lambda)^2 y_2 - (1-\lambda)y_2 + 2\lambda(1-\lambda)x_1 y_1 \leq 0 )

Factor terms with ( x_2 ) and ( y_2 ):
( \lambda(\lambda - 1)x_2 + (1-\lambda)(1-\lambda - 1)y_2 + 2\lambda(1-\lambda)x_1 y_1 \leq 0 )

Simplify the coefficients:
( -\lambda(1-\lambda)x_2 - \lambda(1-\lambda)y_2 + 2\lambda(1-\lambda)x_1 y_1 \leq 0 )

Factor out ( -\lambda(1-\lambda) ) (note that ( \lambda(1-\lambda) \geq 0 ) for ( 0 \leq \lambda \leq 1 )):
( -\lambda(1-\lambda)\left( x_2 + y_2 - 2x_1 y_1 \right) \leq 0 )

Now, recall ( x_1^2 \leq x_2 ) and ( y_1^2 \leq y_2 ). So ( x_2 + y_2 \geq x_1^2 + y_1^2 ). Substitute this into the parentheses:
( x_2 + y_2 - 2x_1 y_1 \geq x_1^2 + y_1^2 - 2x_1 y_1 = (x_1 - y_1)^2 \geq 0 )

Since ( (x_1 - y_1)^2 ) is always non-negative, ( x_2 + y_2 - 2x_1 y_1 \geq 0 ). Multiplying this non-negative value by ( -\lambda(1-\lambda) ) (which is non-positive) gives a result that's ≤ 0. Exactly what we needed!

This means ( z_1^2 \leq z_2 ) holds true.

Since all three conditions are satisfied for the convex combination ( z ), ( z \in X ). By the definition of convex sets, ( X ) is convex.

内容的提问来源于stack exchange，提问作者r_dub_