Convexity of a right triangle

Question

I am having a hard time proving this statement. I usually have an intuition about the direction to take when doing a proof, but I am really having difficulty with this one. Could anyone give me a hint?

$\{(x_1, x_2)\in\mathbb R^2, x_1 \geq\ 0, x_2 \geq\ 0 \land x_1+x_2 \leq\ 1\}$

Thank you!

Update 1

This is what I have. Is this valid?

Step 1: Lower Bound

$\lambda \in \lbrack 0,1 \rbrack, 0 \leq x_i \leq 1$

The Sum of $x_1 + x_1' \leq 1$ and $x_1 \land x_1'$ are nonnegative. Since $(1 - \lambda)$ is the complement of $\lambda$ $\land \lambda \in \lbrack 0,1 \rbrack$, the sum of $x_1\lambda + (1-\lambda)x_1'$ must also be nonnegative.

Step 2: Upper Bound

Since $x_1 + x_1' \leq 1$, their sum is at most 1. If the sum is 1, then $x_1\lambda+(1-\lambda)x_1'$ is at most 1. Therefore, the upper bound is 1.

Step 3:

The same logic applies for $x_2 \land x_2'$.

Answer 1

Hint:

If $(x_1,x_2)$ and $(y_1,y_2)$ are both points with the described properties, can you show that $(tx_1+(1-t)y_1,tx_2+(1-t)y_2)$ for $0\leq y\leq 1$ also has those properties?

If $x_1,y_1\geq 0$, and $0\leq t\leq 1$, then $tx_1+(1-t)y_1$ is the sum of nonnegative numbers, and therefore nonnegative. The same goes for the second component. If $x_1+x_2\leq 1$ and $y+1+y_2\leq 1$, then $$tx_1+(1-t)y_1+tx_2+(1-t)y_2=t(x_1+x_2)+(1-t)(y_1+y_2)\leq t(1)+(1-t)(1)=1.$$

Answer 2

More generally, the polyhedron $\Delta = \{x \in \mathbb{R}^n| Ax \le b\}$ is convex. Your triangle can be written in this form with the choice $A := \begin{bmatrix}-1\hspace{.5em}0\hspace{.5em}1\\0\hspace{.5em}-1\hspace{.5em}1\end{bmatrix}^T$ and $b := (0, 0, 1)^T$.

On the Convexity of $\Delta$. Now, if $(x, y, t) \in \Delta^2 \times [0, 1]$, then \begin{eqnarray} A(tx + (1-t)y)= tAx + (1-t)Ay \le tb + (1-t)b = b, \end{eqnarray} and so $tx + (1-t)y \in \Delta$, showing that $\Delta$ is convex.

Even more generally, the affine pre-image of a convex set is again convex. This would recover the result just established as a special case, since $\Delta$ is the pre-image of the multi-interval $\prod_{i=1}^n (-\infty, b_i]$ under the affine (in fact linear) map $A$.