Problem Statement
Let $g_{1},\dots, g_{M}$ be convex functions and let $b_{1},\dots, b_{M}$ be real numbers. Show that $F:= \{ \boldsymbol{x} \in \mathbb{R}^{d} : g_{i}(\boldsymbol{x}) \leq b_{i}, $ for all $i = 1,\dots , M \}$ is a convex set.
Attempted Proof
We endeavour to prove that a given set, $F:= \{ \boldsymbol{x} \in \mathbb{R}^{d} : g_{i}(\boldsymbol{x}) \leq b_{i} \forall i = 1,\dots , M \}$ is convex. Where $\{g_{i}\}_{i=1}^{M}$ represents convex functions and $\{b_{i}\}_{i=1}^{M}$ represents real integers.
We know, from the definition of a convex set that the feasible set $F:= \{ \boldsymbol{x} \in \mathbb{R}^{d} : g_{i}(\boldsymbol{x}) \leq b_{i}, $ for all $i = 1,\dots , M \}$ is convex on the condition that all the functions $\{g_{i}\}_{i=1}^{M}$ are convex.
Explanation
I am trying to show that the set is convex by taking two points x and y which are in the set and show that for any $\lambda \in [0,1]$, the point $\lambda \boldsymbol{x} + (1-\lambda) \boldsymbol{y} $ is also in the set. Where a point will be in the set if it satisfies all the constraints.
The problem is I am struggling to take my first attempted at a proof and use a more mathematically method to prove this by taking two points in the set as described above. I would like to use the definition of a convex function ,i.e. a convex function $f$ , is one that satisfies for $0 \leq \lambda \leq1$,
\begin{eqnarray*}
f(\lambda \boldsymbol{x} +(1-\lambda)\boldsymbol{y} ) \leq \lambda f(\boldsymbol{x}) + (1-\lambda)f(\boldsymbol{y}) \text{.}
\end{eqnarray*}
I have found similar questions with regards to the sum or quotient of convex functions online but I have not come across a proof of this with a good solution so far. Could someone please show me how they would formulate this proof.
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From convex set to convex function
Best Answer
Let $g_1, g_2, \ldots, g_M$ be convex functions from $\mathbb{R}^d \to \mathbb{R}$ and $b_1, b_2, \ldots, b_M \in \mathbb{R}$. We want to show that $F := \{ x \in \mathbb{R}^d : g_i(x) \leq b_i, \, i \in 1, 2, \ldots, M\}$ is convex.
To start, suppose that $x_1 \in F$ and $x_2 \in F$ are arbitrary points in $F$ and that $t \in (0,1)$. We want to show that $x_3 = tx_1 + (1-t)x_2 \in F$.
First, consider an arbitrary $i \in 1,2,\ldots, M$. By definition of the convexity of $g_i$, we know that $g_i(x_3) = g_i(tx_1 + (1-t) x_2) \leq tg_i(x_1) + (1-t)g_i(x_2)$. By definition of $F$, we know that $tg_i(x_1) + (1-t)g_i(x_2) \leq tb_i + (1-t)b_i = b_i$. Thus we have that $g_i(x_3) \leq b_i$, and we can repeat this argument for all $i$. Thus $x_3 \in F$. Hence $F$ is a convex set.