How can I prove this set is convex?

Question

I'm given the problem:

The function $f:\mathbb R^n\to\mathbb R$ is convex, $a\in\mathbb R^n$, $b\in\mathbb R$: $$\mathcal B = \lbrace x\in\mathbb R^n : f(x) = 0,\langle a,x\rangle\leq b,x\geq 0\rbrace$$

It's easy to check that if $x_1,x_2 \in B$, then $(1-λ)x_1 + λx_2$ satisfies the last two requirements which are $\langle \alpha，x\rangle \le b $, and $x \ge 0$.

But when I tried to verify the first requirement I get stuck.

This is what I get:

$$f((1-\lambda)x_1+\lambda x_2)\leq (1-\lambda)f(x_1)+\lambda f(x_2)$$ Since $f(x_1) = f(x_2) = 0$, we have that: $$f((1-\lambda x_1)+\lambda x_2)\leq (1-\lambda) 0+\lambda 0 = 0$$

but I don't know how can I prove it is equal to $0$.

Can anyone give a hint?

Answer 1

I don't think this is true, unless you impose additional conditions.

Suppose, for example, that $n=1$, $f(x)=(x-2)^2 - 4$, $a=1$ and $b=4$. Then,

$$ \mathcal{B}=\{0,4 \}$$

which is not convex.