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How can I prove this?

I saw a similar question here: (But this was only for when g(x) is ≥0) Prove that a set defined by concave functions on $R^n$ is convex

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Alex Parker
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Let $$S=\{g\colon \mathbb{R}^n\to \mathbb{R}\ \text{s.t}\ g\ \text{concave}\}$$ For $a,b\in[0,1]$ and $x,y\in\mathbb{R}^n$ and $g,f\in S$ we have: \begin{align} (af+(1-a)g)(bx+(1-b)y) &=af(bx+(1-b)y)+(1-a)g(bx+(1-b)y)\\ &\geq a(bf(x)+(1-b)f(y))+(1-a)(bg(x)+(1-b)g(y))\\ &=b(af(x)+(1-a)g(x))+(1-b)(af(y)+(1-a)g(y))\\ &=b(af+(1-a)g)(x) +(1-b)(af+(1-a)g))(y). \end{align}

Then, $af+(1-a)g$ is concave so it belongs $S$ and hence $S$is convex set.