Precisamos provar que \(\forall t\in [0,1]\) e \(\forall (x_1,y_1),(x_2,y_2)\in \mathbb{R}^2\) temos \(f(t(x_1,y_1) + (1-t) (x_2,y_2))\le t f(x_1,y_1) + (1-t) f(x_2,y_2)\).
Então,
\(f(t(x_1,y_1) + (1-t) (x_2,y_2))=\) \(=g(t (x_1,y_1) + (1-t) (x_2,y_2)) + h(t (x_1,y_1) + (1-t) (x_2,y_2)) \)
Usando o fato que \(g\) e \(h\) são convexas, temos
\(g(t (x_1,y_1) + (1-t) (x_2,y_2))\le tg(x_1,y_1) + (1-t) g(x_2,y_2)\)
e
\(h(t (x_1,y_1) + (1-t) (x_2,y_2))\le t h(x_1,y_1) + (1-t) h(x_2,y_2)\)
Logo,
\(f(t(x_1,y_1) + (1-t) (x_2,y_2))\le\)
\(tg(x_1,y_1) + (1-t) g(x_2,y_2) + t h(x_1,y_1) + (1-t) h(x_2,y_2)=\)
\(=t(g(x_1,y_1) + h(x_1,y_1)) + (1-t) (g(x_2,y_2) + h(x_2,y_2))=\)
\(=t f(x_1,y_1) + (1-t) f(x_2,y_2)\)
Portanto, \(f\) é convexa.