Question

Se \(g: \mathbb{R}^2\rightarrow \mathbb{R}\) e \(h: \mathbb{R}^2\rightarrow \mathbb{R}\) são funções convexas, então \(f(x,y) = g(x,y) + h(x,y)\) é convexa?

Answer 1

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.