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,
\(f(t(x_1,y_1) + (1-t) (x_2,y_2))=|t x_1 + (1-t) x_2| + |t y_1 + (1-t) y_2|\)
\(\le |t x_1| + |(1-t) x_2| + |t y_1| + |(1-t) y_2|=\)
\(= t| x_1| + (1-t) |x_2| + t| y_1| + (1-t)| y_2|=\)
\(= t(| x_1|+| y_1|) + (1-t) (|x_2| + | y_2|)=\)
\(= t f(x_1,y_1) + (1-t) f(x_2,y_2)\)