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4.6: Funciones convexas y derivados
https://espanol.libretexts.org/Matematicas/Analisis/Introducci%C3%B3n_al_An%C3%A1lisis_Matem%C3%A1tico_I_(Lafferriere%2C_Lafferriere_y_Nguyen)/04%3A_Diferenciaci%C3%B3n/4.06%3A_Funciones_convexas_y_derivados
\ end {array}\] Ya que $f^{\prime}\left(c_{1}\right) \leq f^{\prime}\left(c_{2}\right)$ , tenemos\[t f\left(x_{t}\right)-t f\left(x_{1}\right)=f^{\prime}\left(c_{1}\right) t(1-t)\left(x_{2}-x_{1}\righ...\ end {array}\] Ya que $f^{\prime}\left(c_{1}\right) \leq f^{\prime}\left(c_{2}\right)$ , tenemos $t f\left(x_{t}\right)-t f\left(x_{1}\right)=f^{\prime}\left(c_{1}\right) t(1-t)\left(x_{2}-x_{1}\right) \leq f^{\prime}\left(c_{2}\right) t(1-t)\left(x_{2}-x_{1}\right)=(1-t) f\left(x_{2}\right)-(1-t) f\left(x_{t}\right) .$ reordenando términos, obtenemos $f\left(x_{t}\right) \leq t f\left(x_{1}\right)+(1-t) f\left(x_{2}\right) .$ Por lo tanto, $f$ es convexo.

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