Suppose $f_1(x)$, $f_2(x)$ are convex. Prove that the vector valued function $f(x)=(f_1(x),f_2(x))$ is convex.
I can't seem to figure this one out no matter how hard I try. Here's my attempt.
Need to show: $(1): f(\lambda(x)+(1-\lambda)y) \leq \lambda f(x)+(1-\lambda)f(y)$.
Since $f_1(x)$ is convex, we have:
$f_1(\lambda(x)+(1-\lambda)y) \leq \lambda f_1(x)+(1-\lambda)f_1(y)$
Since $f_2(x)$ is convex, we have:
$f_2(\lambda(x)+(1-\lambda)y) \leq \lambda f_2(x)+(1-\lambda)f_2(y)$
Now starting with the left-hand side of $(1)$, we have:
\begin{align*}
f(\lambda(x)+(1-\lambda)y)&=(f_1(\lambda(x)+(1-\lambda)y),f_2(\lambda(x)+(1-\lambda)y))\\
&\leq (\lambda f_1(x)+(1-\lambda)f_1(y),\lambda f_2(x)+(1-\lambda)f_2(y))\\
&=(\lambda f_{1}(x), \lambda f_2(x))+((1-\lambda)f_1(y), ((1-\lambda) f_2(y))\\
&=\lambda (f_1(x), f_2(x))+(1-\lambda)(f_1(y),f_2(y))\\
&=\lambda f(x)+(1-\lambda)f(y)
\end{align*}
I'm not sure if the second step where I have the inequality is correct, I use the convexity of $f_1$ and $f_2$ . Overall, I'm just not very confident. Thanks for the help.
Best Answer
Just to answer the question: the proof you gave is correct. (But it's not standard to describe a function $f:\mathbb R^n \to \mathbb R^n$ as "convex".)