What is the definition of convexity from $f : \mathbb{R}^2 \rightarrow \mathbb{R}$?

Question

$f(\lambda x + (1-\lambda y) \leq \lambda f(x) + (1- \lambda) f(y)$. This is the definition of convexity I am used to. If $f$ is a convex function, then $f : \mathbb{R} \rightarrow \mathbb{R}$.

What if I am doing something like $f : \mathbb{R}^2 \rightarrow \mathbb{R}$? Is the definition of convexity different or is it still the same?

So instead of $f(x)$ I had $f(x,y)$. If the definition is different what would it look like then?

Yes, the definition generalizes as you said. http://en.wikipedia.org/wiki/Convex_function#Definition — Timothy Shields, Feb 24 '15 at 21:01
You confuse notation of $x,y$. So assume $x\in R^n$. The convexity $f(x)$ is defined in the same way. — Alexander Vigodner, Feb 24 '15 at 21:06
So if $Z \subseteq \mathbb{R}^2$ is a convex subset with $(x,y) \in Z$, I can take something like $f(x,y)$? Or no? I was thinking the definition would be something like $f(\lambda (x,y) + (1- \lambda) (x',y')) \leq ...$ — Jake Barrows, Feb 24 '15 at 21:13

score 1 · Accepted Answer · answered Feb 24 '15 at 23:06

No, the definition is formally the same, namely \begin{align*} f(\lambda x+(1-\lambda)y)\leq\lambda f(x)+(1-\lambda)f(y),\qquad x,y\in\mathbb R^2,\lambda\in[0,1]. \end{align*} Note that here $x$ and $y$ are points in $\mathbb R^2$ and $\lambda x+(1-\lambda)y$ runs along the line joining $x$ and $y$ when $\lambda$ runs through $[0,1]$.

What is the definition of convexity from $f : \mathbb{R}^2 \rightarrow \mathbb{R}$?

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