The function $f$ is convex when:
$$f(tx+ (1-t)x', ty + (1-t)y') ≤ tf(x,y) + (1-t)f(x', y')$$
for all numbers $0≤t≤1$ and all pairs of points $(x,y)$ and $(x',y')$ in the domain of $f$ (the whole of $R^2$)
How would I interpret this geometrically? Also what is an example of a convex function whose graph is not a plane?
On
Parameter $t$ is a fraction corresponding to a relative position on a line segment.
Value $tx+(1−t)x′$ is a point in the interval $[x,x']$ chosen by $t$: it is $x'$ for $t=0$ and $x$ for $t=1$.
Similary, $ty+(1−t)y′$ is a corresponding point in the $[y,y']$ interval.
If we name the points $A(x,y)$ and $A'(x',y')$, we get a point in the $AA'$ segment $$P=At+A'(1-t)$$
As a result, parameter $t$ on the LHS of the inequality defines a point at corresponding relative position of the $AA'$ line segment.
The RHS displays a similar relative division of the $f$ values' segment, corresponding to $A$ and $A'$.
Then the inequality, as a whole, says: the value of $f$ at any point $P$ of a line segment $AA'$ is not above the corresponding point of a chord between point $(A,f(A))$ and $(A',f(A'))$.
In $1$ dimension: $$ f(tx+(1-t)y)\le tf(x)+(1-t)f(y) $$
$tx+(1-t)y, \; t \in [0;1]$ describes the line segment between $x$ and $y$ and $tf(x)+(1-t)f(y), \; t \in [0;1]$ describes the line segment between $f(x)$ and $f(y)$. So the inequality means that the image of the line segment $tx+(1-t)y$ (the green curve) lies beneath the line segment between the images of $x$ and $y$ (the pink curve).