User:IssaRice/Convex function

a function $f:\mathbf {R} \to \mathbf {R}$ is often defined to be convex if $f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$ for all $x,y\in \mathbf {R}$ and all $t\in [0,1]$.

two ways to think about this:

• if you want to use the idea that the set of wikipedia:convex combinations of two points is the line segment between them: take the points (x, f(x)) and (y, f(y)). then each convex combination has the form $(tx+(1-t)y,tf(x)+(1-t)f(y))$ where t is between 0 and 1. so now we are saying that if we fix some x value tx + (1-t)y along this line segment, then the y value of the y segment, tf(x) + (1-t)f(y), is at least the y value of the function, f(tx + (1-t)y). in other words, the line segment between any two points lies above the function.
• if you want to reason just from middle-school algebra: the line between (x, f(x)) and (y, f(y)) is given by $l(z)=f(x)+{\frac {f(y)-f(x)}{y-x}}(z-x)$. rewrite this as $l(z)=f(x)+(f(y)-f(x)){\frac {z-x}{y-x}}$. if we consider just the line segment between the points, then $z\in [x,y]$. thus ${\frac {z-x}{y-x}}\in [0,1]$. so if we make the substitution $t={\frac {z-x}{y-x}}$ we get $f(x)+(f(y)-f(x))t=(1-t)f(x)+tf(y)$. now you could say x and y are symmetric here, we could swap them around to get the thing in the definition of convex function, or you could substitute again, say s=1-t, to get it in the right form.

t measures what fraction of the distance from x to y we have traversed (just inspect its definition). the x-value for a given value of t is (just solve for z): $z=x+(y-x)t=(1-t)x+ty$. the y-value of the line segment at this point is (1-t)f(x) + tf(y) as we saw above. thus we are saying that the function value at this x-value, f((1-t)x + ty), is below the line segment's y-value, (1-t)f(x) + tf(y).