How do you prove that a convex function is continuous?

You can do a proof by contradiction.

1. Assume f∈RR is convex, but not continuous at some x0∈(a,b). This means that: ∃ϵ>0∀δ>0∃x∈(x0−δ,x0+δ):|f(x)−f(x0)|≥ϵ
2. (1) The area is either I or II.
3. (2) The area is either III or IV.

Is a convex function always continuous?

All measurable convex functions on open intervals are continuous. There exist convex functions which are not continuous, but they are very irregular: If a function f is convex on the interval (a,b) and is bounded from above on some interval lying inside (a,b), it is continuous on (a,b).

How do you determine if a function is convex across an interval?

To find out if it is concave or convex, look at the second derivative. If the result is positive, it is convex. If it is negative, then it is concave. To find the second derivative, we repeat the process using as our expression.

How do you prove a concave function is continuous?

This alternative proof that a concave function is continuous on the relative interior of its domain first shows that it is bounded on small open sets, then from boundedness and concavity, derives continuity. Theorem 1. If f : C → R is concave, C ⊂ Rl convex with non-empty interior, then f is continuous on int(C).

What is a convex continuous function?

A convex function is a continuous function whose value at the midpoint of every interval in its domain does not exceed the arithmetic mean of its values at the ends of the interval. More generally, a function is convex on an interval if for any two points and in and any where , (Rudin 1976, p. 101; cf.

How do you prove concave?

If f is twice-differentiable, then f is concave if and only if f ′′ is non-positive (or, informally, if the “acceleration” is non-positive). If its second derivative is negative then it is strictly concave, but the converse is not true, as shown by f(x) = −x4.

How do you know if an equation is convex?

If f and g are concave and a ≥ 0 and b ≥ 0 then the function h defined by h(x) = af(x) + bg(x) for all x is concave. If f and g are convex and a ≥ 0 and b ≥ 0 then the function h defined by h(x) = af(x) + bg(x) for all x is convex.

How do you prove that a quadratic function is convex?

If f is a quadratic form in one variable, it can be written as f (x) = ax2. In this case, f is convex if a ≥ 0 and concave if a ≤ 0. When f (x1,x2,…,xn) is a function in n variables, its graph is given by the equation xn+1 = f (x1,x2,…,xn) and it can be drawn in a coordinate system of dimension n + 1.