Proving that $\int^2_0fdx\ge1$ for a convex function Suppose an $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous, non-negtative and convex. It is also given that $f(1)=1$.
Prove that $\int^2_0fdx\ge1$.
This is a question from my Calculus II homework.
I know that I should use these properties of convex functions:
$\forall a,b\in \mathbb{R} \forall \lambda \in[0,1]: f(\lambda a+(1-\lambda)b)\le \lambda f(a)+(1-\lambda)f(b)$
$\forall x\forall x_0:f(x)\ge f(x_0)+f'(x_0)(x-x_0)$
but I can't figure out how they would help me calculate the integral of $f$.
Some help would be appreciated!
 A: First, let’s note that we can actually drop the assumption that $f$ is nonnegative and strengthen the conclusion to $\int_0^2 f(x) dx \geq 2$. Technically, we can even drop the assumption that $f$ is continuous, since that actually follows from convexity.
Step 1: Show $\int_0^2 f(x) dx = \int_0^1 (f(1 + x) + f(1 - x)) dx$.
Step 2: apply convexity to conclude $f(1 + x) + f(1 - x) \geq 2$ for all $x \in [0, 1]$.
Step 3: conclude $\int_0^1 (f(1 + x) + f(1 - x)) dx \geq 2$.
A: Suppose $\int^2_0fdx < 1$, then $f$ must have a point below $x+2y = 2$ (otherwise the integral would be greater than 1).
A: If for all $x < 1$ one has $f(x) \geq 1$ then the desired conclusion is immediate since the integral from $0$ to $1$ is already at least $1$.
Suppose there is some $x < 1$ for which $f(x) < 1$. Then if $y > 1$, by convexity the point $(1,f(1)) = (1,1)$ lies on or below the segment connecting $(x, f(x))$ to $(y, f(y))$. Since $f(x) < 1$, this implies $f(y) > 1$. This will hold for all $y > 1$ so this time the integral from $1$ to $2$ will be greater than $1$.
