Prove that convex functions satisfy $(1+\lambda)f(y_0) - \lambda f(y_0-r)\le f(y_0+\lambda r) \le (1-\lambda)f(y_0) + \lambda f(y_0+r)$

Question

$f$ is said to be convex if $$f(\lambda x + (1-\lambda)y) \le \lambda f(x)+(1-\lambda)f(y)$$ for every $x,y\in (a,b)$ and $\lambda \in [0,1]$. If $f$ is convex, $r \in \mathbb R$ and $[y_0-|r|,y_0+|r|] \subset (a,b)$, prove that $$(1+\lambda)f(y_0) - \lambda f(y_0-r)\le f(y_0+\lambda r) \le (1-\lambda)f(y_0) + \lambda f(y_0+r)$$ for every $\lambda \in [0,1]$.

The RHS is easy (by substituting $y=y_0$ and $x=y_0+r$) but I'm not sure how to prove the LHS from the definition of convexity.

I know both inequalities can be proven easily using geometric arguments but I want to see an algebraic proof if it is possible.

Answer 1

The LHS can be written as $\displaystyle\,f(y_0) \le \frac{\lambda}{1+\lambda} f(y_0-r) + \frac{1}{1+\lambda}f(y_0+\lambda r)\,$, which follows from the definition of convexity since:

• $\displaystyle \frac{\lambda}{1+\lambda}, \frac{1}{1+\lambda} \in [0,1]\,$ and $\,\displaystyle \frac{\lambda}{1+\lambda} + \frac{1}{1+\lambda} = 1\,$;

• $\require{cancel} \displaystyle \frac{\lambda}{1+\lambda}(y_0-\bcancel{r})+\frac{1}{1+\lambda}(y_0+\bcancel{\lambda r})=y_0\,$.

Answer 2

Let $\alpha := \lambda/(1+\lambda)$ so that $(1-\alpha) = 1/(1+\lambda)$, both within [0,1].

Then rewriting your expression you have

$f(y) \leq \alpha f(y-r) + (1-\alpha)f(y + \alpha/(1-\alpha)r$

But by definition of convexity, the RHS of the above satisfies

$\alpha f(y-r) + (1-\alpha)f(y + \alpha/(1-\alpha)r \geq f(\alpha(y-r) + (1-\alpha)(y+r \alpha/(1-\alpha))$,

the RHS of which can be simplified to

$f(\alpha y - \alpha r + y - \alpha y + \alpha r) = f(y)$. And we're done.