MHB Proving Convexity of $f: \mathbb{R} \to \mathbb{R}$

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The discussion focuses on proving the convexity of a continuous function \( f: \mathbb{R} \to \mathbb{R} \) given the condition \( f \left( \frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2} \) for all \( x, y \in \mathbb{R} \). The initial approach involves using a parameter \( \lambda \in [0,1] \) to express \( f \) in terms of its values at points \( x \) and \( y \). A key step involves demonstrating that the inequality can be extended to \( f(\lambda x + (1-\lambda)y) \) by proving it holds for dyadic rationals through induction. Finally, the continuity of \( f \) is crucial to extend the result to all \( \lambda \) in the interval, confirming that the function is indeed convex. The discussion emphasizes the importance of continuity in the proof of convexity.
evinda
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Hello! (Wave)

We are given a continuous function $f: \mathbb{R} \to \mathbb{R}$ such that $f \left( \frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2}, \forall x, y \in \mathbb{R}$. I want to show that $f$ is convex.I have tried the following:

Let $\lambda \in [0,1]$.

We have that $f(\lambda x+(1-\lambda)y)=f \left( \frac{2 \lambda x+ 2(1-\lambda)y}{2}\right) \leq \frac{f(2 \lambda x)+f(2(1-\lambda)y)}{2}$.But can we show that the latter is $\leq \lambda f(x)+(1-\lambda)f(y)$ ?Or can't we show it by showing that the definition of convexity is satisfied?

If not, how else can we prove the desired result? (Thinking)
 
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evinda said:
Hello! (Wave)

We are given a continuous function $f: \mathbb{R} \to \mathbb{R}$ such that $f \left( \frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2}, \forall x, y \in \mathbb{R}$. I want to show that $f$ is convex.I have tried the following:

Let $\lambda \in [0,1]$.

We have that $f(\lambda x+(1-\lambda)y)=f \left( \frac{2 \lambda x+ 2(1-\lambda)y}{2}\right) \leq \frac{f(2 \lambda x)+f(2(1-\lambda)y)}{2}$.But can we show that the latter is $\leq \lambda f(x)+(1-\lambda)f(y)$ ?Or can't we show it by showing that the definition of convexity is satisfied?

If not, how else can we prove the desired result? (Thinking)
There does not seem to be any short proof of this result.

As a first step, $f\bigl(\frac14x + \frac34y\bigr) = f\bigl(\frac12\bigl(\frac12(x+y) + y\bigr)\bigr) \leqslant \frac12\bigl(f\bigl(\frac12(x+y)\bigr) + f(y)\bigr) \leqslant \frac12\bigl(\frac12\bigl(f(x) + f(y)\bigr) + f(y)\bigr) = \frac14f(x) + \frac34f(y).$

So the result holds for $\lambda = \frac14$. Now use that same argument to show (by induction on $n$) that the result holds when $\lambda = \dfrac m{2^n}$ ($m=1,2,\ldots, 2^n$).

Finally, use the continuity of $f$ to show that the result holds for all $\lambda \in [0,1]$ (because each such $\lambda$ can be approximated by dyadic rationals). That last step is essential, because it is known that there are discontinuous functions for which the result is false.
 

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