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

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The discussion centers on proving the convexity of a continuous function $f: \mathbb{R} \to \mathbb{R}$ under the condition that $f \left( \frac{x+y}{2}\right) \leq \frac{f(x)+f(y)}{2}$ for all $x, y \in \mathbb{R}$. The proof strategy involves demonstrating that for any $\lambda \in [0,1]$, the inequality $f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y)$ holds. The approach includes using induction on dyadic rationals and leveraging the continuity of $f$ to extend the result to all values of $\lambda$ in the interval.

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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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