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

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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.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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