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

[evinda]

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)

 

[Opalg]

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.

 

[math771]

Hi there! Interesting problem. Let's see if we can work through this together.

First, let's recall the definition of convexity. A function $f$ is convex if for any two points $x$ and $y$ in its domain and any $\lambda \in [0,1]$, we have $f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y)$.

Now, let's take a closer look at the expression we have: $f(\lambda x + (1-\lambda)y) \leq \frac{f(2\lambda x) + f(2(1-\lambda)y)}{2}$. Notice that the only difference between this expression and the definition of convexity is that we have $\frac{1}{2}$ on the right hand side.

But, we can use the given property of $f$ to our advantage. We can rewrite $f(2\lambda x)$ as $f(\lambda x + \lambda x)$, and similarly for $f(2(1-\lambda)y)$. Then, by the given property, we have $f(\lambda x + \lambda x) \leq \frac{f(\lambda x) + f(\lambda x)}{2} = f(\lambda x)$, and similarly for $f(2(1-\lambda)y)$. So, our expression becomes $f(\lambda x + (1-\lambda)y) \leq \frac{f(\lambda x) + f( (1-\lambda)y)}{2}$.

Now, all we have to do is apply the definition of convexity, and we get $f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y)$, which is exactly what we wanted to show.

I hope this helps! Let me know if you have any other questions.