$\lambda f(x)+(1-\lambda)f(y)=f(\lambda x+(1-\lambda)y)$ implies $f $ linear?

Question

Let $X$ be a Banach space. $f:X \to X$ a continuous function. If we assume that $f$ satisfies the following convexity condition: $$\lambda f(x)+(1-\lambda)f(y)=f(\lambda x+(1-\lambda)y),$$ for all $x,y\in X$ and $0\leq \lambda \leq 1$. Does this imply that $f$ is automatically linear ?

Answer 1

If you add $f(0)=0$ then you get $f(\lambda x)=\lambda f(x)$ for $\lambda\in[0,1]$ and then $$0=f\left(\frac{1}{2}x + \frac{1}{2}(-x)\right) = \frac{1}{2}f(x)+\frac{1}{2}f(-x)$$ so $$f(-x)=-f(x).$$

Next, show that $f(ax)=af(x)$ for all $a>1$. This shows it is actually true for all $a\in\mathbb R$.

Then you can show that $$\frac{1}{2}f(x+y)=f\left(\frac{1}{2}x + \frac{1}{2}y\right)=\frac{1}{2} f(x) + \frac12 f(y)$$ so $$f(x+y)=f(x)+f(y).$$