Prove that if $f: \mathbb{R} \to \mathbb{R}$ is convex and bounded from above, then $f$ is constant.
Our teacher showed us this, and asked us to solve the rest. But I'm a bit confused by what he did.
$$\lambda x_1 + (1 – \lambda)x_2 = x_3$$
$$f(x_3) \leq \lambda f(x_1) + (1 – \lambda) f(x_2)$$
$$\lambda \to 1$$
$$f(x_3) \leq f(x_1)$$
$$\lambda(x_1 – x_2) = x_3 – x_2$$
$$\lambda = \frac{x_3 – x_2}{x_1 – x_2} \to \frac{x_3/x_2 – 1}{x_1/x_2 – 1} \to 1$$
$$x_2 \to \infty$$
$$\lambda := \frac{x_3 – x_2}{x_1 – x_2} \to 1$$
$$x_2 \to \infty$$
$$\lambda = \frac{x_2 – x_3}{x_2 – x_1} < 1$$
$$f(x_3) \leq f(x_1)$$
He asked us to do
$$f(x_1) \leq f(x_3)$$
I'm a bit confused by what he did. I know the definition of a convex function as he used at the start. I also know that bounded from above means $f(x) \leq M$ for a number $M$.
But I don't really see what he is trying to prove. It seems he wants to show $ f(x_1) = f(x_3)$ but I don't really see how what he did proves anything. If $\lambda < 1$ like at the end of his proof, our original equation is
$$f(x_3) \leq \lambda f(x_1) + (1 – \lambda) f(x_2)$$
So if $\lambda < 1$ why $f(x_1) \leq f(x_3)$ ? This seems to hold if $\lambda = 1$ but not for $\lambda < 1$. Or am I missing something ?
Best Answer
To address what your teacher was doing, he is fixing $x_1$ and $x_3$, and then noting that as long as $x_2 \neq x_1$, we can set $\lambda = \frac{x_3 - x_2}{x_1 - x_2}$ to get that $\lambda x_1 + (1 - \lambda) x_2 = x_3$.
We can therefore make a smart choice of $x_2$, and therefore $\lambda$, in order to get that $f(x_3) \leq f(x_1)$. To do this, we note that as $x_2 \to \infty$, we have $\lambda \to 1$. Letting $B$ be the upper bound, so $f(x) \leq B$ for all $x \in \mathbb{R}$, we have that $(1 - \lambda) f(x_2) \leq (1 - \lambda) B$. Thus convexity implies that $f(x_3) \leq \lambda f(x_1) + (1 - \lambda) B$. Now, $B$ and $f(x_1)$ are fixed, so it's not too hard to convince yourself that for any $\epsilon > 0$, we can take $\lambda$ close enough to $1$ to get that $f(x_3) \leq (1 - \epsilon) f(x_1) + \epsilon$. Taking $\epsilon \to 0$, this shows that $f(x_3) \leq f(x_1)$.
Something close to this is your teacher's argument. I haven't included all the details, but this should give you enough of the idea if you're in a calculus course, and if you're in a Real Analysis course, then figuring out the remaining details and the places where I've been imprecise is a great exercise :).