Prove that if $f: \mathbb{R} \to \mathbb{R}$ is convex and bounded from above, then $f$ is constant.

Question

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 $$f(x_1) \leq f(x_3)$$

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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 ?

Answer 1

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 :).

Answer 2

Let $f(x)\leqslant B$ for all $x\in\mathbb{R}$, and suppose $f(x_1)<f(x_2)$ for some $x_1,x_2\in\mathbb{R}$. After possibly replacing $f(x)$ with $f(-x)$, we can assume $x_1<x_2$. Thus the slope of the ray going from $(x_1,f(x_1))$ to $(x_2,f(x_2))$ has strictly positive slope, say

$$s:=\frac{f(x_2)-f(x_1)}{x_2-x_1}>0.$$

Now consider the sequence

$$s_n := \frac{f(n)-f(x_1)}{n-x_1}.$$

Since $f$ is bounded above by $B$, we have

$$s_n \leqslant \frac{B-f(x_1)}{n-x_1}.$$

The numerator of this fraction is constant while the denominator is unbounded as $n\to\infty$, so

$$\lim_{n\to\infty} s_n \leqslant \lim_{n\to\infty} \frac{B-f(x_1)}{n-x_1} = 0.$$

In particular, we can find some $n>x_2$ such that $s_n<s$. But then

\begin{align} f(x_2) &= s(x_2-x_1)+f(x_1) \\ &> s_n(x_2-x_1)+f(x_1) \\ &= (f(n)-f(x_1))\frac{x_2-x_1}{n-x_1} + f(x_1)\\ &= \lambda f(n) - (1-\lambda)f(x_1), \end{align} where $\lambda = \frac{x_2-x_1}{n-x_1}\in (0,1)$, which contradicts convexity.

Answer 3

The idea here is that $\ x_1\ $ and $\ x_3\ $ are fixed but arbitrary real numbers with $\ x_1\le x_3\ .$ Then for any $\ x_2\ge x_3\ $ (no matter how large) you can always express $\ x_3\ $ in the form $$ x_3=\lambda x_1+(1-\lambda)x_2 $$ where $\ \lambda=\frac{x_2-x_3}{x_2-x_1}\in(0,1)\ ,$ and $\ \lambda\rightarrow1\ $ as $\ x_2\rightarrow\infty\ .$ From the convexity and boundedness of $\ f\ $ you then get \begin{align} f(x_3)&=f(\lambda x_1+(1-\lambda)x_2)\\ &\le \lambda f(x_1)+(1-\lambda)f(x_2)\\ &\le \lambda f(x_1)+(1-\lambda)M \end{align} for all $\ x_2>x_3\ .$ Now taking limits as $\ x_2\rightarrow\infty\ $ in this inequality gives $\ f(x_3)\le f(x_1)\ $ because $\ \lambda\rightarrow1\ ,$$\ 1-\lambda\rightarrow0\ ,$ and $\ M\ $ is fixed.

To prove the reverse inequality, you use the same idea, but now with $\ x_2<x_1\ ,$$\ x_2\rightarrow-\infty\ ,$ and $$ x_1=\mu x_3+(1-\mu)x_2\ , $$ where $\ \mu=\frac{x_1-x_2}{x_3-x_2}\in(0,1)\ ,$ and $\ \mu\rightarrow1\ $ as $\ x_2\rightarrow-\infty\ .$ Can you finish it off from here?