# Continuous function satisfying $f\left(\dfrac{x+t}{2}\right) \le f(x) + f(t)$ inequality must be $0$

Let $$f$$ a real function defined and continuous on $$[0,1]$$ such that

$$f(0)=f(1)=0$$ $$f\left(\dfrac{x+t}{2}\right) \le f(x) + f(t)$$ for all $$x,t$$

prove that $$f$$ is zero.

My try was proving first that f is nonnegative (no problem) then using the fact that $$f([0,1]) = [0,M]$$ try to prove that $$M$$ must be zero.

by contradiction if I assume that $$M=f(\alpha)>0$$ then continuity of $$f$$ must be positive on a whole neighbourhood of $$\alpha$$. but then I was stuck, trying to draw from here a contradiction.

Any advice would be greatly appreciated.

• I guess, in general, if we replace the two boundary conditions by two general points, then the resulting function must be the straight line between the two points? Oct 26, 2022 at 10:03
• @BenjaminWang: I don't think so. Any non-negative convex function satisfies that inequality. For example, $f(x) = x^2$ is a solution with $f(0) = 0$ and $f(1) = 1$. Oct 26, 2022 at 14:18
• @MartinR thanks. By the way, the question and each answer all have 5 votes. Wow. Oct 26, 2022 at 23:06
• @MartinR The special thing with these boundary conditions is that if $a,b\leq 0$ then also $a+b\leq0$. So even if we replace the boundary condition to $f(0)=f(1) = c > 0$ there can be non constant solutions to the problem.
– Lazy
Oct 27, 2022 at 6:56
• @BenjaminWang Sadly this equilibrum has been broken, as the answers have six upvotes now. But do not fret, I’ve done all I can to restore equilibrum...
– Lazy
Oct 27, 2022 at 18:26

Hint

Let $$f:[0,1]\to \mathbb R$$ be continuous and s.t. $$f\left(\frac{x+t}{2}\right)\leq f(x)+f(t),\tag{P}$$ for all $$x,t\in [0,1]$$.

• Let $$\mathcal D=\bigcup_{n\in\mathbb N}\left\{\frac{k}{2^n}\mid k\in\{0,...,2^n\}\right\},$$ the set of dyadic numbers in $$[0,1]$$. It's a dense set in $$[0,1]$$.

• Using $$\text{(P)}$$, one can prove that $$f(x)\geq 0$$ for all $$x\in [0,1]$$ and that $$f(u)=0$$ for all $$u\in \mathcal D$$.

• Using a density argument, it follows that $$f\equiv 0$$.

You do know that $$f$$ cannot be negative: If $$f(x) < 0$$ then $$f(x/2) \leq f(0)+f(x) = f(x)$$, and thus you can construct a sequence $$x_n\to 0$$ such that $$f(x_n)\leq f(x) < 0$$.

But similarly wedo know that $$f$$ cannot be positive, as we do know: If $$f(x),f(y)\leq 0$$ then also $$f((x+y)/2) \leq 0$$. As we know that $$f(0),f(1) = 0$$ this implies that for each point of the form $$q/2^k$$ ($$0\leq q\leq 2^k$$) we have $$f(q/2^k) \leq 0$$ (this follows from induction over $$k$$):

For $$k=0$$ this is simply $$f(0),f(1)=0$$. For $$k+1$$ we only need to consider $$q$$ odd, so $$q=2l+1$$. Then $$q = (x+y)/2$$ for $$x = l/2^k$$ and $$y=(l+1)/2^k$$. Thus we get this property inductively.

As these points are dense in $$[0,1]$$ we thus immediately get $$f\leq 0$$. Combining this you get $$f\equiv 0$$.

• $f(x) \ge 0$ follows more easily from $f(x) = f\left(\frac{x+x}{2}\right) \le f(x) +f(x) = 2f(x)$. Oct 26, 2022 at 11:25
• @MartinR Oh, that is cool!
– Lazy
Oct 26, 2022 at 14:05
• @MartinR you should post it as an answer. Probably the easiest solution of all. Oct 26, 2022 at 23:05
• @BenjaminWang Problem is though that the OP already has $f\geq 0$, but he struggles with $f\equiv 0$.
– Lazy
Oct 27, 2022 at 6:52

Let $$x\in[0,1]$$. We construct two sequences $$\{a_n\}_{n\in\mathbb{N}}$$, $$\{b_n\}_{n\in\mathbb{N}}$$ the following way:

Let $$a_0=0$$, $$b_0=1$$.

Now fix $$n\in\mathbb{N}$$. By how we construct this sequence (which will be clear soon), we have that $$x\in[a_n,b_n]$$. Set $$\xi=\frac{a_n+b_n}{2}$$. If $$x\in[a_n,\xi]$$, set $$a_{n+1}=a_n$$ and $$b_{n+1}=\xi$$, otherwise set $$a_{n+1}=\xi$$ and $$b_{n+1}=b_n$$.

In particular by this construction, the two sequences are bounded and monotone, so they converge, i.e. $$a_n\to a$$ and $$b_n\to b$$ as $$n\to\infty$$. Now notice that

$$\lvert b_n-a_n\rvert=\frac{1}{2^n},$$

and so letting $$n\to\infty$$, we get that $$a=b$$.

Now notice furthermore that, since you've shown that $$f$$ is non-negative,

$$0\leq f\left(\frac{a_n+b_n}{2}\right)\leq f(a_n)+f(b_n),$$

and so since $$f(a_0)=f(b_0)=0$$, and all terms in the sequences are either on the form $$\frac{a_k+b_k}{2}$$, or $$a_0$$ or $$b_0$$, it follows by the above inequality that $$f(a_n)=f(b_n)=0$$ for all $$n\in\mathbb{N}$$. In particular then,

$$f(a)=\lim_{n\to\infty}f(a_n)=0.$$

Finally, notice that by construction, we always have that $$a_n\leq x\leq b_n$$ for all $$n\in\mathbb{N}$$, and so letting $$n\to\infty$$, we get that (since $$b=a$$) $$x=a$$. Thus

$$f(x)=f(a)=0.$$

Since $$x$$ was arbitrary, this means that $$f(x)=0$$ for all $$x\in[0,1]$$, and so $$f$$ is the zero function.