10

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.

Jam
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daiski
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    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? – Benjamin Wang Oct 26 '22 at 10:03
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    @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 $f(1) = 1$. – Martin R Oct 26 '22 at 14:18
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    ... and $\min(2x, 1)$ is a concave solution with the same boundary conditions. – Martin R Oct 26 '22 at 14:34
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    @MartinR thanks. By the way, the question and each answer all have 5 votes. Wow. – Benjamin Wang Oct 26 '22 at 23:06
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    @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 '22 at 06:56
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    @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 '22 at 18:26
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    When you mean $f$ on $[0,1]$, are you talking about the domain, codomain, or both? – gist076923 Oct 27 '22 at 20:32
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    @gst076923 It seems quite clear to me that here $f:[0,1]\to\mathbb R$. – Lazy Oct 27 '22 at 20:35

3 Answers3

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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 $f\equiv 0$.

Surb
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9

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.

Lorago
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9

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 $f\equiv 0$.

Lazy
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