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.
 A: 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.
A: 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 $f\equiv 0$.
A: 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 $f\equiv 0$.
A: Assume as demonstration $x=0$ and $t=1$ then $f(\frac{x+t}{2})=f(\frac{0+1}{2})=f(0.5)\leq f(x)+f(t)=f(0)+f(1)=0+0=0$.
As prove:
We have $f ∈ C([0,1])$. $C(X)$ is the set of continuous functions on $X$. If this is the case the Intermediate value theorem applies. So $f(0)\leq f(0.5)\leq f(1)$ and for all other $x ∈ [0,1]$ is the same valid.
By this theorem $f==0$ on $[0,1]$.
The demonstration underlying this question is an direct one from the Intermediate value theorem.
Since $[0,1]$ is a closed interval in the reals $\mathbb R$ and $f ∈ C([0,1]).
From the demonstration given the $f([0,1])$ must be an interval since f in continuous and $[0,1]$ is an interval. From Intermediate value theorem we get that with the conditions $f(0)=f(1)=$ and minimum and maximum at the interval edges that this interval is the point $0$ without any further work.
For abbrevation I used $C$ as $C^{1}$ for simple continuous function sets. It would not have mattered if the function would be taken with higher order continuity.
