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.

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.
