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Let $f \in C([0;1])$ be a non-derivable continous convex function with three aligned points, i.e. $\exists \ a \in ]0;1[ : (0,f(0)), (a,f(a)), (1,f(1)) \in L$ for some line $L \subseteq \mathbb{R}^2$.

I need to show that $f$ is affine over $[0;1]$.

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    that is not true. You can easily find a non linear (even non polynomial) function wwith such properties – 2017-01-09
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    Among the missing conditions, you need three **different** points. – 2017-01-09
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    Yes ! i forgot some details... I corrected it ! – 2017-01-10

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This is simply not true. The function, which graph is a broken line, serves as a counterexample. Take the broken line joining the points $(0,1)$, $(0.25,0)$, $(0.75,0)$ and $(1,1)$.

The assertion is true, whenever $a=0$ and $c=1$. By convexity it is easy to see that if there exists $d\in(0,1)$ s.t. $\bigl(d,f(d)\bigr)$ lies below $L$, then $\bigl(b,f(b)\bigr)$ lies below $L$, which contradicts our hypothesis.

With your hypothesis it could be shown that $f$ is linear in the interval $\bigl[\min\{a,b,c\},\max\{a,b,c\}\bigr]$.

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    Thank you for explaining, I indeed forgot a condition (that i just corrected). – 2017-01-10
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    Are you now able to write a proof using my hint? – 2017-01-10
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    Yes I think a proof would be : Suppose there exists $d \in (0;1) : (d,f(d))$ lies below L, then by convexity, f is below D, where D is the line passing through $(0,f(0))$ and $(d,f(d))$. It's easy to see that D is below L on $(0,1)$. Therefore, $(b,f(b))$ should also be below D, which is absurd since it belongs to L which is above D. – 2017-01-10
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    Of course. Nevertheless this argument works whenever $b$\bigl(d,f(d)\bigr)$ and $\bigl(1,f(1)\bigr)$. – 2017-01-10