Show every convex function $f$ is quasiconvex.
A measurable function $f: \mathbb{R}\to \mathbb{R}$ is quasiconvex if for every bounded and open $\Omega=(a,b)\in \mathbb{R}, \xi\in\mathbb{R}, \phi \in W_0^{1,\infty}(\Omega,\mathbb{R})$: $$f(\xi)\leq\frac{1}{|\Omega|}\int_a^bf(\xi + \phi'(x))dx.$$
I know f is convex if $f(\lambda x+(1-\lambda y))\leq \lambda f(x)+(1-\lambda)f(y), \lambda\in(0,1)$ or also $f(y)\geq f(x)+\beta(y-x)$ with $\beta\in \mathbb{R}$. My first idea was Jensen inequality but I gave up this idea and literally have no other idea how to prove it. Thanks for any help!!
The definition of quasiconvexity which you gave is slightly incorrect. Instead of $\phi \in W^{1,\infty}_0(I,\mathbb{R})$ it should read $\phi \in W^{1,\infty}_0(\Omega,\mathbb{R})$, keeping all other entities, specifically $\Omega$, defined as before. (Edit: the question was edited accordingly after this answer was written...).
With this correction, the claim immediately follows from Jensens inequality, which directly implies (assuming $b>a$) $$\frac{1}{b-a}\int_a^b f(\xi+\phi^\prime(x))\, dx\ge f\left(\frac{1}{b-a}\int_a^b \xi+\phi^\prime(x)\, dx \right) $$ Now note that the right hand side of this inequality is just $f(\xi)$, since $\phi$ has support in $[a,b]$ (if this is not clear to you you may want to start with $\phi\in C^\infty_0((a,b))$ and then approximate the $\phi$ from the general case).