let $\phi \subset C[0,1]$ be the set $\phi =\{f\in C^1[0,1]: \int^1_0 |f'(x)|^3dx\le 7\}$ prove that $\phi$ is equicontinous
Def (Equicontunous ):
A collection of functions on $\phi \subset C[0,1]$ Equicontunous if given $\epsilon>0$ there exist $\delta>0$ such that whenever $x,y\in \phi$ satisfy $d(x,y)<\delta, |\phi(x)-\phi(y)|<\epsilon$
how to proceed with this?
sorry I'm not getting any idea
Your definition of equicontinuity is confused. Here is a more precise statement of (uniform) equicontinuity: a family $\mathcal{F}\subset C[0,1]$ is uniformly equicontinuous if for all $\epsilon>0$ there is $\delta>0$ such that, whenever $x,y\in[0,1]$ satisfy $d(x,y)<\delta$, then $d(f(x),f(y))<\epsilon$ for all $f\in \mathcal{F}$.
Now, let $f\in\mathcal{F}$ as defined in your question; suppose $x\leq y$ so the fundamental theorem of calculus gives us the estimate $$|f(y) - f(x)| = \left|\int_x^yf'(t)\,dt\right|\leq \int_x^y |f'(t)|\,dt.$$ From here we need to somehow relate the integral of $|f'|$ to the integral of $|f'|^3$. This should naturally bring to mind the Hölder inequality. We use exponents $p=3$ and $q=3/2$ to find $$\int_x^y |f'(t)|\cdot 1\,dt\leq \left(\int_x^y|f'(t)|^3\,dt\right)^{1/3}\left(\int_x^y 1^{3/2}\right)^{2/3}\leq 7^{1/3}(y-x)^{2/3}.$$ Thus we have $|f(y)-f(x)|\leq 7^{1/3}(y-x)^{2/3}$ for $x\leq y$, which easily implies $|f(y)-f(x)|\leq 7^{1/3}|y-x|^{2/3}$ for all $x,y\in[0,1]$. Since this estimate is true for all $f\in\mathcal{F}$ equicontinuity follows, since for given $\epsilon>0$ we can choose $\delta = (\epsilon/7^{1/3})^{3/2}$.