Let $U \subset \mathbb{R}^n =: E$ be an open convex set and $\phi: U \to \mathbb{R}$ a convex function. Show the function is continuous.
This exercise gives steps to follow to complete problem. I am confused about the first hint in the following image 
Why is it 1) relevant to notice this, and 2) why is true that $\phi$ is continuous at $x_0$ iff $\phi^{~}(x) =: \phi(x_0 + x)$ s continuous at $0$?
1) This may simplify so,e expressions later in the proof (some constants that need not be carried all the way)
2) $\let\phi\varphi \phi$ is continuous at $x_0$ means, loosely speaking, that $x\approx x_0$ implies $\phi(x)\approx \phi(x_0)$. Now with $\tilde\phi(x):=\phi(x_0+x)$ this just means that $x\approx 0$ (and so $x+x_0\approx x_0$) implies $\tilde\phi(x)\approx \tilde\phi(0)$. In other words $\tilde\phi$ is continuous at $0$.