About definition of convex from Rudin's analysis This is from Rudin's Mathematical analysis:

We call a set $E \subset R^k$ convex if $\lambda x+(1- \lambda)y \in E$ whenever $x \in E$, $y \in E$, and $0< \lambda <1$. 

What is the significance of that definition? And how we come up with the equation $\lambda x+(1- \lambda)y \in E$?
 A: The expression $\lambda x+(1-\lambda) y$ parametrizes a straight line segment starting at $y$ and ending at $x$. When $\lambda=0$ you have:
$$0\cdot x+(1-0)\cdot y=y$$
and when $\lambda=1$ you have:
$$1\cdot x+(1-1)\cdot y=x,$$
and for $0<\lambda<1$ the point $\lambda x+(1-\lambda)y$ is somewhere over the segment. The nearer $\lambda$ is to $0$, the nearer $\lambda x+(1-\lambda)y$ is to $y$, and the nearer $\lambda$ is to $1$, the nearer $\lambda x+(1-\lambda)y$ is to $x$. 
The motivation for that, is that a convex set is a set for which given any two poins $x$ and $y$, the line segment through them is also contained in the set.
You can understand the geometric meaning of $\lambda x+(1-\lambda)y$ in the case where you have a subset of $\mathbb R^2$. Try to plot in $\mathbb R^2$ some points of the form $(x_1+(1-\lambda)y_1, x_2+(1-\lambda y_2))$ where $(x_1, x_2)$ and $(y_1, y_2)$ are fixed and $\lambda$ is varying in between $0$ and $1$.
A: $\lambda x + (1-\lambda)y$ is the parameterization of the line connecting $x$ to $y$. A convex set just means all lines stay in the set for every point on the line.
