i have the following problem
show $B_2=\{(x,y)│x^2+y^2≤1\}$ is a convex subset of $\mathbb{R}^2$
my idear if I have understood the definition correctly
construct $(x_1,x_2 ),(y_1,y_2 )\in C$
we then have $x_1+y_1≤1,x_2+y_2≤1$
but what to do next. do i just set fx $x=x_1+y_1,y=x_2+y_2$ and multiply it with $\lambda$ and $(1-\lambda)$ and then inset x and y ind the given inequality or something else
thanks in advance from just started a course in convex optimization
Not quite - here's a hint: The goal is to show that, for any $\lambda\in\left]0,1\right[$, the vector $\lambda(x_1,x_2) + (1-\lambda)(y_1,y_2)$ resides in $B_2$, where that addition is the addition in $\mathbb{R}^2$. So, we'd need to show that $(\lambda x_1 + (1-\lambda)y_1, \lambda x_2+ (1-\lambda)y_2)\in B_2$.
In particular, we have to show that $$(\lambda x_1+(1-\lambda)y_1)^2 +(\lambda x_2+ (1-\lambda)y_2)^2\leq 1$$
given the information that $(x_1,x_2)$ and $(y_1,y_2)$ reside in $B_2$, i.e., $$x_1^2 +x_2^2\leq 1\quad\text{and}\quad y_1^2 +y_2^2\leq 1.$$
Let $x,y\in B_2$ and $\lambda\in\left[0,1\right]$. We have to show that $\lambda x +(1-\lambda)y\in B_2$.
If $x=(x_1,x_2)$ then $x_1^2+x_2^2\leq1$. Similarly if $y=(y_1,y_2)$ then $y_1^2+y_2^2\leq1$. To prove what we want, since $\lambda x +(1-\lambda)y=(\lambda x_1+(1-\lambda)y_1,\lambda x_2+(1-\lambda)y_2)$. We have to prove the following inequality: $$(\lambda x_1+(1-\lambda)y_1)^2+(\lambda x_2+(1-\lambda)y_2)^2\leq1$$ Expanding the LHS we get:
$$\lambda^2(x_1^2+x_2^2)+(1-\lambda)^2(y_1^1+y_2^2)+2\lambda(1-\lambda)(x_1y_1+x_2y_2)$$ By Cauchy-Schwarz $x_1y_1+x_2y_2=(x_1,x_2)(y_1,y_2)\leq(x_1^2+x_2^2)(y_1^2+y^2_2)\leq1$. Hence: $$(\lambda x_1+(1-\lambda)y_1)^2+(\lambda x_2+(1-\lambda)y_2)^2\leq\lambda^2+(1-\lambda)^2+2\lambda(1-\lambda)=(\lambda+1-\lambda)^2=1$$ And $B_2$ is convex.
I don't know if you've ever heard about norms, but you can prove this for $B_n$ in a much easier way using norms and the triangular inequality. See here: https://math.stackexchange.com/a/2045057/962125.
