Define the set $$A=\{\langle x,y\rangle : x\in [-1,1]^d, \: y\in \mathbb{R}^d, \:\|y\|_1\leq c\}$$ for some $c\in \mathbb{R}$. I want to show that is is convex. Take $\langle x,y\rangle, \langle x',y'\rangle\in A$. I want to show that if $\lambda \in [0,1]$ then $$\lambda\langle x,y\rangle+(1-\lambda) \langle x',y'\rangle\in A$$ My problem is that when I do this computation, I can never end up with only one dot product (i.e. an element of $A$). For example I have tried
\begin{align} \lambda\langle x,y\rangle+(1-\lambda) \langle x',y'\rangle&=\langle \lambda x,y\rangle + \langle (1-\lambda) x',y'\rangle\\ &=\langle \lambda x,y\rangle + \langle (1-\lambda) x',y+(y'-y)\rangle\\ &=\langle \lambda x,y\rangle + \langle (1-\lambda) x',y\rangle + \langle (1-\lambda)x',y'-y\rangle\\ &=\langle \lambda x+(1-\lambda)x',y\rangle + \langle (1-\lambda)x',y'-y\rangle\\ \end{align} No matter how many tricks I do, I always seem to end up with $2$ scalar products. How can I solve this?
Edit I know that this is actually a convex polytope and can intuitively see it, but I am having problems showing it
Let's define $X=\{(x,y): x\in[-1,1]^d, y\in\mathbb{R}^d, ||y||_1\leq c\}$ and $f:X\to\mathbb{R} $ given by $f(x,y)=\langle x,y\rangle$.
Now, $f$ is continuous and $X$ is connected because it is the cartesian product of the $d$-dimensional hypercube (which is connected) and a ball of radius $c$ (which is connected too). Thus, $A=f(X)\subseteq \mathbb{R}$ is connected. But the only connected sets in the standard topology of $\mathbb{R}$ are intervals, so $A$ is an interval and thus convex.