I am struggling with the following question:
Let $a \in \mathbb{R}^n $ and $ b \in \mathbb{R}$ and define $ f: \mathbb{R}^n \rightarrow$ $\mathbb{R} $ by
$f(x)=\langle x, a \rangle + b,x \in \mathbb{R}^n$
Now show that for every convex set $Y$ in $\mathbb{R}$, $f^{-1}(Y)= \{x \in\mathbb{R}^n: f(x) \in Y\}$ is a convex set in $\mathbb{R}^n$.
Can I apply any function for the pre-image or how am I supposed to solve this question?
Thank you in advance!
Let $Y$ be a convex set in $\mathbb{R}$. It's actually an interval, but we don't need that fact. Let $X = f^{-1}(Y) \subseteq \mathbb{R}^n$, and pick two points in it: $p$ and $q$.
To show it's convex, all we need to do is show the line between $p$ and $q$ is contained in $X$. All points $r$ on this line are of the form $r = p + t(q - p)$ for some $t \in [0, 1]$. We compute the image of this point: $$ \begin{align*} f(r) &= f(p + t(q - p)) \\ &= \langle p + t(q - p), a \rangle + b \\ &= \langle p, a \rangle + t\langle q - p, a \rangle + b \\ &= f(p) + t\langle q - p, a \rangle \\ &= f(p) + t \left( \langle q, a \rangle - \langle p, a \rangle \right) \\ &= f(p) + t \left( f(q) - f(p) \right) \\ \end{align*} $$
Since $f(r)$ is between $f(p)$ and $f(q)$, $f(r) \in Y$. Thus, $r \in X$, and $X$ is convex.