Prove that in a normalized space the function reaches its infimum.

Question

Let X be a normalized space, $x, y \in X$. Function $\varphi:\mathbb{R}→\mathbb{R}$ is given by the formula $\varphi(t)=||x−ty||$. Prove that it reaches its infimum.
$\textbf{My idea:}$ Consider the geometric interpretation: $ty$ is a straight line. Then our function calculates the distance from point $x$ to the points of the line. Let the infimum be reached at $t_0 \in D$, where $D$ is compact. I want to prove that for all $t$ not belonging to $D$, the distance is greater than for $t$ lying in $D$. If I prove this, we will immediately get the necessary statement, since the norm function is continuous, and on a compact the continuous function reaches its minimum. Help me prove that for all $t$ not belonging to $D$, the distance is greater than for $t$ lying in $D$. I would also be grateful if you suggest another way to solve this problem. Thank you in advance.

Answer 1

A different approach: Suppose $y \neq 0$. $|\phi (t)-\phi (s)| \leq |t-s|\|y\|$, so $\phi$ is continuous. Also, $\phi (t) \ge |t|\|y\|-\|x\| \to \infty$ as $|t| \to \infty$. Any continuous function $\phi: \mathbb R \to \mathbb R$ such that $|\phi(t)| \to \infty$ as $|t| \to \infty$ attains its minimum at some point.

The case $y=0$ is trivial.