I am just wondering why the following equalities are true:
$$\begin{aligned} \sup\limits_{x \in \mathbb{R}^n} \left\{ \langle u, x \rangle - \frac{1}{2}\|x\|^2 \right\} &= \sup\limits_{t \geq0, \|x\| = t} \{ \langle u,x \rangle - \frac{1}{2}\|x\|^2 \}\\ &= \sup\limits_{t \geq0} \{t \cdot\sup\limits_{\|x\| = 1} \langle u, x \rangle - \frac{1}{2} \|x\|^2 \} \\ &= \sup\limits_{t \geq0} \{t \|u\| - \frac{1}{2}\|x\|^2 \}\end{aligned}$$
I have some intuition about them, but I don't get clearly how to prove it analytically.
We have $$ \begin{align} \sup_{x\in\mathbb{R}^n} \Big( \left\langle u,x \right\rangle -\dfrac{1}{2}\|x\|^2 \Big) &= \sup_{x\in\mathbb{R}^n} \Big( \frac{1}{2}\|u\|^2+ \frac{1}{2}\|x\|^2- \frac{1}{2}\|u-x\|^2 -\frac{1}{2}\|x\|^2 \Big) \\ &= \sup_{x\in\mathbb{R}^n} \Big( \frac{1}{2}\|u\|^2- \frac{1}{2}\|x-u\|^2 \Big)\\ &=\frac{1}{2}\|u\|^2. \end{align} $$ However, since $\lim_{t\to+\infty}t\|u\|=+\infty$ if $u\neq 0$, the last expression in your question gives $+\infty$ when $u\neq 0$, showing that the equality in your question is...incorrect.