An equality about an infimum over a vector: $\inf_y = \inf_{\|v\|=1}\inf_t$ I am reading convex optimization lecture notes, where I saw the equality as a part of the derivation:

Here $f$ is a function and $\nabla f$ is the gradient. What seems to happen here is that the author replaces $(y-z) = t v$ with $\|v \| = 1$ and converts infimum over $y$ to double infimum (is this trivial? i.e. infimum over multiplication equals taking a double infimum?). What confuses me further here is, with that, now $z$ depends on $v$ and I don't understand how this works, or I'm overthinking. Could anyone provide a proof for this? Is this a well-known technique?
 A: To understand the step from $y$ to $v$, let's make us aware that the notation in the lecture notes doesn't print from which set the possible $y$'s come from since it's "obvious". It's $ℝ^n$. Writing this explicitly might make it clearer how to get from $y$ to $w = tv$:
$$
\inf_{y ∈ ℝ^n} g(y) = \inf_{w ∈ ℝ^n - z} g(w + z)
$$
Here $ℝ^n - z = \{y - z | y ∈ ℝ^n\}$. The operation $· - z$ is a bijection and by applying the inverse operation to the argument of the $\inf$-ed function it is hopefully clear that we take the infimum over the same values.
We use this for $g(y) = f(z) + ∇f(z)^T(y-z) + \frac{L}2 \|y-z\|^2$. So after cancelling the $z$'s we have $g(w + z) = f(z) + ∇f(z)^Tw + \frac L2 \|w\|^2$.
Concerning your worry about dependence of $z$: it is a number like $2$ or $L$. That's why @Matematleta said "set $z=0$ for convenience". $0$ is one possible value for $z$ but all other possible values work the same way.
Now we replace $w$ by $tv$ with $t ∈ ℝ_{≥ 0}$ and $v ∈ ^{n-1} := \{ν ∈ ℝ^n | \|ν\| = 1\}$. We can find for every $w$ such $v$ and $t$ and other way around as well. (It's not a bijection since $0v = 0v'$ even for $v \neq v'$.)
Now to show that the $\inf$'s are interchangeable, take the definition of $\inf$: one part is that it is a lower bound. That means $$∀α∈ A : \inf_{a ∈ A} g(a) ≤ g(α)$$. The other part is, that the $\inf$ is the highest lower bound, so $$∀ α ∈ A: X ≤ g(α) ⇒ X ≤ \inf_{a∈A} g(a)$$. Use this:
$$
\begin{align}
&∀ ν∈ ^{n-1} :& \inf_{v ∈ ^{n-1}} \inf_{t ∈ ℝ} h(v, t) &≤ \inf_{t ∈ ℝ} h(ν, t)  \\
⇒ &∀ ν ∈ ^{n-1} : ∀τ ∈ ℝ :& \inf_{v ∈ ^{n-1}} \inf_{t ∈ ℝ} h(v, t) &≤ \inf_{t ∈ ℝ} h(ν, t) ≤ h(ν, τ) \\
⇒ &∀τ ∈ ℝ : ∀ ν ∈ ^{n-1} :& \inf_{v ∈ ^{n-1}} \inf_{t ∈ ℝ} h(v, t) &≤ h(ν, τ) \\
⇒ &∀τ ∈ ℝ :& \inf_{v ∈ ^{n-1}} \inf_{t ∈ ℝ} h(v, t) &≤ \inf_{v ∈ ^{n-1}} h(v, τ) \\
⇒ && \inf_{v ∈ ^{n-1}} \inf_{t ∈ ℝ} h(v, t) &≤ \inf_{t ∈ ℝ} \inf_{v ∈ ^{n-1}} h(v, t)
\end{align}
$$
We can do the same argument with $v$ and $t$ exchanged and get
$$
\inf_{v ∈ ^{n-1}} \inf_{t ∈ ℝ} h(v, t) ≤ \inf_{t ∈ ℝ} \inf_{v ∈ ^{n-1}} h(v, t) ≤ \inf_{v ∈ ^{n-1}} \inf_{t ∈ ℝ} h(v, t) \\
⇒ \inf_{v ∈ ^{n-1}} \inf_{t ∈ ℝ} h(v, t) = \inf_{t ∈ ℝ} \inf_{v ∈ ^{n-1}} h(v, t)
$$
