What does the norm function do in the notation of a class of convex functions?

Question

In Yurii Nesterov book "Lectures on Convex Optimization" (Springer, second edition, 2018) there are the following definitions of convex and strongly convex functions (def 2.1.2 and def.2.1.3 respectively):

• A continuously differentiable function $f(\cdot)$ is called convex on a convex set $Q \subseteq \mathbb{R}^n$ (notation $f \in \mathscr{F}^1(Q)$) if for any $x,y \in Q$ we have $$f(y) \ge f(x) + \langle \nabla f(x), y-x\rangle.$$
• A continuously differentiable function $f(\cdot)$ is called strongly convex on $Q \subseteq \mathbb{R}^n$ (notation $f \in \mathscr{S}^1_\mu(Q,\|\cdot\|)$) if there exists a constant $\mu \gt 0$ such that for any $x,y \in Q$ we have $$f(y) \ge f(x) + \langle \nabla f(x), y-x\rangle + \frac{1}{2} \mu \|y-x\|^2.$$

Everything is clear with these definitions, but on p.74 there is the following statement, which confuses me a lot:

Note that the class $\mathscr{S}^1_0(Q,\|\cdot\|)$ coincides with $\mathscr{F}^1(Q,\|\cdot\|)$.

What does the symbol $\|\cdot\|$ mean in the expression $\mathscr{F}^1(Q,\|\cdot\|)$? As you can see, there is no mention of any norm in the definition of a convex function. So why didn't the author just write $\mathscr{F}^1(Q)$ instead of $\mathscr{F}^1(Q,\|\cdot\|)$? At first I thought that this is just a typo. But then on p.73 I saw the similar expression again: $\displaystyle \min_{x \in \mathbb{R}^n} f(x), ~~ f \in \mathscr{F}^1(\mathbb{R}^n,\|\cdot\|)$ (again we see this norm symbol)... Do you have any ideas about this?

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P.S. In this book the author often uses the notation $\mathscr{F}^{1,1}_L(\mathbb{R}^n,\|\cdot\|)$ to denote the class of convex continuously differentiable functions with Lipschitz continuous gradient with respect to the norm $\|\cdot\|$. In this class the norm function is used in the definition of Lipschitz continuous gradient, so here everything is clear for me.

Answer 1

I am writing up the comments in an answer so that this question is marked as answered.

As mentioned in the comments, the definition on convexity depends on the norm through the scalar product (which appears in the definition of convexity) which induces the norm. If we choose the same norm and $\mu = 0$ in the definition for strong convexity, then we get the convexity definition.