Let $\textbf{x}=\left(x_1,\dots,x_n\right)\in \mathbb R^n$. I would like to find the global minimum of function $f(\textbf{x})\in\mathbb R$ on the circle of radius $R$, i.e. $D=\left\{\textbf{x}\mid \|\textbf{x}\|^2\leq R^2 \right\}$.
Generally speaking, if all the second partial derivatives of $f$ exist, I should look for the minimum inside the circle, i.e. $D^0=\left\{\textbf{x}\mid \|\textbf{x}\|^2< R^2 \right\}$ (using Hessian matrix of the stationary points) and those on the border, that is, on the circumference $\partial D=\left\{\textbf{x}\mid \|\textbf{x}\|^2= R^2 \right\}$ (for example applying the method of Lagrange multipliers), and afterward selecting the smallest among them as the global minimum.
I wonder if, generally, this procedure can be substituted by repeating the Lagrange multipliers method on each radius circumference $r<R$.
In other words, $\forall r\in(0,R]$ minimize $f(\textbf{x})$ subject to $\left\{\textbf{x}\mid \|\textbf{x}\|^2= r^2 \right\}$. Let $m_r$ be the resulting minimum given $r$. Then I wonder if finding the $$\min_{r\in(0,R]}\ m_r$$ is equivalent to finding the global minimum among the minimum of $D^0$ and $\partial D$.
If you also consider the degenerate circle $\{0\}=\{x\mid\|x\|^2=0\}$, then, assuming that $f$ is continuous, yes, the answer is affirmative, that is, the smallest value $m$ that $f$ takes on $D$ is $\min_{r\in[0,R]}m_r$. Since $D$ is compact and $f$ is continuous, then $f$ has a minimum $m$ at some point $x_0\in D$. Let $r_0=\|x_0\|$. Then $m_{r_0}=f(x_0)$ and, if $r\in[0,R]\setminus\{r_0\}$, you have $f(x)\geqslant f(x_0)=m$, and therefore $m_r\geqslant m$. So, $\min_{r\in[0,R]}m_r=m$.