I'm trying to follow the argument given on pg 24 of these notes. This is the proof of (2.11)
Here we are considering a convex quadratic form
$$f_{A,b}(h) = \frac{1}{2}h^{T}Ah+ b^{T}h+c$$
and I want to show that if $y$ is a minimizer of $f_{A,b}$ then
$$y^{T}Ay = \lambda^{2}$$ for $\lambda = \max \{b^{T}h : h^{T}Ah \leq 1 \}$.
The argument proceeds by noting that a minimizer $y$ (by taking the derivative) satisfies $$y^{T}Ah = -b^{T}h$$ for all $h \in \mathbb{R}^{n}$. But I'm not sure how the author concludes the statement above. One thing is note is that since the form is convex we have that $y^{T}Ay = -b^{T}y \geq 0$. Therefore $b^{T}y \leq 0$.
First of all, what is written in the lecture note is not necessarily correct.
Let us consider a simple example. Let $A = -2, b= c= 0$, scalar values. Then, $f(h) = -h^2$ clearly is not lower bounded nor attains minimum. However, $\max\{bh; h^TAh \leq 1\}=0$ is finite since $b=0$.
We may eliminate such an example by assuming $A$ is positive semidefinite symmetric matrix. Then We may assume without losing generality that $A$ is a diagonal matrix with diagonal entries $a_i\geq 0$. Then, the condition can be simplified as $\max\{\sum_ib_ih_i; \sum_i a_i^2h_i \leq 1\}$ is finite. Actually, this condition is concerned only when $a_i = 0$. When $a_i=0$, we must have $b_i=0$ for $b_ih_i^2$ to be finite. Thus you can prove the equivalence. (This is a sketch but I think you can fill gaps on your own).
To prove the desired equality, first note that by the observation above we can ignore those $i$'s with $a_i=b_i=0$. We also note that by letting $u_i = \sqrt{a_i}h_i$, we can rewrite as $$ \lambda = \max_i\left\{\sum_i\frac{b_i}{\sqrt{a_i}}u_i; \sum_i u_i^2\leq 1\right\} $$
Therefore, $\lambda$ is equal to $\sqrt{\sum_i \frac{b_i^2}{a_i}}$.
On the other hand, the quadratic form is equal to $$ \frac{1}{2}\sum_i\left(a_ih_i^2 + 2b_ih_i\right)+ c = \frac{1}{2}\sum_i a_i \left(h_i + \frac{b_i}{a_i}\right)^2 + (\text{term without }h) $$
So, it is minimized at $h_i = - \frac{b_i}{a_i}$. Therefore, the minimizer $y$ satisfies $y^TAy = \sum_i \frac{b_i^2}{a_i}$, which is equal to $\lambda^2$.