$f(\mathbf{x})$ is a function of the vector of variables denoted by $\mathbf{x}$.
Let's assume $f(\mathbf{x})>0$.
Also, $f(\mathbf{x})$ is homogenous of degree $h$.
By Euler's homogenous function theorem, then, $$\nabla f(\mathbf{x}) \cdot \mathbf{x} = fh \tag{1*}$$
(where the $i^{th}$ element of the gradient $ \nabla f(\mathbf{x})$ is defined as $[\nabla f(\mathbf{x})]_i = \frac{\partial f}{\partial x_i}$)
Say I want to
$$\max(a f(\mathbf{x})) \:\:\:\: s.t. \:\:\:\: \mathbf{x} \cdot \mathbf{1} = b$$
(where $a$ and $b$ are positive constants and $\mathbf{1}$ is a vector of ones with same length as $\mathbf{x}$)
Then the lagrangian is
$$L=a f(\mathbf{x}) + \lambda (\mathbf{x} \cdot \mathbf{1} - b)$$
And the first order conditions are
$$\nabla L = a \nabla f(\mathbf{x}) + \lambda \mathbf{1}=\mathbf{0} \tag{2*}$$
(where $\mathbf{0}$ is a vector of zeros)
Dotting this through by $\mathbf{x}$ and invoking Euler's Theorem (Eq. $1*$ above) gives
$$\mathbf{x}\cdot L = a h f(\mathbf{x}) + \lambda b = 0$$
Solving for $\lambda$ gives $$\lambda = -\frac{a h f(\mathbf{x})}{b} \tag{3*}$$
Now, to find the second order condition for a maximum I first differentiate the lagrangian again with respect to $\mathbf{x}$. Let this differentiation be denoted $\nabla^2 L$ where $[\nabla^2 L]_{ij} = \frac{\partial}{\partial x_i}\left( \frac{\partial L}{\partial x_j} \right)$. I.e., $\nabla^2 L$ is the Hessian of $L$. Likewise, $\nabla^2 f(\mathbf{x})$ is the Hessian of $f(\mathbf{x})$.
In what follows, it is also important to note that Euler's homogeneous function theorem may be extended to any order of differentiation $n$ as follows:
$$\nabla^n f(\mathbf{x}) \cdot \mathbf{x} = (h - n + 1) \nabla^{n-1} f(\mathbf{x})$$
so that $$\nabla^2 f(\mathbf{x}) \cdot \mathbf{x} = (h-1) \nabla f(\mathbf{x}) \tag{4*}$$
My question is whether or not $\nabla(\lambda)$ vanishes when I take $\nabla^2 L$?
If it does vanish, then this leaves us with
$$\mathbf{x} \cdot \nabla^2 L \cdot \mathbf{x} = a \mathbf{x} \cdot \nabla^2 f \cdot \mathbf{x}$$
which, invoking Eq. $4*$ above, becomes
$$\mathbf{x} \cdot \nabla^2 L \cdot \mathbf{x} = a h(h-1) f(\mathbf{x}) \:\:\: \left\{\begin{matrix} <0, \:\:\: if \:\:\: 0<h<1 \\ >0, \:\:\: if \:\:\: h>1 \:\: or \:\: h<0 \\ = 0, \:\:\: if \:\:\:h=1 \:\: or \:\:\ h = 0 \end{matrix}\right. $$
So that $L$ is maximized only if $h$ is between 0 and 1.
But online resources seem to suggest that the second order conditions should be taken of $L$ as evaluated at the argument that satisfies the first order conditions, in which case, given Eq. $3*$, $\nabla (\lambda)$ would not vanish, leaving us with
$$\nabla^2 L = a \nabla^2 f(\mathbf{x}) + \nabla (\lambda) \mathbf{1}$$
Substituting in for $\lambda$ (Eq. $3*$),
$$\nabla^2 L =a \nabla^2 f(\mathbf{x}) - \frac{a h}{b} Q$$
where $Q$ is a matrix whose rows are all the gradient $\nabla f(\mathbf{x})$, so that $Q\cdot \mathbf{x} = \mathbf{1} h f(\mathbf{x})$ and $\mathbf{x} \cdot Q\cdot \mathbf{x} = b h f(\mathbf{x})$
And now the second order condition becomes
$$\mathbf{x} \cdot \nabla^2 L \cdot \mathbf{x} = a h(h-1) f(\mathbf{x})-a h f(\mathbf{x}) $$
$$\mathbf{x} \cdot \nabla^2 L \cdot \mathbf{x} = a h(h-2) f(\mathbf{x}) \:\:\: \left\{\begin{matrix} <0, \:\:\: if \:\:\: 0<h<2 \\ >0, \:\:\: if \:\:\: h>2 \:\: or \:\: h<0 \\ = 0, \:\:\: if \:\:\:h=2 \:\: or \:\:\ h = 0 \end{matrix}\right. $$
So that $L$ is maximized if $h$ is between 0 and 2.
So which is it?