Given $f_1(\beta) = v^T\beta$ for $\beta \in \mathbb{R}^d$, where $v \in \mathbb{R}^d$ is a fixed vector, then:
$\nabla f_1(\beta) = v\ \text{and}\ \ \mathcal{H}f_1(\beta) = 0_{d \times d}$
solution:
$\frac{\delta f}{\delta \beta_1} = v'(\frac{\delta \beta}{\delta \beta_1}) = v'(1,0,...,0)$ and more generally, $\frac{\delta f}{\delta \beta_i} = v'(\frac{\delta \beta}{\delta \beta_i}) = v'(0,...,1,...,0)$ where the $i^{th}$ entry is 1.
Then $\nabla f(\beta) = v'[\frac{\delta \beta}{\delta \beta_1},\frac{\delta \beta}{\delta \beta_2},...,\frac{\delta \beta}{\delta \beta_d}] = v'[1,...,1] =v$
From above: $\frac{\delta f_1}{\delta \beta_1} = (1,0,...,0)$ then $\frac{\delta f_{1}}{\delta \beta_i} = 0_{1\times d}$ then $\mathcal{H}f(\beta) = [0_{1\times d},...,0_{1\times d}] = 0_{d\times d}$
The notation could be better-in particular, f should not have the subscript 1, lest it be thought that we are taking the partial derivative of f with respect to the first variable. Let's take d=3 for the sake of illustration, and v=[a,b,c]. Then if we set f(x,y,z)=ax+by+cz, $$ \nabla=[a,b,c] $$ and all the 2nd-order partials are 0, so the Hessian matrix is identically 0.