How Every lasso solution gives rise to the same fitted value? I'm reading the paper "The Generalized Lasso Problem and Uniqueness" by Ali and Tibshirani (2019).
They mention this: 

I didn't get the idea of that. As I understand the lasso, the solution depends on $\lambda$, and with every $\lambda$ we get a different set of solutions, so different $\hat\beta$, and as a result $X \hat\beta$ shouldn't give the same fitted values.
What's wrong with my understanding?
 A: The generalized lasso problem given in $(1)$ is
$$
\frac{1}{2}\|y-X\beta\|_2^2 + \lambda\|D\beta\|_1 \to \min_{\beta \in \mathbb{R}^p}
$$
with $y \in \mathbb{R}^n, X \in \mathbb{R}^{n \times p}, D \in \mathbb{R}^{m \times p},$ and $\lambda \in \mathbb{R}_{\geq 0}$.
From this you can see that the minimization considered here is over the vectors ${\beta \in \mathbb{R}^p}$ for a fixed value of $\lambda$.
$\hat{\beta}^{(1)}$ and $\hat{\beta}^{(2)}$ are two different minimizers of $\beta \mapsto \frac{1}{2}\|y-X\beta\|_2^2 + \lambda\|D\beta\|_1$.
A: The result being shown is for a fixed penalty $\lambda$, so it is saying that for any fixed penalty, every solution $\hat{\beta}$ to the LASSO minimisation gives the same $X \hat{\beta}$.  If you instead compare minimisers under different penalty values then it is possible to get different values of $X \hat{\beta}$, just as you intuitively expect.
The proof establishes this by contradiction --- if we assume two minimisers with different values then the proof establishes that any convex combination of the minimisers yields a strictly smaller value of the objective function, which contradicts the initial assertion that these were minimisers.
