1. Let $\widehat\beta^{(1)}_{\lambda}$ and $\widehat\beta^{(2)}_{\lambda}$ be two solutions of (4.4) and set $\widehat\beta=(\widehat\beta^{(1)}_{\lambda}+\widehat\beta^{(2)}_{\lambda})/2$.
We have $\mathcal{L}(\beta)=\|Y-X\widehat\beta\|^2+\lambda|\widehat\beta|_1=\|Y-X(\widehat\beta^{(1)}_{\lambda}+\widehat\beta^{(2)}_{\lambda})/2\|^2+\lambda|(\widehat\beta^{(1)}_{\lambda}+\widehat\beta^{(2)}_{\lambda})/2|_1$
Thanks to the triangle inequality we have $\lambda|(\widehat\beta^{(1)}_{\lambda}+\widehat\beta^{(2)}_{\lambda})/2|_1 \leqslant (\lambda/2)(|\widehat\beta^{(1)}_{\lambda}|_1+|\widehat\beta^{(2)}_{\lambda}|_1)$
Moreover by strict convexity of $\|.\|^2$, since $X\widehat\beta^{(1)}_{\lambda}\neq X\widehat\beta^{(2)}_{\lambda}$ we have $Y-X\widehat\beta^{(1)}_{\lambda}\neq Y-X\widehat\beta^{(2)}_{\lambda}$ and so
(1)Finally,
$\mathcal{L}(\beta)<1/2(\|Y-X\widehat\beta^{(1)}_{\lambda}\|^2+\lambda|\widehat\beta^{(1)}_{\lambda}|_1+\|Y-X\widehat\beta^{(2)}_{\lambda}\|^2+\lambda|\widehat\beta^{(2)}_{\lambda}|_1)$.
The inequality prouve that $\widehat\beta^{(1)}_{\lambda}$ and $\widehat\beta^{(2)}_{\lambda}$ are not solutions of (4.4) when $X\widehat\beta^{(1)}_{\lambda}\neq X\widehat\beta^{(2)}_{\lambda}$.
So $X\widehat\beta^{(1)}_{\lambda}= X\widehat\beta^{(2)}_{\lambda}$ and the fitted value $\widehat{f}_{\lambda}=X\widehat\beta_{\lambda}$ is unique.
2. $\widehat\beta^{(1)}_{\lambda}$ and $\widehat\beta^{(2)}_{\lambda}$ are solutions of (4.4) with $\lambda >0$, so from the optimality Condition (4.2) and the definition of $\partial\mathcal{L}(\beta)$ there exists $\widehat{z}^{(1)}$ and $\widehat{z}^{(1)}$ such that
$-2X^T(Y-X\widehat\beta^{(1)}_{\lambda})+\lambda\widehat{z}^{(1)}=0$
and
$-2X^T(Y-X\widehat\beta^{(2)}_{\lambda})+\lambda\widehat{z}^{(2)}=0$
So, as $X\widehat\beta^{(1)}_{\lambda}= X\widehat\beta^{(2)}_{\lambda}$,
we have
$-2X^TY-2X^TX\widehat\beta^{(1)}_{\lambda}+\lambda\widehat{z}^{(1)}=-2X^TY-2X^TX\widehat\beta^{(1)}_{\lambda}+\lambda\widehat{z}^{(2)}$
Simplifying, we have $\lambda\widehat{z}^{(1)}=\lambda\widehat{z}^{(2)}$.
Since $\lambda>0$ we finally have $\widehat{z}^{(1)}=\widehat{z}^{(2)}$.
3. Set $J=\{j:|\widehat{z}_j|=1\}$ and $\widehat{z}=(2/\lambda)X^T(Y-X\widehat\beta_{\lambda}) \in \partial|\widehat\beta_{\lambda}|_1$.
However $\partial|\widehat\beta_{\lambda}|_1=\{z \in \mathbb{R}^n ; z_j=sign([\widehat\beta_{\lambda}]_j)$ for $[\widehat\beta_{\lambda}]_j \neq0$ and $-1 \leqslant z_j \leqslant 1$ else $\}$.
So if $|\widehat{z}_j| \neq 1$, than $[\widehat\beta_{\lambda}]_j=0$.
And $[\widehat\beta_{\lambda}]_{J^c}=0$.
If $|\widehat{z}_j| = 1$,
as $(\lambda/2)\widehat{z}=X^TY-X^TX\widehat\beta_{\lambda}$,
we have :
$\widehat{z}_j=[X^TY]_j-[X^TX\widehat\beta_{\lambda}]_j$.
But $[X^TY]_j=X^T_jY$ and $[X^T X \widehat\beta_{\lambda}]_j =\sum\limits_{\substack{k\in J}}^{}{X^T_j X_k [\widehat\beta_{\lambda}]_k}= X^T_j X_J [\widehat\beta_{\lambda}]_J$.
We can sum only on J because others terms are equal to zero.
So any $\widehat\beta_\lambda$ to (4.4) fulfills
$[\widehat\beta_{\lambda}]_{J^c}=0$ and $X^T_J X_J [\widehat\beta_{\lambda}]_J=X^T_J Y - (\lambda/2)\widehat{z}_J$.
4. Let $\widehat\beta^{(1)}_{\lambda}$ and $\widehat\beta^{(2)}_{\lambda}$ be two solutions of (4.4), thanks to question 3. $\widehat\beta^{(1)}_{\lambda}$ and $\widehat\beta^{(2)}_{\lambda}$ are equal on $J^c$ (both equal to zero).
So when $X^T_J X_J$ is nonsingular, we have $[\widehat\beta^{(1)}_{\lambda}]_J=(X^T_J X_J)^{-1}X^T_J Y - (\lambda/2)\widehat{z}_J=[\widehat\beta^{(2)}_{\lambda}]_J$.
So when $X^T_JX_J$ is nonsingular, the solution to (4.4) is unique.