I know that it is possible to define the Huber-loss in multiple dimensions, consider $\mathbb R^n$, $n=2$, via infinmal convolution of two functions, namely $f_\varepsilon(x)=\frac{1}{2\varepsilon}\Vert x\Vert_2^2$ and $g(x)=\Vert x\Vert_2$ and the resulting Huber loss looks like
$$H_\varepsilon(x)=\begin{cases} \frac{1}{2\varepsilon}{\lVert x\rVert_2^2} & \text{for } \lVert x\rVert_2 \le \varepsilon \\ \lVert x\rVert_2 - \frac{\varepsilon}{2} & \text{otherwise}. \end{cases}$$
The convex conjugate $H^*_\varepsilon(y) = f^*_\varepsilon(y) + g^*(y) = \delta(y) + \frac\varepsilon2\Vert y\Vert_2^2$ can be calculated using the properties of infinmal convolution and add the convex conjugate of $f_\varepsilon$ and $g$.
I would like to use the Huber loss with an L1-norm, i.e. $$H_\varepsilon(x)=\begin{cases} \frac{1}{2\varepsilon}{\lVert x\rVert_1^2} & \text{for } \lVert x\rVert_1 \le \varepsilon, \\ \lVert x\rVert_1 - \frac{\varepsilon}{2} & \text{otherwise}, \end{cases}$$
which should be possible, right? Furthermore, I additionally would like to calculate the convex conjugate of this second Huber loss involving the L1-norm. I mean, I can't just substitute the L2- with the L1-norm and use the upper properties again, because they hold only for the L2-norm, right?
What I know so far is that the convex conjugate of the L1-norm is the indicator function with the sup-norm, since $\Vert\cdot\Vert_\infty$ is the dual norm of the L1 norm. This also holds for the L2-norm, using the fact that the L2-norm is dual to itself, i.e. indicator function with the L2-norm (upper $\delta$ function). I'm also aware of the convex conjugate of $\frac12\Vert x\Vert_2^2$, which is itself.
Unfortunately, not a lot of literature on the multidimensional Huber loss nor the convex conjugate of this specific case can be easily found. I'm sorry if the answer to this is already somewhere (online) available, I searched for it, but wasn't able to find it.