Dual norm of a truncated and ordered (decreasing order) $\ell_1$-norm

I accidentally derived the dual norm of your dual norm. Let us first derive some conjugate functions: $$ \begin{align} f(x) &= \max\left\{g(x),h(x)\right\} \\ g(x) &= \frac{1}{k} \| x \|_1 \\ h(x) &= \lVert x\rVert_{\infty} \\ g^*(y) &= 0 \text{ if } k \| y \|_\infty \leq 1 \;(\infty \text{ otherwise}) \\ h^*(y) &= 0 \text{ if } \| y \|_1 \leq 1 \;(\infty \text{ otherwise}) \\ \end{align} $$ Using the well-known rule for the conjugate of $\max\{g,h\}$, we get: $$\begin{align} f^*(y) &= \inf_{v,z} \{ (z_1g)^*(v) + (z_2h)^*(y-v) \mid z_1+z_2 = 1, z\geq 0 \}\\ &= 0 \text{ if } \exists v,z : k \| v \|_\infty \leq z_1, \; \| y-v \|_1 \leq z_2, \; z_1+z_2 = 1, z\geq 0 \\ &= 0 \text{ if } \exists v : k \| v \|_\infty +\| y-v \|_1 \leq 1 \;(\infty \text{ otherwise}) \end{align}$$ Using that the convex conjugate of a norm is the indicator of the unit ball for the dual norm, the dual norm of the dual norm is $$||y||^{**} = \inf_v \{ k \| v \|_\infty +\| (y-v) \|_1 \}.$$ This does not equal the norm you started with. Either the truncated L1-norm is not a real norm, or I made a mistake. Maybe you could do the same analysis for the norm itself, or check the above for a possible mistake.