Suppose $f(w,v)$ is strongly-concave in terms of $v$, which conditions could infer $\Phi(w,v)= f(w,v-(\nabla_v f(w,v)/\|\nabla_v f(w,v)\|))$ is strongly-concave in terms of $v$?
From the concavity of $f$, have: $$ f\left(w, \alpha v_1+(1-\alpha) v_2\right) \geqslant \alpha f\left(w, v_1\right)+(1-\alpha) f\left(w, v_2\right)+\frac{\alpha(1-\alpha) \mu}{2}\left\|v_1-v_2\right\|^2, \alpha \in[0,1] $$ Then consider $$ \begin{aligned} &\Phi\left(w, \alpha v_1+(1-\alpha) v_2\right) \\ &=f\left(w, \alpha\left(v_1-(\nabla_{v} f(w,v_1)/\|\nabla_{v_1} f(w,v_1)\|)\right)+(1-\alpha)\left(v_2-(\nabla_{v_2} f(w,v_2)/\|\nabla_{v} f(w,v_2)\|)\right)\right) \\ & \geqslant \alpha \Phi\left(w, v_1\right)+(1-\alpha) \Phi\left(w, v_2\right) \\ & +\frac{\alpha(1-\alpha) \mu}{2}\left\| v_1-v_2\left\|^2+\right\| \nabla_{v} f(w,v_1)/\|\nabla_{v} f(w,v_1)\| - \nabla_{v} f(w,v_2)/\|\nabla_{v} f(w,v_2)\| \|^2-2 \left\langle v_1-v_2, \nabla_{v} f(w,v_1)/\|\nabla_{v} f(w,v_1)\| - \nabla_{v} f(w,v_2)/\|\nabla_{v} f(w,v_2)\| \right\rangle\right] \\ (*) &\geqslant \alpha \Phi\left(w, v_1\right)+(1-\alpha) \Phi\left(w, v_2\right)+\frac{\alpha(1-\alpha) \mu}{2}\left\|v_1-v_2\right\|^2 \end{aligned} $$
In order to get (*), $ 2 \left\langle v_1-v_2, \nabla_{v} f(w,v_1)/\|\nabla_{v} f(w,v_1)\| - \nabla_{v} f(w,v_2)/\|\nabla_{v} f(w,v_2)\| \right\rangle \leq 0 $ should hold.
already have $\left\langle v_1-v_2, \nabla_{v} f(w,v_1) - \nabla_{v} f(w,v_2) \right\rangle \leq 0$
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