When the gradient of a differentiable scalar-valued function is Lipschitz, i.e.,
$$ \|\nabla f(y) - \nabla f(x)\| \leq L\|y-x\| $$
for all $x, y \in \mathbb{R}^n$ and some $L>0$, then it is well-known that the following holds.
$$ f(y) \leq f(x) + \left\langle \nabla f(x), y - x \right\rangle + \frac{L}{2}\|y-x\|^2, \qquad \forall x, y \in \mathbb{R}^n $$
It is straightforward to let $y = x - \gamma \nabla f(x)$ and show the following:
$$ f(y) \leq f(x)-\gamma \left(1-\frac{\gamma L}{2}\right)\|\nabla f(x)\|^2. $$
The above shows that letting $0<\gamma \leq \frac{2}{L}$ would decrease the function value unless $\nabla f(x)=0$ where $\nabla f(x)=0$ is the stationary condition.
General case
Notice $y=x - \gamma \nabla f(x)=P_{\mathbb{R}_n}(x - \gamma \nabla f(x))$ so we can define $y=P_{C}(x - \gamma \nabla f(x))$ where $C$ is any closed convex set, $x \in C$, and $P_{C}$ is the orthogonal projection onto $C$. Then one can write the following:
$$ \begin{aligned} f(y) &\leq f(x)+\frac{1}{\gamma}\langle y-(x-\gamma\nabla f(x)), y-x \rangle -(\frac{1}{\gamma}-\frac{L}{2})\|y-x\|^2 \\ &= f(x)+\frac{1}{\gamma}\langle P_{C}(v)-v, P_{C}(v)-x \rangle -(\frac{1}{\gamma}-\frac{L}{2})\|y-x\|^2 \end{aligned} $$ where $v=x-\gamma\nabla f(x)$. The above shows we can decrease the function value because the inner product is the variational inequality and is negative.
Question
Let $C$ be the unit $\ell_2$-ball where the stationary point is either $\nabla f(x)=0$ or $x=\alpha \nabla f(x)$ for some $\alpha \leq 0$. Is it possible to write an equivalent inequality that includes only $x$ and $\nabla f(x)$ not $v$ which does not decrease when $\nabla f(x)=0$ or $x=\alpha \nabla f(x)$?
My try
When $v \in \bar{B}(0,1)$, then $P_{C}(v)=v$ and we are back in $\mathbb{R}^n$. When $v \notin \bar{B}(0,1)$, then $y=\frac{v}{\|v\|}$ and
$$ \langle P_{C}(v)-v, P_{C}(v)-x \rangle =1-\|v\|+(1-\frac{1}{\|v\|})\|x\|^2-\gamma (1-\frac{1}{\|v\|}) \langle x, \nabla f(x) \rangle $$ and $$ \|y-x\|^2 = 1+\|x\|^2(1-\frac{2}{\|v\|})+\frac{2\gamma}{\|v\|} \langle x, \nabla f(x) \rangle . $$
Remark
The above inequalities are coming from solving $\min_{x \in C} f(x)$ using projected gradient descent where $x_{k+1}=P_C(x_k-\gamma \nabla f(x_k))$. However, here we only do one step. When $C=\mathbb{R}^n$ then $\nabla f(x)=0$. When $C=\bar{B}_2(0,1)$ then $\nabla f(x)=0$ or $x=\alpha \nabla f(x)$ for some negative $\alpha$.
Let $y = P_C(x-\gamma \nabla f(x))$. Then $y$ is a solution of $$ \min_y f(x) +\nabla f(x)(y-x) + \frac1{2\gamma} \|y-x\|^2 + I_C(y), $$ where $I_C$ is the indicator function of $C$. Then $$ f(y) \le f(x) +\nabla f(x)(y-x) + \frac L2 \|y-x\|^2\\ \le \frac L2 \|y-x\|^2 - \frac1{2\gamma} \|y-x\|^2\\ = \frac12(L-\frac1\gamma) \|y-x\|^2. $$ And the function value decreases if $L< \frac1\gamma$.