# Is entropy (or relative entropy) function smooth in the sense that it is gradient is lipschitz continuous?

I wonder if the relative entropy function satisfies the uniform smoothness condition? Let $$x \in \mathbb{R}^n_+$$ be a probability distribution and $$f: \mathcal{X}\rightarrow \mathbb{R}$$ such that $$f(x)=\sum_{i=1}^n x_i log(x_i)$$. The function $$f$$ is called $$L_f-$$smooth if:

$$\begin{equation} f(y) \leq f(x) + \langle \nabla f(x), y-x \rangle + \frac{L_f}{2} \|y-x\|^2, \qquad \forall x,y \in \mathcal{X}, \end{equation}$$ where $$\mathcal{X} = \{x \in \mathbb{R}^n_+: \sum_i x_i = 1 \}$$. I'm unable to show whether or not the (negative) entropy function satisfies this smoothness condition for some $$L_f$$. I have the same question for relative entropy function, i.e., KL-divergence, $$KL(x,y) = \sum_{i=1}^n x_i log(\frac{x_i}{y_i})$$.

To your first question, as to whether the negative entropy $$f$$ satisfies the condition over $$\mathcal{X}$$: it doesn't.
Let's expand the RHS of the inequality. $$\frac{\partial f}{\partial x_{i}}=\log x_{i}+1$$ for all $$i$$, so that $$\langle \nabla f(x), y-x \rangle=\sum_{i=1}^{n}(y_{i}-x_{i})(\log x_{i}+1)=-f(x)+\sum_{i=1}^{n}y_{i}\log x_{i}.$$ Note we use above that $$\sum_{i=1}^{n}(y_{i}-x_{i})=0$$, as $$y$$ and $$x$$ are both probability distributions. The RHS is then $$\sum_{i=1}^{n}y_{i}\log x_{i}+\frac{L_{f}}{2}||y-x||^{2}.$$ Subtracting the sum $$\sum_{i=1}^{n}y_{i}\log x_{i}$$ tells us that the inequality you're hoping for is equivalent to $$KL(y, x) \leq \frac{L_{f}}{2}||y-x||^{2}$$ where $$KL(y,x)=\sum_{i=1}^{n}y_{i}\log \frac{y_{i}}{x_{i}}.$$ Since $$x$$ and $$y$$ are probability distributions, $$||y-x||^{2} \leq (||x||+||y||)^{2} \leq 4n^{2}$$, so we'd at least need $$KL(y,x) \leq 2n^{2}L_{f}$$ for some $$L_{f}$$ if this were to be true. However, it is easy to see that $$KL(y,x)$$ can be made arbitrarily large. For instance, with $$n=2$$, take $$y=\left(\frac{1}{2},\frac{1}{2}\right)$$ and $$x=(x_{1},x_{2})$$ with $$x_{2}=1-x_{1}$$ and $$x_{1} \in [0,1].$$ Then $$KL(y,x) \to \infty$$ as $$x_{1} \to 0$$ (and as $$x_{1} \to 1$$). Similar counterexamples arise for $$n \geq 3.$$
As far as KL-divergence satisfying it, I'm not sure exactly how you want to approach it. The divergence could be considered a function from $$\mathcal{X} \to \mathbb{R}$$ if you fix some $$y \in \mathcal{X}$$ and let $$x \mapsto K(x,y)$$ or a function from $$\mathcal{X} \times \mathcal{X} \to \mathbb{R}$$ where $$(x,y) \mapsto K(x,y).$$