Assume $f(x)$ has an L-Lipschitz continuous gradient say $L$ i.e there is a constant L>0 such that $$\|\nabla f(x) - \nabla f(y)\|_2 \le L\|x-y\|_2$$ for any $x,y$.
Does $f(\beta) = \sum_{i=1}^n Y_i X_i \beta-\log{[1+\exp{(X_i \beta)}]}$ have an L-Lipschitz continuous gradient? $Y_i \in \mathbb R$ ,$X_i, \beta \in \mathbb R^p\ \forall i \in [1..n]$ where $p\ge 1 $.
What I have done: $$\nabla f(\beta) = \sum_{i=1}^n [Y_i-h_\beta(X_i)]X_i^T$$ where $h_B(x) = \frac{1}{1+ \exp(-x \cdot \beta)}$. Basically, it is a sigmoid function.
$||\sum_{i=1}^n (h_{{\beta}_2}(X_i) - h_{{\beta}_1}(X_i))X_i^T||\le L||{\beta}_1-{\beta}_2||$
I would appreciate any help.
First, WLOG $Y_i = 0$. Second, its enough to check that
$$g:\mathbb R\to\mathbb R,\ g(t) = \log(1+\exp(t))$$ has Lipschitz gradient, and it does because its second derivative is bounded. Then the composition of Lipschitz maps is Lipschitz, and your thing is
$$ \nabla f(\beta) = -g'(h(\beta))X_i^T,\quad h(\beta) = X_i \cdot \beta$$