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If I have a function $l(w,x)$, defined as $\mathbb R^n \times \mathbb R^m \mapsto \mathbb R$ that is continuously differentiable and $L$-smooth with respect to $w$, $\|\nabla l(w,x) - \nabla l(w',x)\| \leq L$ for all $w,w' \in \mathbb R^n$, is the function $w \mapsto \int_{x \in \mathbb R^m} l(w,x) d\mathbb P(x)$ always smooth for a probability measure $\mathbb P$? If not, what are some sufficient conditions to guarantee its smoothness?

Interception
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  • I don’t understand your hypotheses. Is $h$ just $C^1$ with that gradient assumption, or is there a genuine smoothness condition? For $C^1$, it should work, provided that you can find one $w$ such that $\int_x{|l(w,x)|dP(x)}$ and $\int_x{|\nabla l (w,x)|dP(x)}$ are finite. – Aphelli Feb 21 '23 at 09:19
  • The gradient assumption is the smoothness by definition. I am not sure why one $w$ is sufficient in your answer. If I expand the definition I need to verify for all $w$ and $w'$ that $|\int_x \nabla_w l(w,x) d\mathbb P(x) - \int_x \nabla_{w'} l(w',x) d\mathbb P(x) |$ by some constant $L'$. I get it that the integral of $l$ should be bounded. Is it possible to get the second hypothesis in your answer from the smoothness of $l$? – Interception Feb 21 '23 at 09:31

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