Proof involving the sum of $L-$smooth functions

Question

Consider a set of functions $\{f_i\}_{i=1}^n$ where each $f_i$ is $L_i$ smooth, where $L_i$ will be the smallest possible number such that $$\|\nabla f(x) -\nabla f(y) \| \leq L_i \|x-y\|$$ for any $x,y$.

Then $F(x) = \frac1n \sum_i f_i(x)$ is also $L$-smooth with $L \leq \frac1n \sum_i L_i$.

I am trying to solve this problem, but I am a little lost. I know that the sum of $L-$smooth functions is also $L-$smooth, but this question is slightly different. Does anyone have any ideas or how to construct this proof?

Thanks!

Answer 1

$$\begin{align}\|\nabla F(x) - \nabla F(y)\| &= \left\|\frac1n\sum_i \nabla f_i(x) - \frac1n\sum_i \nabla f_i(y)\right\|\\ &=\frac1n\left\|\sum_i (\nabla f_i(x)-\nabla f_i(y))\right\|\\ &\le \frac1n \sum_i\left\| \nabla f_i(x)-\nabla f_i(y)\right\|\\ &\le \frac1n \sum_i (L_i\|x-y\|)\\ &=\left(\frac1n \sum_i L_i\right)\|x-y\|\end{align}$$