Given a function $g(\boldsymbol{x})$,
$g(\boldsymbol{x})=log(\sum_{i=1}^{m}e^{\frac{f_{i}(\boldsymbol{x})}{\lambda_{i}}})$, where $\boldsymbol{x} \in \mathbb{R}^{n}$, and $f_{i}$ is a $L_{i}$-smooth function. $\lambda \in \mathbb{R}_{+}^{m}$ and $\sum_{i=1}^{m}\lambda_{i}=1$.
How to prove the smoothness and covexity of $g(\boldsymbol{x})$?
I have an idea, but I'm not sure can I do better. \begin{aligned} &||\nabla_{\mathbf{x}}g(\mathbf{x_1}|\lambda)-\nabla_{\mathbf{x}}g(\mathbf{x_2}|\lambda)|| \\ &=||\sum_{i=1}^m\frac{e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}\nabla f_{i}(\mathbf{x_{1}})-\sum_{i=1}^m\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{2}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{2}})}{\lambda_{i}}}\nabla f_{i}(\mathbf{x_{2}})|| \\ &\leq\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}\nabla f_{i}(\mathbf{x_{1}})-\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{2}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{2}})}{\lambda_{i}}}\nabla f_{i}(\mathbf{x_{2}})|| \\& =\sum_{i=1}^m ||\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}} \nabla f_{i}(\mathbf{x_{1}})-\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}} \nabla f_{i}(\mathbf{x_{2}})+\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}} \nabla f_{i}(\mathbf{x_{2}})-\frac{\lambda_{i}e^\frac{f_{i}(\mathbf{x_{2}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{2}})}{\lambda_{i}}} \nabla f_{i}(\mathbf{x_{2}})|| \\ & \sum_{i=1}^m ||\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}} (\nabla f_{i}(\mathbf{x_{1}})-\nabla f_{i}(\mathbf{x_{2}}))+(\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}-\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{2}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{2}})}{\lambda_{i}}})\cdot \nabla f_{i}(\mathbf{x_{2}})|| \\& \leq \sum_{i=1}^m ||\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}} (\nabla f_{i}(\mathbf{x_{1}})-\nabla f_{i}(\mathbf{x_{2}}))+(\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}-\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{2}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{2}})}{\lambda_{i}}})\cdot ||\nabla f_{i}(\mathbf{x_{2}})|| \ || \\& \sum_{i=1}^m ||\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}} || \ ||\nabla f_{i}(\mathbf{x_{1}})-\nabla f_{i}(\mathbf{x_{2}})||+||\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{1}})}{\lambda_{i}}}-\frac{1}{\lambda_{i}}\frac{e^\frac{f_{i}(\mathbf{x_{2}})}{\lambda_{i}}}{\sum_{i}e^\frac{f_{i}(\mathbf{x_{2}})}{\lambda_{i}}}||\cdot ||\nabla f_{i}(\mathbf{x_{2}})|| \ || \end{aligned}
The first term is smooth because $f_{i}$ is $L_{i}$-smooth, I don't know how to proceed after that, or is there any other way?