I have a non-convex bounded function $h$ that is L-smooth and it has bounded gradient, i.e. $\|\nabla h(x)\| \leq \sigma^2, \forall x \in \mathbb{R}^d$.
Define the function $f(y) = \exp(\alpha \times h(y))$ where $\alpha > 0$. Can we show that $f$ is a smooth function with bounded gradient?
I'm focusing on showing the bounded gradient property as I believe this gives the smoothness using the mean value theorem. I know that $\nabla f(y) = \alpha \nabla h(y) \exp(\alpha \times h(y))$. What I did so far is the following: $$ \|\nabla f(y)\| \leq \alpha \sigma^2 \exp(\alpha \times h(y))$$
Since $h$ is bounded, i.e. $|h(y)| \leq M$ then I can write: $$ \|\nabla f(y)\| \leq \alpha \sigma^2 \exp(\alpha M)$$
Would this be sufficient to say that $f$ is a smooth function with a bounded gradient?
I am unsure the result you want to prove is true. If $h$ is the identity function (which is convex though), $h$ is smooth with bounded gradient and $f$ has not bounded gradient.
We can think of a non convex function which shows that your claim does not hold, for example $h(x)=x$ if $x\geq 0$ and $h(x)=2x$ if $x\leq 0$. I am aware it is not differentiable at $0$ but this gives some intuition that your result cannot be proved.
If you want a differentiable function you can use a quadratic part around $0$, like that:
\begin{equation} h(x)= \begin{cases} x & \text{if } x\geq 1\\ -\dfrac{1}{4}(x^2-6x+1) & \text{if } -1\leq x\leq 1\\ 2x & \text{if } x\leq -1 \end{cases} \end{equation}
Of course for $x\geq 1$, $f$ has not bounded gradient.