Define a function $f: [-b,b]^d\mapsto \mathbb{R}^d (d>1)$ by $$ f(x)=\int_{[-a, a]^d} \frac{u}{1 + \exp(u^\top x)}du $$ Can we show that the image of $f$ is a convex set? I did some simulations and it seems the property holds, but I cannot prove it or find any solution in the literature.
The following picture shows the image of $f$ on a discrete grid over $[-10,10]^2$ where the integration is over $[-10,10]^2$ as well (i.e., a = b = 10).
The following image shows the image of $f$ on a discrete grid (finer than the previous one) over $[-10,10]^2$ where the integration is over $[-1,1]^2$(i.e., a = 1, b = 10).
By the result in Convexity of Nonlinear Image of a Small Ball with Applications to Optimization, we can show that the image of $f$ over any small neighbor area is convex, but what I want is a global property.
I think since the integrant is a smooth and nicely-behaved function, there might be some existing results that contain the property I want, but I cannot find any. I have verified that $f$ is not convex, so it seems that results in convex analysis won't help here.
To give you more information, we can prove some property of $f$ by some elementary calculus, e.g., we can show the property shown by the following figure, where the value indicates the first coordinate $x_1$ of $x$ and the curve is the image of $f$ over $\{x_1\}\times [-b, b]$. But I didn't see how to use such property to show the global image is convex.
Any help or discussion would be appreciated!