Let $f:\mathbb{R}^m\rightarrow \mathbb{R}$ be a smooth function.
I know $f(x)$ is convex if its Hessian ($\frac{\partial^2 f(x)}{\partial x\partial x^T}$) is positive semi-definite.
Now, let $g(t)=f(x_0+t\Delta x)$ ($t\in \mathbb{R}$) for some arbitrary $x_0,\Delta x\in \mathbb{R}^m$.
I have the following questions:
1- Is it correct that if $g(t)$ is convex in $t$ for any $x_0,\Delta x\in \mathbb{R}^m$, then $f(x)$ is convex in $x$?
2- If the answer is no in general, under what conditions it holds? And, if the answer is yes, is there any proof?
Thanks!
Yes: by the definition of convexity, it suffices to show $(1-\lambda)f(a) + \lambda f(b) \ge f((1-\lambda)a+\lambda b)$ for any $\lambda \in [0,1]$ and any $x,y$. If $g$ is as you define it with $x_0:= a$ and $\Delta x := b-a$, then the above inequality is equivalent to $(1-\lambda) g(0) + \lambda g(1) \ge g(\lambda)$, which holds by assumed convexity of $g$.