Is a Smooth and Strongly Convex Function Essentially Quadratic?

Question

Here by a $\beta$-smooth function, I mean $f(y) \leq f(x) + \nabla f(x)^T(y-x) + \frac{\beta}{2}||x-y||^2$. Equivalently, we have $\beta$ Lipschitz derivative or $\nabla^2 f(x) \leq \beta I$.

One observation is that we can bound $\alpha I \leq \nabla^2f \leq \beta I$. Does this mean $f$ is a quadratic function?

Answer 1

This is false. Take any function with bounded second derivative, $|g''(x)|\le M$ (such functions clearly exist, like $g(x)=1/(1+x^2)$). Then, the function $$ f(x)=Mx^2+g(x) $$ is both smooth (according to your definition) and strongly convex, since $f''(x)\ge M$.