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Claim: Let $\chi$ be a compact set. If $f''(x)>0$ for all $x\in\chi$, then $f$ is strongly convex.

This seems to be true, intuitively, as I can't think of a counterexample. All of the examples I've seen for strictly convex functions, with positive second derivatives on the entire domain (e.g. $f(x)=e^x$), that are not strongly convex have second derivatives that become arbitrarily small. If we limit the domain to be a compact set, does strong convexity follow?

Updated: For completeness, the true statement (thanks to the answer below), is as follows

Let $\chi$ be a compact set. If $f$ is twice continuously differentiable, with $f''(x)>0$ for all $x\in\chi$, then $f$ is strongly convex.

Erik M
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If you assumed $f''$ continuous, this would be true. As stated, this is false. To give a counterexample, it suffices to construct a differentiable function $g:[-1,1]\to\mathbb R$ such that $g'>0$ everywhere and $\inf g'=0$. Then $f$ will be its antiderivative.

The idea is as follows. Take two differentiable functions that have the same tangent line at $0$, say $h_1(x) = x-x^2$ and $h_2(x) = x+x^2$. Make $g$ wiggle between them, so that $h_1\le g\le h_2$, and the derivative $g'$ occasionally drops to nearly zero as we approach $x=0$. Note that $g'(0)=1$ is guaranteed by the squeeze between $h_1,h_2$.

One of possible explicit functions along these lines is $$g(x)=x+x^{5/3}+x^2\sin(1/x)$$ Indeed, $$g'(x)=1-\cos(1/x)+\frac53 x^{2/3}+2x \sin(1/x)$$ is strictly positive near $0$, because $x^{2/3}$ beats $2x$ (this is what it's for). Also, $g'(0)=1$. But $g'$ comes arbitrarily close to $0$ near $0$, because $1-\cos(1/x)$ turns to zero at times.