Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a $C^2$, strictly convex function, and $\theta_\epsilon$ the standard approximation to identity $$\theta_\epsilon=\frac{1}{\psi}e^{\dfrac{1}{x^2-\epsilon^2}}, \text{ where } \psi={\int_\mathbb{R}e^{\dfrac{1}{x^2-\epsilon^2}}dx},$$ for $|x|<\epsilon$ and $0$ otherwise.
Is $f\leq f*\theta_\epsilon $? I am wondering this b/c I know we can think of $f*\theta_\epsilon $ as a moving local average of $f$. By convexity we know that $f''\geq 0$, which can be interpreted as how the value of $f$ at a point compares with the average value of $f$ in a neighborhood of that point. ($f''>0 $ meaning the value of $f$ at that point is less than average). And if this is true, does this hold if $f$ is strictly convex but only $C^1$?
Yes. For non-strict inequality $f\le f*\phi$ to hold, you only need $f$ to be convex (no smoothness assumptions) and $\phi$ to be
All of these properties hold in your example. The proof goes:
$$\begin{split} f(x) &= \int f(x)\phi(t)\,dt \le \int \frac{f(x+t)+f(x-t)}{2}\phi(t)\,dt \\ &= \frac12 \int f(x+t)\phi(t)\,dt + \frac12 \int f(x-t)\phi(t)\,dt \\ & = \frac12 \int f(x-t)\phi(t)\,dt + \frac12 \int f(x-t)\phi(t)\,dt \\ & = f*\phi(x) \end{split}$$ where the replacement of $t$ with $-t$ is a change of variables, and this is where the fact that $\phi$ is even is needed.
If you have strict convexity, you can modify the above to obtain strict inequality.