Let us consider $f, g:\mathbb{R}\rightarrow\mathbb{R}^+$ such that $$ f(x) = g(x) + \varepsilon\phi(x) $$ for some $\varepsilon>0$ and $\phi:\mathbb{R}\rightarrow\mathbb{R}^+$ - convex functon.
My question is:
Does convexity of $f(x)$ imply convexity of $g(x)$?
I noted this idea in some proofs where convexity of $g(x)$ is shown by proving convexity of $f(x)$ and then saying that taking $\varepsilon\rightarrow 0$ we can conclude that $f(x)$ is indeed convex. Is it obvious?
If we have convexity of $f$ (here $f$ depends on $\varepsilon > 0$) for all $\varepsilon >0$ small, then the answer is YES, $g$ becomes convex. Indeed, convexity of $f_\varepsilon$ means that for any $x,y \in \mathbb{R}$ and any $\lambda \in [0,1]$ we have $$ g(( 1 - \lambda) x + \lambda y) + \varepsilon \phi(( 1 - \lambda) x + \lambda y) \leq (1-\lambda) [ g(x) + \varepsilon \phi(x)] + \lambda [ g(y) + \varepsilon \phi(y)]. $$ Since the inequality holds for all $\varepsilon > 0$ small enough, taking $\varepsilon \to 0$ implies the convexity of $g$.
However, if we have the convexity of $f_\varepsilon$ for a single value of $\varepsilon >0$ then $g$ does not need to be convex. As a counterexample take $g(x) = \sin x$ and $\phi(x) = 10x^2 $. Then for $f(x) = g(x) + \varepsilon \phi(x)$ we have $$ f''(x) = -\sin x + 10\varepsilon >0, $$ hence $f$ is convex for any $\varepsilon > 1/10$ but $g$ is not.