I have two convex functions $f$ and $g$ defined on $[0, \frac34], [\frac14,1]$ respectively such that they agree on the common domain $[\frac14, \frac34].$ How can I show that the (well defined) function $$F(x)=
\begin{cases}
f(x), & \text{if $x\in[0, \frac34]$} \\
g(x), & \text{if $x\in[\frac14,1]$}
\end{cases}$$ is convex on $[0,1]$?
I tried to obtain a contradiction assuming the epigraph of $F$ is not convex, but couldn't find one. Does anyone has a good idea? I do not need a complete solution, but only a good starting.
Thank you in advance for all your helps.
2026-03-27 23:22:31.1774653751
Extension of a convex function
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It depends on how much background you can assume. One characterization of convexity is that the right-hand side derivative is increasing (a.k.a. "nondecreasing"); see Convex Analysis by Rockafellar. With that characterization, your result is clear.