Piecewise convexity and global convexity

A differentiable function is convex in an interval $I$ if and only if $f'$ is increasing in $I$. Now we have that $$f'(x)\leq f'(z)\leq f'(y)$$ for any $0\leq x<z<y\leq 1$. Can you take it from here and show that the proposition is true?


A differentiable function $f$ is convex on an interval $(a,b)$ if and only if its derivative $f'$ is increasing.

Therefore, your assumptions imply that $f'$ is increasing on $(0,z)$ and on $(z,1)$. Now your assumption that $f'$ is continuous immediately gives you that $f'$ is increasing on $(0,1)$ and therefore convex.

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Real Analysis

Convex Analysis

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