Piecewise convexity and global convexity

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be continuous on $\left[0,1\right]$. Consider $z\in\left(0,1\right)$ and suppose that $f$ is differentiable and convex on $\left[0,z\right]$ and $\left[z,1\right]$. If $f'$ (i.e., $\frac{df}{dx}$) is continuous at $z$, then $f$ is convex on $\left[0,1\right]$. - Is this proposition true? While I could not create a counter-example, I am finding it difficult to generate a clean proof as well.

Answer 1

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.

Answer 2

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?

Answer 3

Yes it's true. $f$ is convex iff $f'$ is monotonically non-decreasing, but it works on $[0, z)$ and $(z, 1]$. But $f'$ is continuous at $z$ so it can't be less than 0 otherwise there would be a neighborhood where it is negative and thus not convex.