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Prove that $$ f(x)=\begin{cases} \frac{2}{3}x^2+\frac{2}{3}x-\frac{1}{12} &\quad x<-0.5 \\ -0.25 &\quad x\geq -0.5 \\ \end{cases} $$ is convex over $\mathbb{R}$.

As far as I understand I cannot use second derivative because the function is non-differentiable (at $x=-0.5$).

If both $x,y \geq0.25$ or $x,y < 0.25$ this is easy (both cases are convex). But I couldn't find an algebric approach to prove the case where $x<-0.5$ and $y \geq -0.5$.

f(x) graph

Please advise.

Thank you.

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    $\begingroup$ The function is differentiable, and its derivative is non-decreasing, which should be enough for convexity. $\endgroup$
    – dxiv
    Jun 1 at 7:47
  • $\begingroup$ @dxiv You are absolutely right, didn't notice. Thank you. $\endgroup$
    – Dennis
    Jun 1 at 12:01
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Thanks to @dxiv.

I didn't notice that the function $f(x)$ actually is differentiable. $$ f'(x)=\begin{cases} \frac{4}{3}x+\frac{2}{3} &\quad x<-0.5 \\ 0 &\quad x\geq -0.5 \\ \end{cases} \quad f''(x)=\begin{cases} 4/3 &\quad x<-0.5 \\ 0 &\quad x\geq -0.5 \\ \end{cases} $$

One can see that $f'_+(-0.5)=0=f'_-(-0.5)$ (left and right derivatives are equal), so $f$ is continuously differentiable over $\mathbb{R}$.

From Wikipedia:

  • differentiable function of one variable is convex on an interval if and only if its derivative is monotonically non-decreasing on that interval
  • A twice differentiable function of one variable is convex on an interval if and only if its second derivative is non-negative there
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