- #1
mathmari
Gold Member
MHB
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Hey!
- Could you give me an example of a strong concave function $f:[0,3]\rightarrow \mathbb{R}$ that is not continuous? (Wondering)
We have that $f''(x)<0$.
Since the function has not to be continuous, the derivatives are neither continuous, are they? (Wondering)
Is maybe the following one such function?
$f(x)=\left\{\begin{matrix}
-x^2 & \text{ if } x \in (0,3]\\
1 & \text{ if } x=0\\
\end{matrix}\right. $
Then $f'(x)=\left\{\begin{matrix}
-2x & \text{ if } x \in (0,3]\\
0 & \text{ if } x=0\\
\end{matrix}\right.$ and so $f''(x)=\left\{\begin{matrix}
-2 & \text{ if } x \in (0,3]\\
0 & \text{ if } x=0\\
\end{matrix}\right.$
(Wondering)
But this function isn't strong concave, is it? (Wondering) - We have the set $\{(x,y) \in \mathbb{R}^2 \mid y\leq f(x)\}$, where $y=f(x)$ is a concave function. What can be said about that set? (Wondering)
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