MHB Can Discontinuity and Non-Derivability Exist in Strongly Concave Functions?

[mathmari]

Hey!

1. 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)
2. 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)

 

[I like Serena]

We have that $f''(x)<0$.

Hey mathmari! (Smile)

I think that's not true.
The definition of a strictly concave function is that every line segment with end points on the graph lies strictly below the graph. See wiki.
Consider for instance $f(x)=-x^4$, which has $f''(0)=0$, but which is strictly concave. (Nerd)

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)

Isn't $f'(0)$ undefined since there is no distinct point within $\epsilon=1/2$? (Wondering)

And no, it's not concave since the line segment from x=0 to x=3 is not completely below the graph.

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)

Isn't it a convex set? That is, every line segment with end points in the set is entirely inside the set? (Wondering)

 

[Opalg]

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. $

Your example is almost correct. Did you perhaps mean to write $f(x)=\begin{cases}x^2 & \text{ if } x \in (0,3]\\ \color{red}-1 & \text{ if } x=0 \end{cases}$ ?

 

[mathmari]

Hey mathmari! (Smile)

I think that's not true.
The definition of a strictly concave function is that every line segment with end points on the graph lies strictly below the graph. See wiki.
Consider for instance $f(x)=-x^4$, which has $f''(0)=0$, but which is strictly concave. (Nerd)

Isn't $f'(0)$ undefined since there is no distinct point within $\epsilon=1/2$? (Wondering)

And no, it's not concave since the line segment from x=0 to x=3 is not completely below the graph.

Ah ok... To check if indeed every line segment with end points on the graph lies strictly below the graph, do we have to see the graph? (Wondering)

Your example is almost correct. Did you perhaps mean to write $f(x)=\begin{cases}x^2 & \text{ if } x \in (0,3]\\ \color{red}-1 & \text{ if } x=0 \end{cases}$ ?

If we have $x^2$ instead of $-x^2$ isn't the function convex and not concave? (Wondering)

Or did you mean $f(x)=\begin{cases}-x^2 & \text{ if } x \in (0,3]\\ -1 & \text{ if } x=0 \end{cases}$ ? (Wondering)

Then we have [desmos="0,3,-9,1"]-1 x=0;-x^2;[/desmos]

 

[Opalg]

If we have $x^2$ instead of $-x^2$ isn't the function convex and not concave? (Wondering)

Or did you mean $f(x)=\begin{cases}-x^2 & \text{ if } x \in (0,3]\\ -1 & \text{ if } x=0 \end{cases}$ ? (Wondering)

Then we have

Haha, yes. I made the same mistake of missing out a minus sign. (Blush)

So the example should have been $f(x)=\begin{cases}-x^2 & \text{ if } x \in (0,3]\\ -1 & \text{ if } x=0 \end{cases}$.

 

[mathmari]

Haha, yes. I made the same mistake of missing out a minus sign. (Blush)

So the example should have been $f(x)=\begin{cases}-x^2 & \text{ if } x \in (0,3]\\ -1 & \text{ if } x=0 \end{cases}$.

Isn't $f'(0)$ undefined since there is no distinct point within $\epsilon=1/2$? (Wondering)

We have that $f$ is continuous at $(0,3]$, since it is a polynomial function.
At $x=0$ we have the following: $$\lim_{x\rightarrow 0^{-}}f(x)=\lim_{x\rightarrow 0^{-}}(-1)=-1 \\ \lim_{x\rightarrow 0^{+}}f(x)=\lim_{x\rightarrow 0^{+}}(x^2)=0$$
So, the function is not continuous at $x=0$. Therfore, it is neither differentiable at $x=0$, right?
So, do we have to write the derivative only for $x\in (0,3]$ ? (Wondering)
And

To check if indeed every line segment with end points on the graph lies strictly below the graph, do we have to see the graph? (Wondering)

Or do we have to use the definition? (Wondering)

 

[I like Serena]

We have that $f$ is continuous at $(0,3]$, since it is a polynomial function.
At $x=0$ we have the following: $$\lim_{x\rightarrow 0^{-}}f(x)=\lim_{x\rightarrow 0^{-}}(-1)=-1 \\ \lim_{x\rightarrow 0^{+}}f(x)=\lim_{x\rightarrow 0^{+}}(x^2)=0$$
So, the function is not continuous at $x=0$. Therfore, it is neither differentiable at $x=0$, right?
So, do we have to write the derivative only for $x\in (0,3]$ ? (Wondering)

Isn't $[0,3]$ the domain of $f$? (Wondering)
If so then:
$$\lim_{x\to 0^-} f(x) = \text{undefined}$$
since $f(x)$ is not defined for $x<0$.

Still, since $f(0)=-1\ne 0 = \lim_{x\to 0^+} f(x)$, $f$ is indeed not continuous.
And that implies it's not differentiable either at $x=0$.

And Or do we have to use the definition? (Wondering)

I believe so yes.
We can use the derivative and/or the 2nd derivative only if the function is differentiable (twice) on all of its domain.
Still, that does mean that from the fact that $f''(x)<0$ on the interval $(0,3]$, that $f$ is strictly concave on $(0,3]$.
That leaves verifying that any line segment that has one end point at $(0,-1)$ is strictly below the graph. (Thinking)

 

[mathmari]

Isn't $[0,3]$ the domain of $f$? (Wondering)
If so then:
$$\lim_{x\to 0^-} f(x) = \text{undefined}$$
since $f(x)$ is not defined for $x<0$.

Still, since $f(0)=-1\ne 0 = \lim_{x\to 0^+} f(x)$, $f$ is indeed not continuous.
And that implies it's not differentiable either at $x=0$.

Ah ok... (Thinking)

We can use the derivative and/or the 2nd derivative only if the function is differentiable (twice) on all of its domain.

So, can use the above function and consider only the domain $(0,3]$ or can we not use that function at all? (Wondering)

Still, that does mean that from the fact that $f''(x)<0$ on the interval $(0,3]$, that $f$ is strictly concave on $(0,3]$.

Could you give me an example of a function that satisfy the condition $f''(x)<0$ but is not strictly concave? (Wondering)

Still, that does mean that from the fact that $f''(x)<0$ on the interval $(0,3]$, that $f$ is strictly concave on $(0,3]$.
That leaves verifying that any line segment that has one end point at $(0,-1)$ is strictly below the graph. (Thinking)

Why do we have to take any line segment that has one end point at $(0,-1)$ ? (Wondering)

 

[I like Serena]

So, can use the above function and consider only the domain $(0,3]$ or can we not use that function at all? (Wondering)

Could you give me an example of a function that satisfy the condition $f''(x)<0$ but is not strictly concave? (Wondering)

Why do we have to take any line segment that has one end point at $(0,-1)$ ? (Wondering)

Yes, we can use the function on $(0,3]$.

If a function is differentiable twice everywhere on its domain and if $f''(x)<0$, it is strictly concave.
The reverse does not always hold true.
A function does not even have to be continuous to be strictly concave (as in your example).
And a strictly concave function can still have $f''(x)=0$ for some $x$ (as in the example $f(x)=-x^4$ for $x=0$).

We have to prove that any line segment with end points on the graph lies completely below the graph.
Since we already know that $f$ is strictly concave on $(0,3]$, this is true for any line segment between points $(a,f(a))$ and $(b,f(b))$ with $0<a<b\le 3$.
That leaves proving it for $0=a<b\le 3$. (Thinking)

 

[mathmari]

Isn't $[0,3]$ the domain of $f$? (Wondering)
If so then:
$$\lim_{x\to 0^-} f(x) = \text{undefined}$$
since $f(x)$ is not defined for $x<0$.

Still, since $f(0)=-1\ne 0 = \lim_{x\to 0^+} f(x)$, $f$ is indeed not continuous.
And that implies it's not differentiable either at $x=0$.

We have that a function $f$ is continuous at a point $x_0$

• $x_0$ is defined
• $\displaystyle{\lim_{x\rightarrow x_0}}f(x)$ exists
• $\displaystyle{\lim_{x\rightarrow x_0}f(x)=f(x_0)}$

In this case we have the following:

• $0$ belongs to the domain.
• Left Integral: $\displaystyle{\lim_{x\to 0^-} f(x) = \text{undefined}}$, since $f(x)$ is not defined for $x<0$.
  Right Integral: $\displaystyle{\lim_{x\to 0^+} f(x) =\lim_{x\to 0^+} (-x^2)=0 }$
• Value of the function: $f(0)=-1$
  Since $\displaystyle{f(0)=-1\ne 0 = \lim_{x\to 0^+} f(x)}$, the function is not continuous at the point $x=0$.

Is this correct? Could I improve something? (Wondering)

We have to prove that any line segment with end points on the graph lies completely below the graph.
Since we already know that $f$ is strictly concave on $(0,3]$, this is true for any line segment between points $(a,f(a))$ and $(b,f(b))$ with $0<a<b\le 3$.
That leaves proving it for $0=a<b\le 3$. (Thinking)

Ah ok... We have the following:
\begin{align*}&f(ta+(1-t)b)>tf(a)+(1-t)f(b) \Rightarrow f(t\cdot 0+(1-t)b)>tf(0)+(1-t)(-b^2) \\ &\Rightarrow f((1-t)b)>-t-(1-t)b^2 \Rightarrow -((1-t)b)^2>-t-(1-t)b^2 \Rightarrow -(1-t)^2b^2>-t-(1-t)b^2 \\ &\Rightarrow -(1-t)^2b^2+(1-t)b^2>-t \Rightarrow (1-t)b^2(-(1-t)+1)>-t \Rightarrow (1-t)b^2(-1+t+1)>-t \\ &\Rightarrow (1-t)b^2t>-t\Rightarrow (1-t)b^2t+t>0\Rightarrow [(1-t)b^2+1]t>0\end{align*}
We have that $t\in (0,1)$, so it holds that $1-t>0$, $t>0$ and since $b>0$ and so $b^2>0$, the inequality $[(1-t)b^2+1]t>0$ holds, so it implies that all the line segments between the points $(0,-1)$ and $(b,f(b))$ are completely under the graph.

Is this correct? Could I improve something? (Wondering)

 

[mathmari]

How could we draw the function $f$ ? At desmos the function $f(x)=-1$ for $x=0$ is not appeared. Or have I done something wrong? Do we have to use an other program? (Wondering)

Isn't it a convex set? That is, every line segment with end points in the set is entirely inside the set? (Wondering)

We have that $y=f(x)$ is a concave function. That means that every line segment with end points on the graph of that function lies strictly below the graph. In that set there are all the points that are below the graph of the function, right? (Wondering)
So, for all $x$, $y=f(x)$ is below the graph. That means that all points of the line segment belongs to that set, right? (Wondering)

 

[mathmari]

How could we draw the function $f$ ? At desmos the function $f(x)=-1$ for $x=0$ is not appeared. Or have I done something wrong? Do we have to use an other program? (Wondering)

I found my mistake!

[desmos="-1,3,-10,1"]\left(0,-1\right);-x^2;[/desmos]

 

[I like Serena]

I found my mistake!

How about:
[desmos="-1,4,-10,1"]\left(0,-1\right);-x^2\left\{0<x\le3\right\};\left(3,-9\right);[/desmos]

Or:
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[samples=51,
grid=both,
restrict x to domain=0:3,
xtick={0,1,...,3},
ytick={0,-1,...,-9}
]
\addplot[blue, ultra thick, only marks] coordinates {(0,-1) (3,-9)} ;
\addplot[blue, ultra thick] (x,-x^2) ;
\end{axis}
\end{tikzpicture}

(Wondering)

 

[mathmari]

Ah ok! (Nerd) Is then at the following graph:

[desmos="-10,10,-10,10"]\left(0,-1\right);-x^2\space\left\{0<x\le3\right\};-\frac{8}{3\space}x-1\space\left\{0<x\le3\right\};[/desmos]

the convex set the one that is constructed by all the elements between the two functions, or not? (Wondering)

 

[I like Serena]

Ah ok! (Nerd)

Is then at the following graph:

the convex set the one that is constructed by all the elements between the two functions, or not? (Wondering)

[desmos="-1,4,-10,1"]\left(0,-1\right);-\frac{8}{3\space}x-1\le y\le -x^2\left\{0<x\le 3\right\};[/desmos]
This is indeed a convex set.
All line segments with their end points in the set lie completely inside the set. (Happy)

 

[I like Serena]

We have that a function $f$ is continuous at a point $x_0$

• $x_0$ is defined
• $\displaystyle{\lim_{x\rightarrow x_0}}f(x)$ exists
• $\displaystyle{\lim_{x\rightarrow x_0}f(x)=f(x_0)}$

In this case we have the following:

• $0$ belongs to the domain.
• Left Integral: $\displaystyle{\lim_{x\to 0^-} f(x) = \text{undefined}}$, since $f(x)$ is not defined for $x<0$.
  Right Integral: $\displaystyle{\lim_{x\to 0^+} f(x) =\lim_{x\to 0^+} (-x^2)=0 }$
• Value of the function: $f(0)=-1$
  Since $\displaystyle{f(0)=-1\ne 0 = \lim_{x\to 0^+} f(x)}$, the function is not continuous at the point $x=0$.

Is this correct? Could I improve something? (Wondering)

Aren't they "limits" instead of "integrals"? (Wondering)

Oh, and I'd make it $\displaystyle{f(0)=-1\ne 0 = \lim_{x\to 0^+} f(x) = \lim_{x\to 0} f(x)}$, since we need the "regular" limit instead of the one-sided limit - they just happen to be the same in this case. (Nerd)

Ah ok... We have the following:
\begin{align*}&f(ta+(1-t)b)>tf(a)+(1-t)f(b) \Rightarrow f(t\cdot 0+(1-t)b)>tf(0)+(1-t)(-b^2) \\ &\Rightarrow f((1-t)b)>-t-(1-t)b^2 \Rightarrow -((1-t)b)^2>-t-(1-t)b^2 \Rightarrow -(1-t)^2b^2>-t-(1-t)b^2 \\ &\Rightarrow -(1-t)^2b^2+(1-t)b^2>-t \Rightarrow (1-t)b^2(-(1-t)+1)>-t \Rightarrow (1-t)b^2(-1+t+1)>-t \\ &\Rightarrow (1-t)b^2t>-t\Rightarrow (1-t)b^2t+t>0\Rightarrow [(1-t)b^2+1]t>0\end{align*}
We have that $t\in (0,1)$, so it holds that $1-t>0$, $t>0$ and since $b>0$ and so $b^2>0$, the inequality $[(1-t)b^2+1]t>0$ holds, so it implies that all the line segments between the points $(0,-1)$ and $(b,f(b))$ are completely under the graph.

Is this correct? Could I improve something? (Wondering)

We're starting with what we need to prove. I don't think we're supposed to that.
Don't we need a chain of reasoning in the opposite direction? (Worried)

I think we should state $f(ta+(1-t)b)\overset{?}{>}tf(a)+(1-t)f(b)$, with a question mark, indicating that's what we need to prove.
We could start with the last expression and then work our way back, with the arrows in the opposite direction... (Thinking)

 

[mathmari]

This is indeed a convex set.
All line segments with their end points in the set lie completely inside the set. (Happy)

(Happy)

Aren't they "limits" instead of "integrals"? (Wondering)

Oh yes... (Blush)

Oh, and I'd make it $\displaystyle{f(0)=-1\ne 0 = \lim_{x\to 0^+} f(x) = \lim_{x\to 0} f(x)}$, since we need the "regular" limit instead of the one-sided limit - they just happen to be the same in this case. (Nerd)

Ah ok... (Nerd)

We're starting with what we need to prove. I don't think we're supposed to that.
Don't we need a chain of reasoning in the opposite direction? (Worried)

I think we should state $f(ta+(1-t)b)\overset{?}{>}tf(a)+(1-t)f(b)$, with a question mark, indicating that's what we need to prove.
We could start with the last expression and then work our way back, with the arrows in the opposite direction... (Thinking)

Do you mean that we should write it as follows? (Wondering) We want to prove that $f(ta+(1-t)b)>tf(a)+(1-t)f(b)$ for $a=0$, so $f((1-t)b)>tf(0)+(1-t)f(b)$, i.e. $-((1-t)b)^2>-t-(1-t)b^2$.

We have that $1-t>0$ and $t>0$ and $b>0$, so $b^2>0$.
We have that $(1-t)b^2+1>(1-t)b^2>0$. Since $t>0$, we get $[(1-t)b^2+1]t>0$.
Therefore, we have the following:
$$(1-t)b^2t+t>0 \Rightarrow (1-t)b^2(t+1-1)>-t \Rightarrow (1-t)b^2(-(1-t)+1)>-t \\ \Rightarrow -(1-t)^2b^2+(1-t)b^2>-t \Rightarrow -(1-t)^2b^2>-t-(1-t)b^2$$Or should we write it in an other way? (Wondering)