Find a function so that the composition is continuous

[mathmari]

Hey!

For which real constants $a,b,c$ is the following function $f$ continuous on $\mathbb{R}$ ?
$$f(x)=\begin{cases}1+x^2 & \text{ for } x\leq 1, \\ ax-x^3 & \text{ for } 1<x< 2 \\ b & \text{ for } x=2\\ cx^2 & \text{ for } x>2\end{cases}$$
For $a=1, b=-6, c=-\frac{3}{2}$ the function $f$ is discontinuous at $x_0=1$. Are there continuous functions $g:\mathbb{R}\rightarrow \mathbb{R}$, such that $f\circ g$ or $g\circ f$ is continuous on whole $\mathbb{R}$ ?

We have to check the continuity at the points $x=1$ and $x=2$.

We consider the point $x=1$:

We have that $\lim_{x\rightarrow 1^-}f(x)=\lim_{x\rightarrow 1^-}(1+x^2)=2$, $\lim_{x\rightarrow 1^+}f(x)=\lim_{x\rightarrow 1^+}(ax-x^3)=a-1$ and $f(1)=1+1^2=2$.

So that the function is continuous at $x=1$ it must hold $a-1=2\Rightarrow a=3$.

We consider the point $x=2$:

We have that $\lim_{x\rightarrow 2^-}f(x)\overset{ a=3 }{ = } \lim_{x\rightarrow 2^-}(3x-x^3)=6-8=-2$, $\lim_{x\rightarrow 2^+}f(x)=\lim_{x\rightarrow 2^+}(cx^2)=4c$ and $f(2)=b$.

So that the function is continuous at $x=2$ it must hold $-2=4c=b$. So we get $b=-2$ and $c=-\frac{1}{2}$.

Is everything correct?

For $a=1, b=-6, c=-\frac{3}{2}$ we get the function $$f(x)=\begin{cases}1+x^2 & \text{ for } x\leq 1, \\ x-x^3 & \text{ for } 1<x< 2 \\ -6 & \text{ for } x=2\\ -\frac{3}{2}x^2 & \text{ for } x>2\end{cases}$$

How can we find (or check if there exist) continuous functions $g$ such that $f\circ g$ or $g\circ f$ is continuous on whole $\mathbb{R}$ ?

 

[I like Serena]

Is everything correct?

Hey mathmari! (Smile)

Yep. (Nod)

For $a=1, b=-6, c=-\frac{3}{2}$ we get the function $$f(x)=\begin{cases}1+x^2 & \text{ for } x\leq 1, \\ x-x^3 & \text{ for } 1<x< 2 \\ -6 & \text{ for } x=2\\ -\frac{3}{2}x^2 & \text{ for } x>2\end{cases}$$

How can we find (or check if there exist) continuous functions $g$ such that $f\circ g$ or $g\circ f$ is continuous on whole $\mathbb{R}$ ?

What do we get if we apply the definition of continuity to $g\circ f$ for some unknown function $g$? (Wondering)

 

[mathmari]

What do we get if we apply the definition of continuity to $g\circ f$ for some unknown function $g$? (Wondering)

Let $a$ be an arbitrary point.

We want that $f(g(x))$ is continuous at each $a$, so $f$ is continuous at each $g(a)$.

We know that $g$ is a continuos function. So it will be continuous at each $a$.
That means that for all $\epsilon>0$ there is some $\delta>0$ such that $$|x-a|<\delta \Rightarrow|g(x)-g(a)|<\epsilon$$

Since $f$ is continuous at $g(a)$ we have that for all $\tilde{\epsilon} > 0$ there is some $\tilde{\delta}$ such that $$|g(x) - g(a)| < \tilde{\delta} \Rightarrow |f(g(x))-f(g(a))|<\tilde{\epsilon}$$

For $\epsilon=\tilde{\delta}$ we have the following:

For all $\epsilon > 0$ there is some $\delta > 0$ and a $\tilde{\delta}> 0$ such that $$|x-a| < \delta\Rightarrow |g(x)-g(a)|<\tilde{\delta}\Rightarrow |f(g(x)) - f(g(a))|<\epsilon$$
right? (Wondering)

 

[I like Serena]

Let $a$ be an arbitrary point.

We want that $f(g(x))$ is continuous at each $a$, so $f$ is continuous at each $g(a)$.

We know that $g$ is a continuos function. So it will be continuous at each $a$.
That means that for all $\epsilon>0$ there is some $\delta>0$ such that $$|x-a|<\delta \Rightarrow|g(x)-g(a)|<\epsilon$$

Since $f$ is continuous at $g(a)$ we have that for all $\tilde{\epsilon} > 0$ there is some $\tilde{\delta}$ such that $$|g(x) - g(a)| < \tilde{\delta} \Rightarrow |f(g(x))-f(g(a))|<\tilde{\epsilon}$$

For $\epsilon=\tilde{\delta}$ we have the following:

For all $\epsilon > 0$ there is some $\delta > 0$ and a $\tilde{\delta}> 0$ such that $$|x-a| < \delta\Rightarrow |g(x)-g(a)|<\tilde{\delta}\Rightarrow |f(g(x)) - f(g(a))|<\epsilon$$
right? (Wondering)

Let's not jump to $\epsilon$-$\delta$ definitions of limits. (Worried)What we need for continuity of $g\circ f$ is that $\lim_{x\to x_0} g(f(x)) = g(f(x_0))$.
More specifically in our case that means that:
$$\lim_{x\to 1^+} g(f(x)) = \lim_{x\to 1^-} g(f(x)) = g(f(1))$$
What do we require from $g$ for that to be true? (Wondering)

 

[mathmari]

What we need for continuity of $g\circ f$ is that $\lim_{x\to x_0} g(f(x)) = g(f(x_0))$.
More specifically in our case that means that:
$$\lim_{x\to 1^+} g(f(x)) = \lim_{x\to 1^-} g(f(x)) = g(f(1))$$
What do we require from $g$ for that to be true? (Wondering)

$g$ has to continuous at $f(1)=2$, right? (Wondering)

 

[I like Serena]

$g$ has to continuous at $f(1)=2$, right? (Wondering)

Not quite, since f is not continuous at 1. (Thinking)

 

[mathmari]

Not quite, since f is not continuous at 1. (Thinking)

Do we maybe take as $g$ a constant function? Then we would have that $g(x)=C$ and $g(f(x))=C$, which is continuous everywhere? (Wondering)

 

[I like Serena]

Do we maybe take as $g$ a constant function? Then we would have that $g(x)=C$ and $g(f(x))=C$, which is continuous everywhere? (Wondering)

But $f$ is not continuous is it?
So if $g$ is a constant function, $g\circ f$ won't be continuous either would it?

Suppose we substitute the relevant parts of $f$ into the limit expression... (Thinking)

 

[mathmari]

Suppose we substitute the relevant parts of $f$ into the limit expression... (Thinking)

Do you mean the following?

$$\lim_{x\to 1^+} g(f(x)) =\lim_{x\to 1^+} g(x-x^3) \\ \lim_{x\to 1^-} g(f(x)) = \lim_{x\to 1^-} g(1+x^2) $$

So, it must hold that $$\lim_{x\to 1^+} g(x-x^3)= \lim_{x\to 1^-} g(1+x^2)$$ right? (Wondering)

 

[I like Serena]

Indeed. And since g is continuous, we can evaluate those limits, can't we? (Wondering)

 

[mathmari]

Indeed. And since g is continuous, we can evaluate those limits, can't we? (Wondering)

Ah yes. So, we get $$\lim_{x\to 1^+} g(x-x^3)= \lim_{x\to 1^-} g(1+x^2)\Rightarrow g\left (\lim_{x\to 1^+}(x-x^3)\right )= g\left (\lim_{x\to 1^-} (1+x^2)\right )\Rightarrow g(0 )= g(2)$$ right?

Such a function is for example $g(x)=x\cdot (x-2)$, isn't it?

(Wondering)

 

[I like Serena]

Ah yes. So, we get $$\lim_{x\to 1^+} g(x-x^3)= \lim_{x\to 1^-} g(1+x^2)\Rightarrow g\left (\lim_{x\to 1^+}(x-x^3)\right )= g\left (\lim_{x\to 1^-} (1+x^2)\right )\Rightarrow g(0 )= g(2)$$ right?

Such a function is for example $g(x)=x\cdot (x-2)$, isn't it?

Yep.

And I'm just realizing that I was wrong before. If $g$ is a constant function, $g\circ f$ will also be a constant function and therefore continuous. (Blush)

 

[mathmari]

Yep.

And I'm just realizing that I was wrong before. If $g$ is a constant function, $g\circ f$ will also be a constant function and therefore continuous. (Blush)

Ah ok!

If $g$ is a constant, $g(x)=C$. then we have that $g(f(x))=C$ and a constant function is everywhere continuous, right?

Is it possible that $f\circ g$ is continuous for some function $g$ ?

Can we take here also a constanst function? When $g(x)=C$ then we have that $f\circ g(x)=f(g(x))=f(C)$, which is also constant, right?

(Wondering)

 

[I like Serena]

Yep.

 

[mathmari]

But can $C$ be also $1$ ?

Then we would have $f\circ g(x)=f(g(x))=f(1)=2$, or isn't $f(1)=2$ ? (Wondering)

 

[I like Serena]

Sure. That's all correct. (Nod)

 

[mathmari]

Do you mean the following?

$$\lim_{x\to 1^+} g(f(x)) =\lim_{x\to 1^+} g(x-x^3) \\ \lim_{x\to 1^-} g(f(x)) = \lim_{x\to 1^-} g(1+x^2) $$

So, it must hold that $$\lim_{x\to 1^+} g(x-x^3)= \lim_{x\to 1^-} g(1+x^2)$$ right? (Wondering)

Ah yes. So, we get $$\lim_{x\to 1^+} g(x-x^3)= \lim_{x\to 1^-} g(1+x^2)\Rightarrow g\left (\lim_{x\to 1^+}(x-x^3)\right )= g\left (\lim_{x\to 1^-} (1+x^2)\right )\Rightarrow g(0 )= g(2)$$ right?

Such a function is for example $g(x)=x\cdot (x-2)$, isn't it?

(Wondering)

Could we do that also for $f\circ g$ ?

$$\lim_{x\to 1^+} f(g(x)) =\lim_{x\to 1^+} (g(x)-g^3(x))=g(1)-g^3(1) \\ \lim_{x\to 1^-} g(f(x)) = \lim_{x\to 1^-} (1+g^2(x))=1+g^2(1)$$

So we have to find a function $g$ such that $g(1)-g^3(1)=1+g^2(1)$.

Which function could that be? (Wondering)

 

[I like Serena]

Could we do that also for $f\circ g$ ?

$$\lim_{x\to 1^+} f(g(x)) =\lim_{x\to 1^+} (g(x)-g^3(x))=g(1)-g^3(1) \\ \lim_{x\to 1^-} g(f(x)) = \lim_{x\to 1^-} (1+g^2(x))=1+g^2(1)$$

So we have to find a function $g$ such that $g(1)-g^3(1)=1+g^2(1)$.

Which function could that be? (Wondering)

I'm afraid that for $f\circ g$ it's a little more complicated.
In this case it's not $x$ that has to approach $1^+$, but $g(x)$. (Worried)

Suppose that there is some $x_0$ such that $g(x_0)=1$ and $g$ is increasing in $x_0$.

Then
$$\lim_{{x\to x_0}^+} f(g(x)) = \lim_{y\to 1^+} f(y) =\lim_{y\to 1^+} (y-y^3)=0 \\
\lim_{{x\to x_0}^-} f(g(x)) = \lim_{y\to 1^-} f(y) =\lim_{y\to 1^-} (1+y^2)=2
$$
That meanst that $f\circ g$ would not be continuous in $x_0$ then.

So $g$ can't have a function value of $1$ where $g$ is monotone as well.
However, $g$ might still have a function value of $1$ if it is a (global) extreme. (Thinking)

 

[mathmari]

I'm afraid that for $f\circ g$ it's a little more complicated.
In this case it's not $x$ that has to approach $1^+$, but $g(x)$. (Worried)

Suppose that there is some $x_0$ such that $g(x_0)=1$ and $g$ is increasing in $x_0$.

Then
$$\lim_{{x\to x_0}^+} f(g(x)) = \lim_{y\to 1^+} f(y) =\lim_{y\to 1^+} (y-y^3)=0 \\
\lim_{{x\to x_0}^-} f(g(x)) = \lim_{y\to 1^-} f(y) =\lim_{y\to 1^+} (1+y^2)=2
$$
That meanst that $f\circ g$ would not be continuous in $x_0$ then.

So, is $f\circ g$ continuous only when $g$ is a constant function? (Wondering)

 

[I like Serena]

So, is $f\circ g$ continuous only when $g$ is a constant function?

$f$ has jump discontinuities in $1$ and $2$.
That means that $g$ cannot 'cross' those values.
But for instance $g(x)= 3 + x^2$ will work.
And $g(x)= 2 + x^2$ will also work since $\lim_{x\to 2^+} f(x) = f(2)$. (Thinking)

 

[mathmari]

$f$ has jump discontinuities in $1$ and $2$.
That means that $g$ cannot 'cross' those values.
But for instance $g(x)= 3 + x^2$ will work.
And $g(x)= 2 + x^2$ will also work since $\lim_{x\to 2^+} f(x) = f(2)$. (Thinking)

Ah I see! Thank you so much! (Yes)