MHB Can Derivatives Exist at Discontinuities of a Piecewise Function?

[mathmari]

Let's pick a point close to 0 that is in $D$.
Say $x=0.095 \in (\frac 1{10}, \frac 1{11})$.
Then $f'(x)=0$ isn't it?

No matter how close we pick $x$ to $0$, as long as it's inside an interval of the form $(\frac 1{k+1}, \frac 1{k})$, which is where $f'(x)$ exists, we have $f'(x)=0$ don't we?
Consequently, the limit is $0$ instead of $1$.

Does this mean that for each $x\in D$ we have that $f'(x)=0$ since $f(x)$ is constant?

 

[I like Serena]

Does this mean that for each $x\in D$ we have that $f'(x)=0$ since $f(x)$ is constant?

Yep. (Nod)

That is, $f(x)$ is constant in each of those disjointed open intervals, although the constant is different on each of those intervals. (Nerd)

 

[mathmari]

Yep. (Nod)

That is, $f(x)$ is constant in each of those disjointed open intervals, although the constant is different on each of those intervals. (Nerd)

Does this mean that every limit of $f'(x)$, when $x\in D$, will be zero?

I mean:

$$\lim_{\substack{x\rightarrow x_0+0, \\ x\in D}} f'(x) =\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}} f'(x) =0$$ for $n\in \mathbb{N}$.

(Wondering)

 

[I like Serena]

Does this mean that every limit of $f'(x)$, when $x\in D$, will be zero?

I mean:

$$\lim_{\substack{x\rightarrow x_0+0, \\ x\in D}} f'(x) =\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}} f'(x) =0$$ for $n\in \mathbb{N}$.

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

But then if $f_+'(x_n), f_-'(x_n)$ and so $f'(x_n)$ (for $n\in \mathbb{N}_0$) exist, do they not have to be equal to $0$ ? (Wondering)

 

[I like Serena]

But then if $f_+'(x_n), f_-'(x_n)$ and so $f'(x_n)$ (for $n\in \mathbb{N}_0$) exist, do they not have to be equal to $0$ ?

Not necessarily.As we have seen, $f_+'(x_0)=f_-'(x_0)=f'(x_0)=1$, while $f'(x)=0$ everywhere else where $f'(x)$ is defined.
That is, $f'$ has a removable discontinuity in $x=0$. (Nerd)That's because for $f_\pm'(x_n)$, we evaluate $\frac{f(x_n+h)-f(x_n)}{h}$, where $f(x_n+h)$ and $f(x_n)$ can fall into different intervals.
And for the limit with $x\in D$, we evaluate $\frac{f(x+h)-f(x)}{h}$, where $f(x+h)$ and $f(x)$ are in the same interval. (Thinking)

 

[mathmari]

As we have seen, $f_+'(x_0)=f_-'(x_0)=f'(x_0)=1$, while $f'(x)=0$ everywhere else where $f'(x)$ is defined.
That is, $f'$ has a removable discontinuity in $x=0$. (Nerd)That's because for $f_\pm'(x_n)$, we evaluate $\frac{f(x_n+h)-f(x_n)}{h}$, where $f(x_n+h)$ and $f(x_n)$ can fall into different intervals.
And for the limit with $x\in D$, we evaluate $\frac{f(x+h)-f(x)}{h}$, where $f(x+h)$ and $f(x)$ are in the same interval. (Thinking)

Why are $f(x+h)$ and $f(x)$ in the same interval when $x\in D$ ? (Wondering)
I have also a question about the case $n=1$.

I'm a bit unsure about $f'(x_1)$. It seems to me that it does exist and is equal to $f_-'(x_1)$.
That's because if we calculate $f'(x)$ according to $\varepsilon$-$\delta$ definition of a limit, that's what we'll find. (Thinking)

Is this because $x_1^+$ is not in the interval that we consider? (Wondering)

The first limit explicitly excludes $x_n=1$, so we cannot calculate that limit.
We can calculate that second limit though.
Note that all $x_n\not\in D$, but that does not prevent us from calculate the limit to it. (Thinking)

We have that $\displaystyle{\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}}f'(x)}$ is not defined, since $1^+\notin D$, right?

For $x\rightarrow 1^-$, we have that $x\in D$, or not? That would mean that $f'(x)=0$, right?
So, we get $\displaystyle{\lim_{\substack{x\rightarrow x_1-0\\ x\in D}}f'(x)=\lim_{\substack{x\rightarrow x_1-0\\ x\in D}}0=0}$, right?

(Wondering)

 

[I like Serena]

Why are $f(x+h)$ and $f(x)$ in the same interval when $x\in D$ ? (Wondering)

$x$ is inside an open interval.
And for $h$ small enough, $x+h$ must be in that same interval.
So when we approach the limit close enough, we have $x$ and $x+h$ both in an interval where the function values are the same. (Thinking)

I have also a question about the case $n=1$.

Is this because $x_1^+$ is not in the interval that we consider?

Yes.
The problem is that different texts define a limit slightly differently.
Some texts require that the upper limit and the lower limit must both exist for the limit to exist.
But certainly the generalized definitions of a limit do not require this.

See wiki that begins explaining the first simplified form, which is actually only for functions $\mathbb R\to \mathbb R$, so it would effectively not even apply to our problem.
And that definition is then inconsistent with the later form in the sections Functions on metric spaces and Functions on topological spaces. (Nerd)

We have that $\displaystyle{\lim_{\substack{x\rightarrow x_n+0, \\ x\in D, x_n\neq 1}}f'(x)}$ is not defined, since $1^+\notin D$, right?

For $x\rightarrow 1^-$, we have that $x\in D$, or not? That would mean that $f'(x)=0$, right?
So, we get $\displaystyle{\lim_{\substack{x\rightarrow x_1-0\\ x\in D}}f'(x)=\lim_{\substack{x\rightarrow x_1-0\\ x\in D}}0=0}$, right?

Yep. (Nod)

 

[mathmari]

$x$ is inside an open interval.
And for $h$ small enough, $x+h$ must be in that same interval.
So when we approach the limit close enough, we have $x$ and $x+h$ both in an interval where the function values are the same. (Thinking)

Ah ok!

The problem is that different texts define a limit slightly differently.
Some texts require that the upper limit and the lower limit must both exist for the limit to exist.
But certainly the generalized definitions of a limit do not require this.

See wiki that begins explaining the first simplified form, which is actually only for functions $\mathbb R\to \mathbb R$, so it would effectively not even apply to our problem.
And that definition is then inconsistent with the later form in the sections Functions on metric spaces and Functions on topological spaces. (Nerd)
)

I see! So, we have that $f'(x_1)$ exists and this derivative is equal to $f'(x_1)=f_-'(x_1)=0$. At the case $x_0=0$ we have that the derivative is discontinuous since the limit is equal to $0$, $\displaystyle{\lim_{\substack{x\rightarrow x_0+0, \\ x\in D}}f'(x)=\lim_{\substack{x\rightarrow x_0-0\\ x\in D}}f'(x)=0}$ but the value of the derivative is $f'(x_0)=1$. At the case $n\in \mathbb{N}\setminus \{1\}$ we have that the limit is equal to zero, but the value of $f'(x_n)$ doesn't exist. Does this mean again that the derivative is discontinuous in $x_n$ ? Or would we have this meaning only if $f'(x_n)$ would exist and would be different from $0$ ? (Wondering)

 

[I like Serena]

Ah ok!

I see! So, we have that $f'(x_1)$ exists and this derivative is equal to $f'(x_1)=f_-'(x_1)=0$.

At the case $x_0=0$ we have that the derivative is discontinuous since the limit is equal to $0$, $\displaystyle{\lim_{\substack{x\rightarrow x_0+0, \\ x\in D}}f'(x)=\lim_{\substack{x\rightarrow x_0-0\\ x\in D}}f'(x)=0}$ but the value of the derivative is $f'(x_0)=1$.

Yep. (Nod)

At the case $n\in \mathbb{N}\setminus \{1\}$ we have that the limit is equal to zero, but the value of $f'(x_n)$ doesn't exist. Does this mean again that the derivative is discontinuous in $x_n$ ? Or would we have this meaning only if $f'(x_n)$ would exist and would be different from $0$ ? (Wondering)

Continuity and discontinuity are only defined at points where the function (in our case the derivative) is defined.
So indeed, we can only say that $f'$ is discontinuous in $x_n$ if it exists and has a limit that is different. (Happy)

 

[mathmari]

Continuity and discontinuity are only defined at points where the function (in our case the derivative) is defined.
So indeed, we can only say that $f'$ is discontinuous in $x_n$ if it exists and has a limit that is different. (Happy)

I see! Thank you so much! (Happy)