Strong differentiability condition

[jostpuur]

Assume that a point $x$ is an interior point of domain of some function $f:[a,b]\to\mathbb{R}$, and assume that the limit

$$\lim_{(\delta_1,\delta_2)\to (0,0)} \frac{f(x+\delta_2)-f(x+\delta_1)}{\delta_2-\delta_1}$$

exists. What does this imply?

Well I know it implies that $f'(x)$ exists, but does it imply more? Does it imply that the derivative of $f$ exists in some neighbourhood of $x$ and is continuous? Where's a counter example?

To be more presice, we can define a set

$$\mathcal{D}=\big\{(\delta_1,\delta_2)\;\big|\; |\delta_1|<D,\; |\delta_2|<D,\; \delta_1\neq \delta_2\big\}$$

with some small $D$, and then consider the given expression as a mapping

$$\mathcal{D}\to\mathbb{R}$$

So the domain is a square with a thin diagonal removed. Then we seek its limit at the center point $(0,0)$.

 

[Stephen Tashi]

Well I know it implies that $f'(x)$ exists,.

What about the function defined by
$f(x) = x^2$ for $x \ne 0$
$f(0) = 3$

What's $f'(0)$ ?

 

[jostpuur]

I just managed to prove that if the mentioned limit exists, $f$ will necessarily be continuous on some interval $]x-D,x+D[$. So not only will the function be continuous at the point $x$, but also in some interval around it.

That was interesting. Next, I'm interested in the differentiability...

What about the function defined by
$f(x) = x^2$ for $x \ne 0$
$f(0) = 3$

What's $f'(0)$ ?

With this function the mentioned limit does not exist. The question is that what are the consequences of the existence of the limit.

 

[jostpuur]

I just understood that Tashi must have assumed that I had been demanding $\delta_1\neq 0$ and $\delta_2\neq 0$ during the limit. There is no such assumption. Only $\delta_1\neq \delta_2$. See the explanation about the domain $\mathcal{D}$ in my opening post.

 

[HallsofIvy]

I just understood that Tashi must have assumed that I had been demanding $\delta_1\neq 0$ and $\delta_2\neq 0$ during the limit. There is no such assumption. Only $\delta_1\neq \delta_2$. See the explanation about the domain $\mathcal{D}$ in my opening post.

Tashi was NOT "assuming" that- that is part of the definition of "limit". It is you that have right to say that $\delta_1$ and $\delta_2$ are or can be 0.

Given that definition, of derivative, it is NOT true that f must be continuous or differentiable in the usual sense at the point.

 

[jostpuur]

Tashi was NOT "assuming" that- that is part of the definition of "limit". It is you that have right to say that $\delta_1$ and $\delta_2$ are or can be 0.

You are referring to some stupid high schoold definition that nearly nobody is interested about. This is the real definition: Assume $X$ is some topological space, $x\in X$ some point in it, and $x_1,x_2,x_3,\ldots \in X$ is some sequence in it. We say that the sequence converges to $x$ if for all open $U\subset X$ such that $x\in U$ there exist a $N_U\in\mathbb{N}$ with the property

$$n\geq N_U\quad\implies\quad x_n\in U$$

Now the domain of parameters $(\delta_1,\delta_2)$ is

$$\mathcal{D}=\big\{(\delta_1,\delta_2)\;\big|\; |\delta_1|<D,\; |\delta_2|<D,\; \delta_1\neq \delta_2\big\}$$

which is naturally a subset of $\mathbb{R}^2$, so we'll set $X=\mathbb{R}^2$ and use the ordinary topology there. It would make no sense to add some additional constraints such as $\delta_1\neq 0$ and $\delta_2\neq 0$ to the general definition.

Just to be clear I want to ask one thing. I'm sure we agree that the sequence

$$1,\frac{1}{2},\frac{1}{3},\ldots$$

converges to zero. What about the constant sequence

$$0,0,0,\ldots ?$$

IMO this converges to the zero too. Do you insist that it does not converge to zero?

 

[jostpuur]

I managed to prove one more thing. Assume that a function $f:\;]a,b[\to\mathbb{R}$ is differentiable in its domain. Fix some point $x\in\;]a,b[$. Now the derivative $f'$ is continuous at point $x$ iff the limit
$$\lim_{(\delta_1,\delta_2)\to (0,0)}\frac{f(x+\delta_2)-f(x+\delta_1)}{\delta_2-\delta_1}$$
exists.

Very interesting result! The easier direction comes with the intermediate value theorem. More difficult direction can be proven with an antithesis and some sequence $\lambda_n\to x$ that the antithesis provides.

 

[jostpuur]

Just to be clear I want to ask one thing.

hmm.. ok I think it could be that my question was little provocative since it seems that limits

$$\lim_{n\to\infty}x_n$$

and

$$\lim_{x\to x_0} f(x)$$

could be given deliberately different kind of definitions. If you have some very authorative text that says that the latter limit "must be" taken under the constraint $x\neq x_0$, I believe you. Anyway, I have two comments: First, if a definition is unnatural, it is ok to defy it with more logical new definitions. Second, the definitions that you are referring to probably would imply the constraint $(\delta_1,\delta_2)\neq (0,0)$, and not $\delta_1\neq 0$ and $\delta_2\neq 0$? I made it clear in the beginning that we are assuming $\delta_1\neq \delta_2$ anyway.

 

[jostpuur]

oh dear, I just realized I have still not made it 100% unambiquously clear what I meant with the limit then . Well here comes the definition. We say that a real number $L$ is the limit denoted as

$$\lim_{(\delta_1,\delta_2)\to (0,0)}\frac{f(x+\delta_2)-f(x+\delta_1)}{\delta_2-\delta_1}$$

if for all $\epsilon>0$ there exists an open set $U_{\epsilon}\subset\mathbb{R}^2$ such that $(0,0)\in U_{\epsilon}$ and

$$\Phi(U_{\epsilon}\cap\mathcal{D})\subset \;]L-\epsilon,L+\epsilon[$$

where the function $\Phi$ has been defined by the formula

$$\Phi:\mathcal{D}\to\mathbb{R},\quad\quad \Phi(\delta_1,\delta_2) = \frac{f(x+\delta_2)-f(x+\delta_1)}{\delta_2-\delta_1}$$

Do you believe me now?

 

[Stephen Tashi]

First, if a definition is unnatural, it is ok to defy it with more logical new definitions.

You don't get any points for being humble, but at least you are objective enough to catch some of your own mistakes.

It's ok to make new definitions, but you need to state them. I don't think you'll persuade the mathematical world that a new definition of limit is needed.

I made it clear in the beginning that we are assuming $\delta_1\neq \delta_2$ anyway.

You only made a statement that defined a mapping with the condition on it's domain that $\delta_1 \ne \delta_2$ You didn't provide a complete statement that defined a new kind of limit.

 

[jostpuur]

Check my definition in the post #9. Does it look like familiar or "a new kind of limit"? I would understand if it looks "new" to a high school student.

 

[Stephen Tashi]

Do you believe me now?

You clearly stated the definition of a new kind of limit. I don't see that the existence of this new kind of limit implies anything different that the existence of the (usual kind of) limit that defines the derivative f'(x). Do have an example where one limit exists and the other doesn't?

 

[jostpuur]

I don't have a topology book myself, so I would need go to library to check what the real definitions are precisely. But anyway I have hard time believing that my definition could be new. It repeats the idea that small environments must be mapped into small environments. Can you tell what your definition is? It hasn't been made clear either in this thread.

I don't see that the existence of this new kind of limit implies anything different that the existence of the (usual kind of) limit that defines the derivative f'(x).

This works:

$$f(x) = \left\{\begin{array}{ll}<br /> x^2\sin\Big(\frac{1}{x}\Big),\quad & x \neq 0 \\<br /> 0,\quad & x= 0 \\<br /> \end{array}\right.$$

Now

$$\lim_{\delta_2\to 0}\lim_{\delta_1\to 0 }\frac{f(\delta_2)-f(\delta_1)}{\delta_2-\delta_1}<br /> = \lim_{\delta_2\to 0} \delta_2\sin\Big(\frac{1}{\delta_2}\Big) = 0$$

so if the limit exists, it cannot be anything else than zero. Then define sequences $\delta_1^1,\delta_1^2,\delta_1^3,\ldots$ and $\delta_2^1,\delta_2^2,\delta_2^3,\ldots$ by setting

$$\delta_1^n = \frac{1}{2\pi n+ \frac{\pi}{2}},\quad\quad \delta_2^n = \frac{1}{2\pi n - \frac{\pi}{2}}.$$

Now $(\delta_1^n,\delta_2^n)\to (0,0)$ when $n\to\infty$, but on the other hand

$$\frac{f(\delta_2^n)-f(\delta_1^n)}{\delta_2^n-\delta_1^n}<br /> = \frac{\frac{1}{(2\pi n - \frac{\pi}{2})^2}\sin\big(2\pi n - \frac{\pi}{2}\big)<br /> -\frac{1}{(2\pi n + \frac{\pi}{2})^2}\sin\big(2\pi n + \frac{\pi}{2}\big)}<br /> {\frac{1}{2\pi n - \frac{\pi}{2}} - \frac{1}{2\pi n + \frac{\pi}{2}}}<br /> = \cdots = -\frac{2}{\pi} + O\Big(\frac{1}{n^2}\Big)$$

so we see that if the limit exists, it cannot be anything else than $-\frac{2}{\pi}$. According to these remarks, the limit does not exist.

 

[Stephen Tashi]

Can you tell what your definition is? It hasn't been made clear either in this thread.

For a real valued function $f$ defined on a metric space, $\lim_{x \rightarrow a } f(x) = L$ means for each real number $\epsilon > 0$ there exists a real number $\delta$ such that $0 < | x - a| < \delta$ implies $| f(x) - L | < \epsilon$ (where the $|...|$ denotes the norm function that defines the metric).

If we want to know the definition for a topological space, we can consult http://en.wikipedia.org/wiki/Limit_of_a_function. Note it makes a statement about $f(U \cap \Omega - \{p\})$, not about $f(U \cap \Omega)$.

 

[jostpuur]

It is my opinion that the definitions are bad, because it makes no sense to artificially remove the limit point from the allowed domain through which the convergence happens, but it doesn't matter now because of this:

the definitions that you are referring to probably would imply the constraint $(\delta_1,\delta_2)\neq (0,0)$, and not $\delta_1\neq 0$ and $\delta_2\neq 0$?

 

[Stephen Tashi]

Returning to the question in the original post:

Does it imply that the derivative of $f$ exists in some neighbourhood of $x$ and is continuous? Where's a counter example?

What about
$f(x) = 0$ if $x$is a rational number.
$f(x) = x^2$ if $x$ is an irrational number.

 

[jostpuur]

First choose some sequence $\delta_1^1,\delta_1^2,\delta_1^3,\ldots$ such that all $\delta_1^n$ are irrational and $0<\delta_1^n<\frac{1}{n}$. With fixed $n$ we can choose a rational $\delta_2^n$ to be as close to $\delta_1^n$ as we want, so we can choose a rational sequence $\delta_2^1,\delta_2^2,\delta_2^3,\ldots$ such that $0<\delta_2^n<\frac{1}{n}$ and

$$\Big|\frac{(\delta_1^n)^2}{\delta_2^n - \delta_1^n}\Big| \geq n$$

 

[pwsnafu]

It is my opinion that the definitions are bad, because it makes no sense to artificially remove the limit point from the allowed domain through which the convergence happens

It makes perfect sense to remove p. Take the function f(0) = 1, and f(x) = 0 elsewhere. Suppose we wanted to take the limit at x=0. We know the limit is zero, so if V = (-1/4, 1/4) we must exclude $f(p) = 1 \notin V$

 

[jostpuur]

If the limit

$$\lim_{x\to x_0}f(x)$$

exists, and if the limit

$$\lim_{n\to\infty} x_n = x_0$$

holds, it would be reasonable to expect

$$\lim_{n\to\infty} f(x_n) = \lim_{x\to x_0} f(x)$$

too. The standard definition implies that this result does not always hold, which means that the standard definition is bad.

 

[pwsnafu]

it would be reasonable to expect

$$\lim_{n\to\infty} f(x_n) = \lim_{x\to x_0} f(x)$$

too.

What‽ For an arbitrary topological space? No, it would not be reasonable to expect that.

 

[jostpuur]

Why do you emphasize that my assumption wouldn't be reasonable on an arbitrary topological space? According to the standard definition, the assumption isn't even true on the real line.

The formula I wrote looks natural. Therefore it would be natural to expect it to be true.

 

[pwsnafu]

Why do you emphasize that my assumption wouldn't be reasonable on an arbitrary topological space? According to the standard definition, the assumption isn't even true on the real line.

The formula I wrote looks natural. Therefore it would be natural to expect it to be true.

Err, it is true on the real line.

 

[jostpuur]

Err, it is true on the real line.

You are making a mistake by not being careful with the definitions. Consider your own example:

It makes perfect sense to remove p. Take the function f(0) = 1, and f(x) = 0 elsewhere. Suppose we wanted to take the limit at x=0. We know the limit is zero, so if V = (-1/4, 1/4) we must exclude $f(p) = 1 \notin V$

Then define a sequence $x_1,x_2,x_3,\ldots$ so that $x_n=0$ for all $n$.

Now

$$\lim_{x\to 0}f(x) = 0$$

and

$$\lim_{n\to\infty} x_n = 0$$

but still

$$\lim_{n\to\infty} f(x_n) = 1 \neq 0 = \lim_{x\to 0 }f(x)$$

 

[pwsnafu]

You are making a mistake by not being careful with the definitions. Consider your own example:
Then define a sequence $x_1,x_2,x_3,\ldots$ so that $x_n=0$ for all $n$.

Now

$$\lim_{x\to 0}f(x) = 0$$

and

$$\lim_{n\to\infty} x_n = 0$$

but still

$$\lim_{n\to\infty} f(x_n) = 1 \neq 0 = \lim_{x\to 0 }f(x)$$

Except the definition states $x_n \neq a$ for all $n$.

Edit: a = 0 in this example.

 

[jostpuur]

Are you trying to tell me that

$$\lim_{n\to \infty} 0 = 0$$

would be wrong?

Or are you trying to tell me that

$$\lim_{n\to\infty} f(0) = 1$$

would be wrong?

I'm sure you are not trying to tell me that

$$\lim_{x\to 0}f(x) = 0$$

would be wrong, because you yourself stated this as something we know anyway.

If all these claims are right, then what I wrote was right, and there is nothing to be complained about.

 

[pwsnafu]

From Wikipedia Limit of a function in terms of sequences:

The limit of f is L if and only if for all sequences $x_n$ (with $x_n$ not equal to $a$ for all $n$) converging to $a$, the sequence $f(x_n)$ converges to $L$.

The entire point of defining limits is so that L is not influenced by what f is doing at x=a. You're example started with $x_n=0$ which is a sequence that is not considered.

 

[jostpuur]

... it would be reasonable to expect

$$\lim_{n\to\infty} f(x_n) = \lim_{x\to x_0} f(x)$$

too. The standard definition implies that this result does not always hold, which means that the standard definition is bad.

The formula I wrote looks natural. Therefore it would be natural to expect it to be true.

Err, it is true on the real line.

You made a mistake here, and you should admit it (at least to yourself) at some point.

 

[Stephen Tashi]

$\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ wouldn't exist if $\frac{f(x+h)-f(x)}{h}$ were required to satisfy some inequality when $h = 0$, so I think the traditional definition of limit will prevail due to people's interest in derivatives.

----------------


For the purposes of this thread, let's explicitly define what a "limit with side conditions" means.
Let $f$ be a function from one metric space to another. Let $S$ be a set of points in the first metric space (the one containing the domain of $f$). Let $a$ be a point in first metric space.

Define $\lim_{x \rightarrow a, x \notin S} f(x) = L$ to mean that for each number $\epsilon > 0$ there exists a number $\delta$ such that $( 0 < |x-a| < \delta$ and $x \notin S$) implies $| f(x) - L | \lt \epsilon$ where the norms $|...|$ are those of the respective metric spaces.


The idea in the original post is consider a real valued function $f(x)$ defined on the real number line and assume the existence of the limit a related real valued function defined on the plane:

$$\lim_{(\delta_1,\delta_2)\to (0,0), (\delta_1,\delta_2) \notin S} \frac{f(x+\delta_2)-f(x+\delta_1)}{\delta_2-\delta_1} = L$$

where $S$ is the of points in the plane $\{(\delta_1,\delta_2) | \delta_1 = \delta_2\}$.

For a given $\epsilon > 0$ we can find a $\delta$ such that $( 0 < |(\delta_1,\delta_2) - (0,0)| < \delta$ and $\delta_1 \ne \delta_2)$ implies $| \frac{ f(x+\delta_1) - f(x + \delta_2)}{\delta_1 - \delta_2} - L | \lt \epsilon$.

I suggest trying to show that this implies that there exists an interval $I$ on the real number line , $I = (x - \delta, x + \delta)$ and a constant $K$ such that for each pair of points $x_1, x_2$ in $I$ , $| x_1 - x2 | \lt K |f(x_1) - f(x_2)|$. One candidate for $K$ might be $|L| + \epsilon$.

This would show that $f(x)$ is Lipschitz continuous in $I$. (http://en.wikipedia.org/wiki/Lipschitz_continuity) So known results about Lipschitz continuity could be used.

 

[pwsnafu]

You made a mistake here, and you should admit it (at least to yourself) at some point.

If you mean unconstrained sequences, sure.

Anyway, I've explained to you why we delete the point in the definition. I urge you to think about why it is important to do that.

 

[jostpuur]

$\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$ wouldn't exist if $\frac{f(x+h)-f(x)}{h}$ were required to satisfy some inequality when $h = 0$, so I think the traditional definition of limit will prevail due to people's interest in derivatives.

The problem with the standard definition of limit of a function is that it is obviously defined with a heavy focus on the derivative, and it is not clear why the definition should remain like that in more general situations. Especially when a more reasonable generalization would be available. The most reasonable definition would be that if $X,Y$ are topological spaces, and $\phi:\mathcal{D}\to Y$ is a function where $\mathcal{D}\subset X$, we examine the condition $\phi(U\cap\mathcal{D})\subset V$ with open $U\subset X$ and $V\subset Y$.

Notice that if a function is defined by a formula

$$\delta\mapsto \frac{f(x+\delta)-f(x)}{\delta}$$

the domain will obviously be something like $\mathcal{D}=\;]-D,0[\;\cup\;]0,D[$. So there will be no reason to write conditions such as $\phi((U\cap\mathcal{D})\backslash\{0\})\subset V$, because the origin is outside the domain anyway.

Is there any situation, where we actually have a reason to forbid some points from inside the domain during limit?