Punctual and uniform convergence

[Kernul]

Homework Statement

$I$ a set of real number and $f_k : I \rightarrow \mathbb{R}$ a succession of real functions defined in $I$.
We say that $f_k$ converges punctually in $I$ to the function $f : I \rightarrow \mathbb{R}$ if
$$\lim_{k \to \infty} f_k(x) = f(x), \hspace{1cm} \forall x \in I$$
Which is like saying:
$$\forall \epsilon > 0,\hspace{1mm} \forall x \in I,\hspace{1mm} \exists \nu_{\epsilon, x} \in \mathbb{R} : |f_k(x) - f(x)| < \epsilon \hspace{1cm} \forall k > \nu_{\epsilon, x}$$

Homework Equations

The Attempt at a Solution

What is the $\nu_{\epsilon, x}$? Is it the same in the normal succession of real numbers? Could someone make some examples?
And I've read that for the uniform convergence, this $\nu_{\epsilon, x}$ depends only from $\epsilon$, so $\nu_{\epsilon}$, and that the uniform convergence implies the punctual convergence but not vice versa. Any example with these too?

 

[fresh_42]

Homework Statement

$I$ a set of real number and $f_k : I \rightarrow \mathbb{R}$ a succession of real functions defined in $I$.
We say that $f_k$ converges punctually in $I$ to the function $f : I \rightarrow \mathbb{R}$ if
$$\lim_{k \to \infty} f_k(x) = f(x), \hspace{1cm} \forall x \in I$$
Which is like saying:
$$\forall \epsilon > 0,\hspace{1mm} \forall x \in I,\hspace{1mm} \exists \nu_{\epsilon, x} \in \mathbb{R} : |f_k(x) - f(x)| < \epsilon \hspace{1cm} \forall k > \nu_{\epsilon, x}$$

Homework Equations

The Attempt at a Solution

What is the $\nu_{\epsilon, x}$? Is it the same in the normal succession of real numbers? Could someone make some examples?
And I've read that for the uniform convergence, this $\nu_{\epsilon, x}$ depends only from $\epsilon$, so $\nu_{\epsilon}$, and that the uniform convergence implies the punctual convergence but not vice versa. Any example with these too?

You may consider the sequence $f_k := \max \{k-k^2\cdot \,\vert \,x-\frac{1}{k}\,\vert \, , 0\}$ and $f(x)=0$ as its limit.

Covergence here is $f_k \longrightarrow f$. That is $f_k(x) \longrightarrow f(x)$. Now the $\nu$ define in a way how fast this is because they (usually) become bigger the smaller the $\epsilon$ is. So the difference between punctual and uniform convergence is basically whether this "speed of convergence" is the same at all points $x$, in which case we call it uniform, or whether this "speed of convergence" depends on where we are on the $x-$axis, in which case it is only punctual.

 

[hilbert2]

One example of punctual converge is how the function $f_{n}(x) = e^{-x^{2n}}$ approaches the function "$g : g(x) = 1$ if $|x|<1$, $g(x) = 1/e$ if $|x| = 1$, $g(x) = 0$ if $|x|>1$" when the natural number $n$ grows without bound. If you set $\epsilon = 0.05$, there will always be a point $x$ for which $f_{n}(x) = 0.9$ and outside the chosen bounds, no matter how large $n$ is.

An example of uniform convergence is how the sequence $f_{n}(x) = x + 1/n$ approaches $g(x) = x$ when $n\rightarrow \infty$

 

[Mark44]

punctual convergence

I've never seen it described as "punctual" convergence. The English words point and punctual are derived from Latin "punctum," but the usual meaning of punctual is "on time," as in a punctual arrival of a student to a class.

The type of convergence discussed in this thread is called "pointwise" convergence in English.

 

[Kernul]

You may consider the sequence $f_k := \max \{k-k^2\cdot \,\vert \,x-\frac{1}{k}\,\vert \, , 0\}$ and $f(x)=0$ as its limit.

Covergence here is $f_k \longrightarrow f$. That is $f_k(x) \longrightarrow f(x)$. Now the $\nu$ define in a way how fast this is because they (usually) become bigger the smaller the $\epsilon$ is. So the difference between punctual and uniform convergence is basically whether this "speed of convergence" is the same at all points $x$, in which case we call it uniform, or whether this "speed of convergence" depends on where we are on the $x-$axis, in which case it is only punctual.

Oh, I get the $\nu$ part, but why the limit is $f(x) = 0$?

One example of punctual converge is how the function $f_{n}(x) = e^{-x^{2n}}$ approaches the function "$g : g(x) = 1$ if $|x|<1$, $g(x) = 1/e$ if $|x| = 1$, $g(x) = 0$ if $|x|>1$" when the natural number $n$ grows without bound. If you set $\epsilon = 0.05$, there will always be a point $x$ for which $f_{n}(x) = 0.9$ and outside the chosen bounds, no matter how large $n$ is.

An example of uniform convergence is how the sequence $f_{n}(x) = x + 1/n$ approaches $g(x) = x$ when $n\rightarrow \infty$

Why if setting $\epsilon = 0.05$, we have $f_n(x) = 0.9$?

I've never seen it described as "punctual" convergence. The English words point and punctual are derived from Latin "punctum," but the usual meaning of punctual is "on time," as in a punctual arrival of a student to a class.

The type of convergence discussed in this thread is called "pointwise" convergence in English.

Yes, you're right. I'm sorry but I'm not a native English speaker. When I translated from my native language (Italian) to English, I thought it was correct.

 

[fresh_42]

Oh, I get the $\nu$ part, but why the limit is $f(x) = 0$?

$f_k(x) = max \{k-k^2\,\vert \,x-\frac{1}{k}\,\vert \,\, , \,0\}$.
So the only values $f_k(x) \neq 0$ are in the interval $]0,\frac{2}{k}[$ which is getting smaller by increasing $k$. Hence we have only to consider the point $x_0=0$. But $\lim_{k \rightarrow \infty}f_k(x_0)=0$ because you can always find a zero in a small neighborhood of $x_0$, although the values at exactly $x_0$ are large.

 

[hilbert2]

Why if setting $\epsilon = 0.05$, we have $f_n(x) = 0.9$?

The function $f_n$ goes through all values in the interval $]0,1]$ no matter what the number $n$ is, because it's a continuous function. Therefore we know that for any $n$, there are points where the function's value differs from the limit $g(x)$ by more than the number $\epsilon = 0.05$ (note that the function $g$ always has value 1, 1/e or 0). This is just a single example that is sufficient to prove that the function sequence $f_n$ is not uniformly convergent.

 

[Stephen Tashi]

The order of quantifiers is important. "...for each ... there exists..." can have a different meaning than "...there exists... for each...".

For example:
"For each instrument X there exists a virtuoso Y such that Y plays X "
versus
"There exists a virtuoso Y such that for each instrument X, Y plays X".

The first statement is a modest claim. The second statement asserts the existence of a fantastic musician who is a virtuoso at playing every instrument.

In pointwise convergence, the definition uses the order "...for each x ...there exists a $\nu$..."

In uniform convergence, the definition uses the order "...there exists a $\nu$ ...for each x...".

The sequence $f_k(x)$ converges pointwise to $f(x)$ on the interval $I$ is the more modest claim that:
For each $\epsilon > 0$ and for each $x \in I$, there exists a $\nu$ such that if $k \gt \nu$ then $|f(x) - f_k(x)| \lt \epsilon$.

The sequence $f_k(x)$ converges uniformly to $f(x)$ on the interval $I$ is the grander claim that:
For each $\epsilon > 0$ there exists a $\nu$ such that for each $x \in I$, if $k \gt \nu$ then $|f(x) - f_k(x)| \lt \epsilon$.