# A question on behavior of a function which is a limit of a sequence of functions converging under some norm

I think that convergence under any norm does not guarantee a desired behavior (continuity or differentiability) for the limit function. So my question is,

Is there a norm under which any sequence of functions with all of its elements having the same behavior (continuity and differentiability), if converge to some function $f$ then it is guaranteed that the function $f$ is having exactly the same behavior as of the functions of the sequence ?

PS : Here all the functions map real line to real line, all defined on the same open interval of real line.

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A uniformly convergent limit of continuous functions is continuous. The norm is the sup-norm, but differentiability need not be respected. See en.wikipedia.org/wiki/Uniform_convergence – Arturo Magidin Apr 18 '11 at 15:04
You may wish to read about Sobolev spaces. – Willie Wong Apr 18 '11 at 15:06
I am beginning to think that it is a broad question, I'd appreciate if someone gives me an example of such a norm which is obvious if it exists. – Rajesh Dachiraju Apr 18 '11 at 15:12
you can give $C^{\infty}(\mathbb{R},\mathbb{R})$ a topology using the family of semi-norms $\{\sum_{k=1}^n|f^{(k)}| : n\in\mathbb{N}\}$ (probably not anything that you're looking for) – yoyo Apr 18 '11 at 20:46
You refer to a discussion on MathOverflow. You were given an answer involving a Banach space. From the ensuing discussion, it is not clear to me whether you know what a Banach space is. If you don't know what a Banach space is, then I'd suggest that you find out - it may answer many of your questions. If you do know what a Banach space is, my apologies (but then what's this about "strange reasons" for "the existence of a norm"?). – Gerry Myerson Apr 19 '11 at 13:08

You should listen to Pietro Majer; he gives good advice.

The problem that is fundamental is that you have a misunderstanding of what it means for a sequence $f_n$ to converge to $f$ in norm (judging by your comments here, at this linked question, and over here).

To have convergence, you need a space $X$, a norm on the space $\|\cdot \|$, a sequence of elements in the space $(f_n)\subset X$, and the limiting element $f$. We say that the sequence of elements converge to $f$ in norm if $$\lim_{n\to\infty} \|f_n - f\| = 0$$

Now. If you have a function space $X$ which by definition all elements in it has the desired continuity/differentiability properties, then what you are asking about is utterly trivial: for a sequence $(f_n)\to f$, since all players are already in this function space $X$, they all already have the desired properties. So there is nothing to prove at all.

But what about statements about "the continuous functions are dense in $L^1$" or "the smooth functions are dense in the continuous functions"? The key here is that you are using a different norm.

Let $(X, \|\cdot\|)$ be some normed linear space, let $Y$ be a subspace of $X$ endowed with a norm $[\cdot]$ which generates a different topology on $Y$ than the one induced from the topology on $X$ generated by $\|\cdot\|$. An example would be $X = C^r([0,1])$ with the norm $\|f\| = \sum_{i = 0}^r \sup_{x\in [0,1]} \left|\frac{d^i}{dx^i} f(x)\right|$, and $Y = C^{r+1}([0,1])$. It is easy to check that $Y$ is a linear subspace of $X$ (finite linear combinations do not destroy differentiability). Take the second norm to be $[f] = \sum_{j = 1}^{r+1} \sup_{x\in [0,1]} \left|\frac{d^j}{dx^j} f(x)\right|$. The two norms are incompatible.

It is under this set-up that $Y$ can be simultaneously dense in $X$ (under the norm $\|\cdot\|$) and have the property that functions converging in $Y$ (under the $[\cdot]$ norm) preserve the differentiability properties.

Another way to think about this is the completeness of a vector space with respect to a given norm.

$Y$, as above, with the norm $[\cdot]$ is complete. $Y$, as above, with the norm induced from $\|\cdot\|$ is not complete. So if $(f_n)$ is a sequence in $Y$ that is Cauchy with respect to $[\cdot]$, we have that there exists an element $f$ of $Y$ such that $f_n\to f$. But if $(g_n)$ is a sequence in $Y$ that is Cauchy with respect to $\|\cdot\|$, we cannot guarantee that there exists an element $g$ of $Y$ such that $f_n\to f$. In fact, it is quite likely that such an element would properly belong to $X\setminus Y$, where $X$ now can be seen as the norm-completion of $Y$ under the norm $\|\cdot\|$.

As another example, $C^r$ is dense in $L^1$ under the $L^1$ norm. $C^r$ is certainly not dense in $L^1$ under the norm $\|\cdot\|$ I wrote above.

Remember, density of function space (or equivalently, convergence of sequences) do not just depend on the sequence and the abstract "set" they live in. To have a notion of density (or convergence) you need a topology. And to the same set you can give vastly different topologies.

So as a closing remark to the question you actually asked: the space $X = C^r([0,1])$ of $r$-times continuously differentiable functions with the norm $\|\cdot\|$ I wrote above, is an example of a space where norm convergence preserves continuity and differentiability. It is even a Banach space. For more details about this I highly recommend you consult a textbook in functional analysis. The book by Yoshida is quite good, and so is the book of Rudin (provided you have already studied his Real and Complex Analysis).

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I thank Willie Wong for giving such a nice answer which indeed cleared some of my misconceptions. I am giving my views on the question especially considering the context in which this question is being asked. I also give some comments on the question mentioned in the motivation part and also the answer given there. I am posting this in the space for answer due to absence of a better place. Please let me know if there is a better alternative. Also request you to comment upon any mistakes in this post.

Outline

We try to give a more general view of the problem and argue that it is important especially in the context (given in the motivation part of the question) in which the question is being asked. In the motivation part of the question a problem was cited as a link and here we refer to it as Problem A. We also discuss some possible approaches to Problem A.

In Section 1 we give some definitions based on continuity and differentiability of a function.In Section 2 we give two theorems which are fundamental tools in the current context. In Section 3 we consider a norm denoted as $\|\cdot\|$ defined on the linear space $X = C^r([0,1])$ and explain how it is tacitly designed to make the normed linear space $(X, \|\cdot\|)$ complete. We also attempt to give a proof for this using the theorems given in section 2. Section 4 is about convergence of functions. In Section 5 we give an example of a normed linear space whose norm is of the form $\|\cdot\|$ which need not necessarily be complete. We also point out some limitations of convergence under such a norm. In Section 6 we give some closing remarks.

Section 1

Here we give some definition based on continuity and differentiability of a function.

1. Continuity in an interval :

The following two statements are equivalent.

1. A function $f$ is continuous in an interval $S$.
2. A function $f$ is continuous at every point $x \in S$.
2. Differentiability in an interval :

The following two statements are equivalent.

1. A function $f$ is differentiable in an interval $S$.
2. A function is differentiable at every point $x \in S$.
3. $\mathcal{C}^k$ in an interval :

A function which is atleast $k$ times continuously differentiable at every point $x \in S$ is said to be $\mathcal{C}^k$ in $S$.

The collection of all functions $f$, such that $f$ has support in $S$ and is atleast $k$ times continuously differentiable at every point $x \in S$, is denoted as $\mathcal{C}^k(S)$.

4. Degree of differentiability :

1. The degree of differentiability of a function $f$ at a point $x$ is $k$. $k \in \{0\}\cup\mathbb{N}\cup\{\infty\}$.
2. A function $f$ is continuous at a point $x$ and the maximum number of times $f$ is continuously differentiable at $x$ is $k$.
5. Differentiability map :

Let a function $f : D \to R$ where $D\subseteq\mathbb{R}$ and $R\subseteq\mathbb{R}$. Let $C_f$ be the set of all points $x \in D$ such that $f$ is continuous at $x$.

The differentiability map of the function $f$ is defined as the mapping $dm_f : C_f \to \{0\}\cup\mathbb{N}\cup\{\infty\}$ such that $dm_f(x) = k$, where $k$ is the degree of differentiability of $f$ at the point $x$.

Hence associated with every function $f : \mathbb{R} \to \mathbb{R}$ there is a differentiability map $dm_f : C_f \to \{0\}\cup\mathbb{N}\cup\{\infty\}$ with $C_f$ defined as above.

6. Differentiability class :

The collection of all functions $f : \mathbb{R} \to \mathbb{R}$ which have $dm : \mathbb{R} \to \{0\}\cup\mathbb{N}\cup\{\infty\}$ as the differentiability map is called the differentiability class $D_{dm}$ of the differentiability map $dm$.

Note that there are differentiability maps whose differentiability classes are empty sets.

Section 2

Two theorems on uniform convergence to continuity and to differentiability are given here

Theorem 1:

Let $\{f_n\}$ be a sequence of functions which are continuous in an interval $S$. If the sequence of functions converges uniformly to $f$ in an interval $S$, then the limit function $f$ is continuous in the interval $S$.

Theorem 2:

Let $\{f_n\}$ be a sequence of functions which are differentiable in an interval $S$. If the sequence converges uniformly to a function $f$ in the interval $S$, and the sequence $\{f^'_n\}$, consisting of derivatives of $\{f_n\}$ in the interval $S$, also converges uniformly to a function $g$ in the interval $S$, then $f$ is differentiable in the interval $S$ and its derivative is $g$.

These theorems can be found here.

A slightly different version of Theorem 2 which tells about the differentiability of the limit function of a sequence of functions, at a specific point, is given below

Theorem 2A:

Let $\{f_n\}$ be a sequence of continuous functions defined in an interval $S$ and the sequence converges uniformly to a function $f$ in the interval $S$. Consider a point $x \in S$ and if there exists a neighbourhood $N$ of $x$ in $S$ such that the functions f_n $n = 1,2,3...$ are all differentiable in $N$ with derivatives $f^'_n$ defined in the interval $N$, and the sequence $\{f^'_n\}$ converges uniformly to a function $g$ in the interval $N$, then the function $f$ is differentiable at $x$ and its derivative at $x$ is given by $g(x)$.

Section 3

We now consider the linear space $X = C^r([0,1])$ with a norm $\|f\| = \sum_{i = 0}^r \sup_{x\in [0,1]} \left|\frac{d^i}{dx^i} f(x)\right|$ denoted as $\|\cdot\|$ defined over it. we now show how the norm $\|\cdot\|$ defined over $X$ is tacitly designed by keeping in mind the theorems given in Section 2, to make the normed linear space $(X, \|\cdot\|)$ complete. We do this by giving a proof for the completeness of $(X, \|\cdot\|)$ using the theorems in Section 2.

Consider the case r=0.

If any sequence $\{f_n\}$ where $f_n \in X$ is a Cauchy sequence under the given norm then it can be easily seen that it converges uniformly (note here that it is point-wise) to some function lets say $l$. From Theorem 1 and given that $f_n \in X$ it can be readily seen that the limit function $l \in X$.

Consider the case r > 0.

Consider a Cauchy sequence $\{f_n\}$, $f_n \to f$, $f_n \in C^r([0,1])$ and hence (from what followed earlier) $f \in C([0,1])$. The norm $\|\cdot\|$ is tacitly designed such that the sequences $\{f_n^k\}$, for $k = 1,2,...r$ are all Cauchy under the given norm.Also $f_n^k \in C([0,1])$ for all $k = 1,2,...r$. Consider the sequence $\{f_n^'\}$ Now we can see from what has been proved earlier for the case r = 0, that the sequence $\{f_n^'\}$ converges uniformly to some function $m \in C([0,1])$. Now using Theorem 2 it readily follows that $f \in C^1([0,1])$. It can be proved that $f \in C^r([0,1]$ by using the Theorem 2 inductively and the fact that the sequences $\{f_n^k\}$, for $k = 1,2,...r}$ are all Cauchy under the given norm.

Hence the normed linear space $(X, \|\cdot\|)$ is complete. The theorems mentioned in Section 2 play a crucial role in proving this.

Section 4

Consider the functions of the form $f : \mathbb{R} \to \mathbb{R}$, in order to define the convergence of a sequence of such functions, there are two ways of doing so, which are mentioned below.

1. Point-wise convergence : In order to define point-wise convergence we use the fact that $\mathbb{R}$ forms a metric space under the metric $d(x,y) = |x-y|$ where $x,y \in \mathbb{R}$.See the definitions of Point-wise convergence and Uniform convergence.

2. Convergence under a norm : To define the convergence of a sequence of functions under a norm, we need a space $X$ whose elements are functions and which contains all the functions in the given sequence, with a norm defined on it.

A sequence $\{f_n\}$ where $f_n \in X$ is said to be converging sequence If there exists a function $f \in X$ also called and the limit of the sequence, such that $\lim_{n\to \infty} \|f_n - f\| = 0$. This is also explained in the previous answer by Willie Wong.

Section 5

We now give an example of a linear space which when equipped with a norm of the form $\|\cdot \|$ need not necessarily be complete.

Let $dm : (0,1) \to \{0\}\cup\mathbb{N}\cup\{\infty\}$ be a differentiability map and $D_{dm}$ be its differentiability class such that $D_{dm}$ is nonempty.

Let $k = \min\limits_{x} dm(x)$. Let $X = \mathcal{C}^k((0,1))$. Therefore the linear space $D_{dm} \subset X$. Now we define a norm denoted by $\|\cdot \|$ given as $\|f\| = \sum_{i = 0}^k \sup_{x\in [0,1]} \left|\frac{d^i}{dx^i} f(x)\right|$, on the spaces $X$ to form two normed linear spaces $(X, \|\cdot\|)$ and $(D_{dm}, \|\cdot\|)$.

We know that $(X, \|\cdot\|)$ is complete. We however note that we note that $(D_{dm}, \|\cdot\|)$ need not necessarily be complete. We can also easily see that the approach of constructing a function $f \in D_{dm}$ as a limit function of a sequence of functions $\{f_n\}$, where $f_n \in X$, converging under the norm $\|\cdot\|)$ may not be such a good idea, as such a convergence only guarantees that the limit function belongs to $X$ but cannot guarantee for the limit function to have a desired differentiability map and hence cannot guarantee that the limit function belongs to $D_{dm}$.

On the other hand to construct a function $f \in D_{dm}$, the approach of constructing $f$ as a limit of sequence of functions $\{f_n\}$ converging under pointwise convergence seems to be a promising idea because the Theorem 2A, given in section 2, gives the sufficient conditions on the functions $f_n$ for the limit function to have a desired degree of differentiability at specified points in its domain.

Section 6

Here we give some closing remarks and also comment upon the answer given to Problem A and why such an approach is not a good idea.

In the answer the function $f$ is given as $f=\sum_{j < q} f_j + \sum_{j\geq q} f_j$. It is said that the first sum does what is needed at the points $p/q$, and the series converges in $C^q_c(\mathbb{R})$. Let us denote the first term as $f_1$ and let the series convergence to some $f_2 \in C^{\infty}_c(\mathbb{R}) \subset C^q_c(\mathbb{R})$. Consider a rational point $x_o = P/Q$ where $P,Q \in \mathbb{N}$ and $P \le Q$ and $P/Q$ in simplest form and $Q \ge q$. We can easily see that $Cf_1(x_o) = \infty$ and $Cf_2(x_o) = \infty$ and hence $Cf(x_o) = \infty$ which violates condition 2 (EDIT: its condition 1 instead of condition 2) of the problem. EDIT 2 : Please see a clarification to answer to Problem A given there.Now it is clear that the answer given there is true. But still this discussion (except for these comments on the answer) is valid in a more general case.

The approach of convergence of a sequence of functions under the norm of the form $\|\cdot \|$ given above is not suitable to get a limit function $f$ with a desired differentiability map.

On the other hand the approach of pointwise convergence is more fundamental and Theorem 2A gives the sufficient conditions on the functions in the sequence, to get a desired degree of differentiability for the limit function at specified points in its domain.

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I don't think this kind of monologue is suitable for this site. – the L Apr 22 '11 at 14:24
@anonymous : I don't think there is anything in it that is irrelevant to the original question. – Rajesh Dachiraju Apr 22 '11 at 15:17

Convergence under any norm does not guarantee a desired behavior

Uniform Convergence

A sequence of functions $\{f_n(x)\}$ converges uniformly converges to a function $f(x)$ on an interval I if, for every $ϵ>0$ there exists an $N$ such that $|f(x)−fn(x)|<ϵ$ for all $x∈I$ and all $n>N$.

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