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Let $\{f_n\}$ be a sequence of smooth functions which converges to a function $f$. If the convergence is not uniform at a point $a$ the $f$ is discontinuous at $a$. Is there any different type of convergence where if it happens at $a$ then $f$ is continuous at $a$ but is not differentiable.

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Let $\{f_n\}$ be a sequence of smooth functions which converges to a function $f$.If there is a discontinuity in $f$ at some point $a$ then the convergence is nonuniform.Is there any different type of convergence needed for $f$ to be continuous at some point $a$ but not differentiable ?

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    @ Rajesh: I am wondering about the first half of your statement. We know that "If the convergence is uniform, then the limit function is continuous". But the first half of your statement is the converse. I am wondering if it is true...2010-11-20
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    I think it is....that is the only way a sequence of continuous functions converge to a function which has a discontinuity2010-11-20
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    It is not true that a nonuniform limit of smooth functions must be discontinuous. E.g., take a sequence of smooth functions $f_n$ with $f_n(1/n)=1$ and the support of $f_n$ contained in $[0,2/n]$.2010-11-20
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    For the actual question, I'm not sure what you're asking. Uniform limits of smooth functions are continuous but not typically differentiable. (For them to be differentiable you can require uniform convergence of the derivatives as well.)2010-11-20
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    @Rajesh D: "the only way a sequence of continuous functions to converge..." is the statement of the *converse* of what you stated. There is a difference between "The only way for P to occur is if Q happens" (that is, Q is *necessary*) and "If Q happens, then P happens" (which means Q is *sufficient*). They mean different things.2010-11-20
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    @ Jonas Meyer: Nice example! It has made my Friday night :)2010-11-20
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    @Sivaram: Thanks. Incidentally, apparently the comment notifications don't work if there's a space between "@" and the name.2010-11-20
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    @Jonas Meyer,@Arturo: I was in a hurry this morning and i couldn't see that i was making a converse statement to justify one. Thanks Arturo for the comment. Although there was confusion..in my mind i was in an impression that my second sentence in the question holds, i.e, "A nonuniform convergence of a sequence of functions always results in a discontinuity in the limit function". Thank you Jonas for the counter example but i am still not able to get what is the limit function in your example.Please throw some light on this.2010-11-20
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    @Rajesh D: The limit function is the zero function.2010-11-20
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    @Jonas Meyer: To note a peculiar point in your counter example, the $Sup|f_n-f|$ is always constant.2010-11-20
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    @Rajesh D: Yes. Well, actually, it doesn't have to be. $\sup|f_n|$ could be an arbitrary positive number greater than or equal to $1$, given the vagueness of my description. For example, we could also require $f_n(\frac{1}{2n})=n$.2010-11-20
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    @Rajesh D: Yes, I'm saying that we can add a third condition to the previous conditions on $f_n(1/n)$ and the support of $f_n$. The point of the counterexample is that $f_n\to 0$ pointwise but $\sup_x|f_n(x)|\geq 1$ for each $n$ (and they have supports all contained in a common compact set). But if one were to write the counterexample in more detail, taking $1$ to be the maximum value would make sense.2010-11-20
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    @Jonas Meyer: What if i put a condition that $f_n$ does not have a compact support ? Would there still be any counter examples ?2010-11-20
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    @Rajesh: Yes. For example, you could take $f_n$ to have support equal to $[n,\infty)$ with $f_n(x)=1$ for $x\geq n+1$.2010-11-20
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    @Jonas Meyer: Ok. This time, are there any counter examples when $f_n$ is supported almost everywhere.2010-11-20
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    @Jonas Meyer: You are taking a bump like function and you are pushing it here and there.....very funny technique...too technical but important !2010-11-20
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    @Jonas Meyer: LOL !!!2010-11-20
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    @Rajesh D. Yes, you can have the support of each $f_n$ equal to $\mathbb{R}$. For example, require $f_n(x)=1/n$ if $|x|\leq n$, $f_n(x)=1$ if $|x|\geq n+1$, and $1/n\leq f_n(x)\leq 1$. (You should try to come up to counterexamples for your next question.)2010-11-20
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    @Jonas Meyer: Pushing the transition region and keeping a 1 over $n$ in the constant region...you are always using peicewise bump like functions and constant function to construct $f_n$...Sounds cool and funny !!!2010-11-20
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    @Jonas Meyer: All you are saying is "I can push it all the time and get you what you want !" A very fundamental idea in Real Analysis !!2010-11-20
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    @Jonas Meyer: OK !!! To note a peculiar point in your counter examples, the $Sup|fn−f|$ is always a convergent sequence in all your examples !!!...2010-11-21
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    @Rajesh: OK, but that is not an essential feature, only a simplification. As mentioned earlier, without much more work one could choose $\sup_x|f_n-f|$ to be an unbounded sequence, and one could actually choose it to be any sequence of nonnegative numbers that does not converge to $0$. There is no need to do so, hence the simplifications for these examples.2010-11-21
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    @Jonas Meyer: yeah got it. when you are translating everything along the $x$-axis towards the boundaries, there is no need to bother about $sup|f_n-f|$.2010-11-21

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P[a,b], the set of polynomials on [a,b] (obviously a subset of the smooth functions on [a,b]) is dense in C[a,b]. In addition the set of functions that are differentiable at (at least) a single point in [a,b] are of the first category in C[a,b].

In some sense "most" convergent sequences of smooth functions converge to nowhere differentiable functions.

Semi-related, but you might in interested in the fact that $C^k(\Omega)$, $\Omega \subset \mathbb{R}^n$ open, can be made into a Fréchet space for any $k = 1,2,\ldots, \infty$. See Wikipedia's article on Fréchet spaces, the topology on $C^k(\Omega)$ being uniform convergence of $f_n$ and all of its derivatives up to order $k$ (q.v. multiindex)

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    I can't be bothered to edit this again so a correcting comment will have to do. The last sentence should say that the usual metric on $C^k(\Omega)$ is the same topology as uniform convergence of $(D^\alpha f_n)$ for all multi-indices $\alpha$ s.t. $|\alpha| \leq k$ *on compact subsets of* $\Omega$.2010-11-20