Convergence and differentiability

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.

EDIT

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... –  user17762 Nov 20 '10 at 4:54
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 $[0,2/n]$. –  Jonas Meyer Nov 20 '10 at 5:06
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.) –  Jonas Meyer Nov 20 '10 at 5:10
@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. –  Arturo Magidin Nov 20 '10 at 5:13
@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.) –  Jonas Meyer Nov 20 '10 at 19:13

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)
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 $\Omega$. –  kahen Nov 20 '10 at 9:35