continuitypointwise-convergencereal-analysisuniform-continuityuniform-convergence
The pointwise limit of a sequence of uniformly continuous functions needn't be continuous. I've been wondering about the converse:
Is a continuous function $f$ necessarily the pointwise limit of some sequence of uniformly continuous functions $(f_n)_{n\in\mathbb{N}}$?
I'm having trouble establishing or disproving this.
The example you give has a point-wise limit of 0, which is not a discontinuous function. Some standard examples of these kinds of functions are ones that kind of smoothly interpolate between two values, with the interpolating region getting smaller and smaller. For example, consider the sequence:
$$ f_n(x) = \tan^{-1}(nx) $$
This is plotted for some various values of $n$ below, on the region $x\in [-1,1]$
Along with Suugaku's answer, this should give you the intuition you need about point-wise limits and continuity.
Let $f_n = 1_{[n,\infty)}$ (that is $f_n(x) = \begin{cases} 1, & x \ge n\\0, & \text{otherwise} \end{cases}$).
Then $f_n(x) \to 0$ for all $x$, but $f_n$ does not converge to 0 uniformly.
Another example, let $g_n(x) = \begin{cases} 0, & x < 0 \\ x^n, & x \in [0,1]\\1, & x > 1 \end{cases}$. Then $g_n(x) \to g(x)$, where $g(x) = \begin{cases} 0, & x < 1 \\1, & x \ge 1 \end{cases}$, which is not continuous. Furthermore, the convergence is not uniform.
Best Answer
If the domain of $f$ is $\Bbb R$, then the answer is yes. The proof is quite easy: let $f_n: \Bbb R \to \Bbb R$ defined by $$f_n(x)= \begin{cases} f(-n) & x<-n \\ f(x) & -n \le x \le n \\ f(n) & x>n \end{cases}$$ these are uniformly continuous and they pointwise converge to $f$.