Question: Define a sequence $\{f_j\}_{j=1}^{\infty} \subset C^0[a,b]$ converging with respect to the $L^p$ norm to $f \in C^0[a,b]$. Define $\{x \in [a,b]:f(x)=\lim_{j \rightarrow \infty} f_j(x) \}$ be the set of pointwise convergence of the sequence, denoted A. Let $B=[a,b] \setminus A$ and $C=(a,b) \setminus A$. Find example such that $B=\{a\}, C=\{x_0\}$ for some $x_0 \in (a,b)$, and $\overline{C}=[a,b]$.
For $B=\{a\}$, is it that I have to choose x=a such that $f(a) \neq \lim_{j \rightarrow \infty} f_j(a)$, i.e. not converging pointwisely with respect to $L^p$ norm? Not sure what a good candidate would be, should I think of a particular function that is not pointwise convergence for x=a?
For $C=\{x_0\}$ , $x_0 \in (a,b)$, I believe this is similar to $B=\{a\}$ with some adjustments, but have not thought of what to proceed.
For $\overline{C}=[a,b]$,I think I need to make sure that for all $x_0 \in (a,b)$, there is $f(x_0)=\lim_{j \rightarrow \infty} f_j(x_0)$.
Have not made further progress and I believe there are lot of observations I have not spotted, would appreciate some ideas. Thanks in advance.
For $B=\{a\}$ take $f_j(x)=(-1)^{j} (\frac {b-x} {b-a})^{j}, f \equiv 0$.
For $C=\{x_0\}$ take $f_j(x)=(-1)^{j} \min \{(\frac {x-a} {x_0-a})^{j},(\frac {b-x} {b-x_0})^{j} \}, f \equiv 0$.
For $\overline C =[a,b]$ divde $[a,b]$ into $2^{n}$ intervals of equal length, say $I_{1,n},I_{2,n},...,I_{2^{n},n}$. Vary $n$ and arrange these intervals in a single sequence with increasing denominators. By taking eh characterstic functions of these intervals you get a sequence $(f_j)$ which converges to $0$ in $L^{p}$ but does not converge at any point. For sure, the functions here are not continuous. But it is not hard to modify them and I will let you think about it.