## If $\|f_n\|_{p}\leq n^{-2}, f_n\in L^{p},\forall n\in\mathbb{N}$, does pointwise convergence of $f_n$ follow for a.e. $x\in\mathbb{R}$?

Let $$p\in[1,\infty)$$ and $$(f_n)_{n\in\mathbb{N}}\subset L^{p}(\mathbb{R})$$ such that $$\|f_n\|_{p}\leq n^{-2}$$ for all $$n\in\mathbb{N}.$$ Does $$(f_n)_{n\in\mathbb{N}}$$ necessarily converge pointwise a.e.? (Proof or counterexample.)

$$\textbf{Attempt:}$$ I think one can construct a counterexample to this. I was considering the following construction done by Saz in this post: https://math.stackexchange.com/a/3132844/595519.

I provide the details below:

Consider the following sequence of intervals $$[-1,0], [0,1],\left[-2,-\tfrac{3}{2}\right], \left[-\tfrac{3}{2},-1\right], \left[-1,-\tfrac{1}{2}\right],\left[-\tfrac{1}{2},0\right],\left[0,\tfrac{1}{2}\right],\left[\tfrac{1}{2},1\right],\left[1,\tfrac{3}{2}\right],\left[\tfrac{3}{2},2\right],\left[-3,-\tfrac{8}{3}\right],\left[-\tfrac{8}{3},-\tfrac{7}{3}\right],\dots,\left[\tfrac{7}{3},\tfrac{8}{3}\right],\left[\tfrac{8}{3},3\right],\dots$$ Denote the $$n$$th interval in the sequence by $$I_n.$$ By construction, it follows that $$m(I_n)\rightarrow 0$$ as $$n\rightarrow\infty.$$ Now denote the characteristic function of the $$n$$th interval above by $$1_{I_n},$$ and put $$f_n(x)=n^{-3}\cdot m(I_n)^{-1/p}\cdot1_{I_n}(x)$$. It follows that $$\left(\int_{\mathbb{R}}|f_n(x)|^p\,dx\right)^{1/p}=n^{-3},$$ and it follows that $$\|f_n-0\|_{p}\rightarrow 0$$ as $$n\rightarrow\infty,$$ so $$f_n\rightarrow 0$$ in the $$L^p$$-norm.

However, I note that the sequence $$(f_n)_{n\in\mathbb{N}}$$ I have constructed above fails, since it actually does converge pointwise to $$f\equiv 0,$$ and even worse, it does so everywhere.

Is there any way to save this example, or should I scratch it and try something else? Finally, is there a way to use decaying exponentials in this problem to make the sequence decrease fast enough?

Thank you for time and appreciate any feedback.

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