I'm having some trouble understanding the difference between convergence in the sense of Lp norms (p is finite) and L infinity norms
The original problem I have is listed below:
Q. Let $1\leq p < \infty $ and assume $f \in L^p(R)$ Prove that $$\lim_{x\to\infty}\int_{x}^{x+1} f(t) dt = 0 $$
I have shown this equation using Holder's inequality for finite p, the next question is, does the limit also apply if $p = \infty $ (In other words, $f \in L^\infty)$?
My guess is that, if we assume that $f(t) = 1$ then $$\lim_{x\to\infty} \textbf{1}_{(x,x+1)}(t) \neq 0 $$Since $f(t) = 1 $ everywhere.
Hence, the limit would not be equal to zero in the L infinity space.
Would appreciate any comments/errors in my approach!
You are right and have detected what sets the space $L^\infty$ apart from $L^p$ for $p<\infty$.
Indeed, $L^p(\mathbb R)$ spaces for finite $p$ are characterised in terms of a function's decay far away from the origin, and the limit in your question is just a necessary condition for a function to be in $L^p(\mathbb R)$.
However, $L^\infty$ fucntions need only to be bounded almost everywhere, which does not rule out constant functions. Notice that $f(t)=1$ belongs to $L^\infty(\mathbb R)$ but does not belong to $L^p(\mathbb R)$ for any $1\leq p < \infty$.