Let {${f_n}$} be a sequence in $L^p$ which converges in the mean of order $p$ to $f$ in $L^p$.Then prove-
- If the sequence {$f_n$} converges in the mean of order $p$ to $g$,then $f=g$ almost every where.
- The sequence {$f_n$} is a $p-mean$ cauchy sequence.
- $\lim_{n\to \infty}||f_n||_p = ||f||_{p}$,in particular the sequence {$f_n$} is bounded with respect to the norm $||.||_{p}$
What I tried for first part:
using contradiction:Suppose $f≠g$ a.e then $f$ & $g$ are not equivalent functions.Which means $f$ and $g$ will not belong to same $Lp$ space simultaneously.which contradicts the hypothesis that $f$ converges to $g$ in mean of order $p$.Hence,f=g a.e.
Is it correct?
While collecting the key points regarding above theorem from many books,ive encountered several examples which justifies the above theorem but i'm not getting how to prove it generally.
I've given it a large amount of time.So it would be better if someone please provide the proof of the above theorem.
Your attempted solution does not really make sense. The idea is to follow the same proof that limits (in $\mathbb R$, for example) are unique, but replace equality with a.e. equality. Specifically, fix $\varepsilon>0$. By definition of $L^p$ convergence, there exists $n_1$ such that $\|f_n-f\|_p<\frac\varepsilon2$ for all $n\ge n_1$, and there exists $n_2$ such that $\|f_n-g\|_p<\frac\varepsilon2$ for all $n\ge n_2$. Let $n=\max\{n_1,n_2\}$. Then $$\|f-g\|_p=\|f-f_n+f_n-g\|_p\le\|f-f_n\|_p+\|f_n-g\|_p<\tfrac\varepsilon2+\tfrac\varepsilon2=\varepsilon.$$ Since $\varepsilon>0$ was arbitrary, it follows that $\|f-g\|_p=0$, so $\int|f-g|^p=\|f-g\|_p^p=0$. This implies $|f-g|^p=0$ a.e. and so $f=g$ a.e.
Note how closely this matches the usual proof that limits in $\mathbb R$ are unique. See if you can similarly mimic the respective proofs for your other two questions.