What Does Norm Convergence Mean in $L^p$ Spaces?

[mathmari]

Hey!

If $f_n, f \in L^p, 1\leq p < +\infty$ and $f_n \rightarrow f$ almost everywhere, and $||f_n||_p \rightarrow ||f||_p$, then $f_n\rightarrow f$ as for the norm.

Could you give me some hints how to show it?? (Wondering)

What does convergence as for the norm mean?? (Wondering)

 

[Opalg]

Hey!

If $f_n, f \in L^p, 1\leq p < +\infty$ and $f_n \rightarrow f$ almost everywhere, and $||f_n||_p \rightarrow ||f||_p$, then $f_n\rightarrow f$ as for the norm.

Could you give me some hints how to show it?? (Wondering)

What does convergence as for the norm mean?? (Wondering)

This is an elusive result. I found a proof of it http://math.ucsd.edu/~lni/math240/Hand-out1.pdf (Theorem 0.1). The proof relies on a generalised version of the dominated convergence theorem, which is proved here. Another ingredient in the proof is the fact that $(a+b)^p \leqslant 2^{p-1}(a^p + b^p)$ for positive numbers $a,b$. You can deduce that from the fact that the function $f(x) = x^p$ is convex for $x>0$, and so $f\bigl(\frac12\!(a+b)\bigr) \leqslant \frac12\!\!\bigl(f(a) + f(b)\bigr).$

"Convergence as for the norm" means $\|\,f_n - f\|_p \to0$ as $n\to\infty$.

 

[mathmari]

This is an elusive result. I found a proof of it http://math.ucsd.edu/~lni/math240/Hand-out1.pdf (Theorem 0.1). The proof relies on a generalised version of the dominated convergence theorem, which is proved here. Another ingredient in the proof is the fact that $(a+b)^p \leqslant 2^{p-1}(a^p + b^p)$ for positive numbers $a,b$. You can deduce that from the fact that the function $f(x) = x^p$ is convex for $x>0$, and so $f\bigl(\frac12\!(a+b)\bigr) \leqslant \frac12\!\!\bigl(f(a) + f(b)\bigr).$

"Convergence as for the norm" means $\|\,f_n - f\|_p \to0$ as $n\to\infty$.

Ahaa... So can I formulate it as followed?? (Wondering)From Fatou`s lemma we have that $$\int \lim \inf [2^{p-1}(|f_n|^p+|f|_p)-|f_n-f|^p]d\mu \leq \\ \lim \inf \int [2^{p-1}(|f_n|^p+|f|^p)-|f_n-f|^p]d\mu \\ \Rightarrow 2^{p-1}\int \lim \inf (|f_n|^p+|f|^p)d\mu+\int \lim \inf (-|f_n-f|^p)d\mu \leq 2^{p-1}(\lim \inf \int |f_n|^pd\mu +\lim \inf \int |f|^pd\mu )+\lim \inf (-\int |f_n-f|^p d\mu) \\ \Rightarrow 2^{p-1}[\int \lim \inf |f_n |^pd \mu+\int \lim \inf |f|^pd \mu]-\int \lim \sup |f_n-f|^pd \mu \leq 2^{p-1}[\lim \inf \int |f_n|^pd \mu+\lim \inf \int |f|^p d\mu]-\lim \sup \int |f_n-f|^pd \mu \ \ \ \ \ (*) $$

Knowing that $||f_n||_p\rightarrow ||f||_p \Rightarrow \left ( \int |f_n|^p\right )^{1/p}\rightarrow \left ( \int |f|^p\right )^{1p}$ , we have that $\lim \inf |f_n|^p=|f|^p$

Therefore, $$(*)\Rightarrow 2^{p-1}\int (|f|^pd \mu+\int |f|^pd\mu)-\int \lim \sup |f_n-f|^pd \mu \leq 2^{p-1}(\lim \inf \int |f_n|^pd \mu +\int |f|^pd \mu)-\lim \sup \int |f_n-f|^pd \mu \\ \Rightarrow 2^{p-1}\int |f|^pd \mu-\int \lim \sup |f_n-f|^pd \mu \leq 2^{p-1}\lim \inf ||f_n||^p_p-\lim \sup \int |f_n-f|^pd \mu \\ \Rightarrow 2^{p-1}||f||^p_p-\int \lim \sup |f_n-f|^pd \mu \leq 2^{p-1} ||f||^p_p-\lim \sup \int |f_n-f|^pd \mu \\ \Rightarrow \lim \sup \int |f_n-f|^pd \mu \leq \int \lim\sup |f_n-f|^pd \mu \\ \Rightarrow \lim \sup ||f_n-f||^p_p \leq \int \lim \sup |f_n-f|^pd \mu =0, \text{ since } f_n\rightarrow f \text{ almost everywhere } $$

So, we conclude that $||f_n-f||_p\rightarrow 0$.

 

[Opalg]

Yes, that looks good. (Yes) (Rock)

I struggled for a long time, unsuccessfully, to prove this result by using the dominated convergence theorem. Then I discovered that online reference to a generalisation of the DCT (proved in the same way as the standard DCT, from Fatou's lemma), in which instead of a single dominating function there is a convergent sequence of them.

 

[blue_raver22]

Convergence as for the norm means that a sequence of functions, denoted as $f_n$, converges to a function $f$ in the norm defined by $||\cdot||_p$, where $p$ is a real number between 1 and infinity. This norm is a measure of the size or magnitude of a function, and it is defined as the $p$-th root of the integral of the absolute value of the function raised to the power of $p$. In simpler terms, it measures how much the function deviates from zero.

To show that $f_n$ converges to $f$ as for the norm, we need to prove that the norm of the difference between $f_n$ and $f$ approaches zero as $n$ approaches infinity. This can be done by using the triangle inequality and the fact that the norm is a continuous function.

First, we can write $||f_n-f||_p$ as $||f_n-f_n+f-f||_p$. Using the triangle inequality, we can split this into $||f_n-f_n||_p + ||f-f_n||_p$. Since $f_n$ converges to $f$ almost everywhere, we know that $f_n$ and $f$ have the same integral, which implies that $||f_n-f_n||_p = 0$. Therefore, we are left with $||f-f_n||_p$.

Next, using the continuity of the norm, we can write $||f-f_n||_p$ as $||f-f_n||_p \leq ||f-f_n||_p + ||f_n||_p - ||f||_p$. Since $||f_n||_p \rightarrow ||f||_p$, we can choose $n$ large enough such that $||f_n||_p - ||f||_p < \epsilon$, where $\epsilon$ is a small positive number. This means that $||f-f_n||_p \leq ||f-f_n||_p + ||f_n||_p - ||f||_p < \epsilon$. As $n$ approaches infinity, the right-hand side of the inequality approaches zero, which means that $||f-f_n||_p$ also approaches zero. This proves that $f_n$ converges to $f$ as for the norm.

In conclusion, convergence as for