If $f_n, f \in L^p, 1\leq p &lt; +\infty$ 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??   What does convergence as for the norm mean??   EDIT:  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 $||f_n-f||_p\rightarrow 0$.
7520
Let $1\leq p &lt; \infty$. Suppose that    $\{f_k\} \subset L^p$ (the domain here does not necessarily have to be finite),  $f_k \to f$ almost everywhere, and  $\|f_k\|_{L^p} \to \|f\|_{L^p}$.    Why is it the case that $$\|f_k - f\|_{L^p} \to 0?$$   A statement in the other direction (i.e. $\|f_k - f\|_{L^p} \to 0 \Rightarrow \|f_k\|_{L^p} \to \|f\|_{L^p}$ ) follows pretty easily and is the one that I've seen most of the time. I'm not how to show the result above though.
115740
I've been working on the following problem:      Show that if $f\in C[a, b]$ , $f\ge 0$ on $[a, b]$, then $\left(\int_a^b f(x)^n \,dx\right)^{1/n}$ converges when $n\to\infty$ and the limit is $\max_If$ with $I=[a, b]$.   This is my solution:  For  Weierstrass $f$ has maximum, $\exists \  \xi : f(\xi)=M$; and as $f$ is defined on $[a,b]$, $f$ is U.C., then:  $\forall \epsilon &gt;0  \ \exists \delta &gt;0: \forall x,y \in [a,b]: |x-y|&lt; \delta \Rightarrow |f(x)-f(y)|&lt; \epsilon$  Let $[a,b]=\bigcup_{k=1}^m I_k$, with $mis(I_k)&lt;\delta$, and $M_k=max_{\bar(I_k)}  f(x)$ $(\int_a^b f(x)^n \  dx)^{1/n}=(\sum_{k=1}^m M_h^n mis(I_k))^{1/n}=(M_1^n mis(I_1)+...+M^n mis(I_j)+...M_m^n mis(I_m))^{1/n}=$ =$M ((M_1/M)^n mis(I_1)+...+mis(I_j)+...+(M_m/M)^n mis(I_m))^{1/n}$  Then:  $(\int_a^b f(x)^n \  dx)^{1/n} \ge M$  $(\int_a^b f(x)^n \  dx)^{1/n} \le M(b-a)^{1/n}$  $\Rightarrow \exists \  \lim_n \ (\int_a^b f(x)^n \  dx)^{1/n}=M=\max_I \ f$
27560
I was reading the wikipedia article about convolution and I found this one:     If $f \in L^1(\mathbb{R}^n)$ and $g\in L^p(\mathbb{R}^n)$, then $||f\ast g||_p \leq ||f||_1||g||_p$.   But to $f\ast g$ be well-defined, $y\mapsto f(x-y)g(y)$ must be of $L^1$. Right? But why? How do I prove this?
69189
Can anybody please help me to prove this:  Let $p$ be greater than or equal to $1$. Show that for the space $\ell_p=\{(u_n):\sum_{n=1}^\infty |u_n|^p&lt;\infty\}$ of all $p$-summable sequences (with norm $||u||_p=\sqrt[p]{\sum_{n=1}^\infty |u_n|^p}\ )$, there is an inner product $&lt;\_\,|\,\_&gt; $ s.t. $||u||^2=&lt;u\,|\,u&gt;$ if and only if $p=2$.
120929
We know that if $f\in L^1(\mathbb{R})$, then $\|f(\cdot+1/n)-f(\cdot)\|_{L^1}\to 0$ as $n\to \infty$, which implies that there exists a subsequence $f_{n_k}=f(x+\frac{1}{n_k})$ such that $f_{n_k}\to f$ a.e.   My problem is that  Is there any $f\in L^1(\mathbb{R})$  function such that the almost convergence result  $$ f(x)=\lim_{n\to\infty}f(x+1/n)\quad \text{a.e.} $$  is NOT valid?   I intend to believe that there exits such function. Otherwise, there would be a beautiful statement such that $f(x)=\lim_{n\to\infty}f(x+1/n)$ a.e. for any $f\in L^1(\mathbb{R})$.   Thank you for the long commenting below @David C. Ullrich. This question has been reduce to find a compact set $K$ such that for any $x\in K$, there are infinite $n$ such that $x+\frac{1}{n}\not\in K$.
121959
For every $p\in [1,\infty)$, consider the set $\ell^p$ of real sequences $x=(x_n)=(x_1,x_2,\ldots)$ such that $\sum_{n=1}^\infty |x_n|^p&lt;\infty$. This is a Banach space with the norm $\Vert x\Vert_p=(\sum|x_n|^p)^{1/p}$.  I have the following problem:     Is it true that $\ell^p=\bigcup_{1\leq q&lt;p}\ell^q$ for $p&gt;1$?   I don't think it is true, but I'm having problems to show this. Clearly, the statement is true for some $p&gt;1$ iff it is true for every $p&gt;1$.  I'm trying to find a sequence $(a_n)$ such that $\sum |a_n|^s$ converges iff $s\geq 1$. I've tried many sequences (such as $(n^{1+1/n})$, involving $\log$, etc...), but none of them worked.
93707
In a finite measure space, let $\{f_{n}\}$ be a sequence of measurable functions. Show that $f_{n}  \rightarrow f$ in measure if and only if every subsequence $\{f_{n_{k}}\}$ contains a subsequence $\{f_{n_{k_{j}}}\}$, that converges almost everywhere to f.   I can prove from converge in measure to converge almost everywhere, but I don't know how to write it down for the other direction.
87077
Suppose $\mu(X)&lt;\infty$. Show that if $\ f\in\ L^q(X,\mu)$ for some $1\le q&lt;\infty$ then $\ f\in\ L^r(X,\mu)$ for $1\le r&lt;q$   In the solution the assistant wrote this:  If $1\le r&lt;q$ then $q/r\ge 1$. Let p be the dual exponent to $q/r$ (i.e. $\frac{1}{q/r}+\frac{1}{p}=1)$  then by Hölder Inequality  $\int_{X}|f|^rd\mu\le\underbrace{(\int_{X}|f|^qd\mu)^{r/q}}_{??????}(\int_X1^pd\mu)^{1/p}&lt;\infty$  for me the middle part doesn't make sense (it is not the hölder inequality) can somebody show it ?  instead of this can't we say (only in this case) that   $\int_{X}|f|^rd\mu\le (\int_{X}|f|d\mu)^r$ then  $(\int_{X}|f|d\mu)^r\le ((\int_{X}|f|^qd\mu)^{1/q}(\int_X1^pd\mu)^{1/p})^r&lt;\infty$  ?
52346
How can I prove that the space $l^{p}$ equipped with the norm (for $x=(x_{n}) \in l^{p})$: $\|x\|_{p}=(\displaystyle\sum_{n}|x_{n}|^{p})^{1/p}$ Is a separable space? (i.e. showing that there is a countable dense set in $l_{p}$). I saw a proof in which they started with first letting $x=(x_{n})$ be an element of $l^{p}$ and for any $\epsilon &gt;0$, then they chose $N$ such that: $(\displaystyle\sum_{n&gt;N}|x_{n}|^{p})&lt;(\frac{\epsilon}{2})^{p}$ However, I'm not sure where does this come from. As well, I'm not sure how to proceed from this step. Thank you for your help!
67021
Suppose $f\in L^{p}(\mathbb{R}^{n}) \cap L^{q}(\mathbb{R}^{n})$. How can I prove that for any $p \lt r \lt q$,  $$ \lVert f \rVert_{r} \leq (\lVert f \rVert_{p})^{(1/r-1/q)/(1/p-1/q)} (\lVert f \rVert_{q})^{(1/r-1/q)/(1/p-1/q)}\:? $$ I tried using Hölder to show that $f$ is in $L^{r}$ but I'm completely lost on how to go about it...
28355
Inclusion of $L^p$ spaces for functions has been discussed .  Does this apply to $l^p$ space of sequences similarly?   I tried to show the following: For $1\leq p&lt;q&lt;\infty$, $l^q\subset l^p$ By using Hölder inequality but it doesn't seem to work.  My question is that is this true? If yes, what's the right way to prove it and what's a good counter example for showing $l^p\subset l^q$ is not true? Thanks.
15358
Let $(X,\mathcal{M},\mu)$ be a measure space. We say that $\{f_n\}$ converges to $f$ in measure if, for any $\epsilon&gt;0$   $$ \lim\limits_{n \to \infty} \mu\Big(\{x\in X: |f_n(x)-f(x)| \ge \epsilon  \}\Big)=0.$$         Show that $\{f_n \}$ converges to $f$ in measure if $$\lim\limits_{n\to \infty} \int_X \frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu(x)=0.$$   Suppose $\mu(X) &lt; \infty$. Show that    $$\lim\limits_{n\to \infty} \int_X \frac{|f_n(x)-f(x)|}{1+|f_n(x)-f(x)|}d\mu(x)=0$$   if $\{ f_n \}$ converges to $f$ in measure.
30915
I'm sure I'm having a notational misunderstanding. Anyway, suppose $(f_n)$ is a sequence of continuous functions from a metric space $X$ to $\mathbb{R}$. So, if $(f_n)$ converges uniformly to a function $f$, then  $$\lim_{n \to \infty} \inf_{x \in X} f_n(x) = \inf_{x \in X} \lim_{n \to \infty} f_n(x)$$  and  $$\lim_{n \to \infty} \sup_{x \in X} f_n(x) = \sup_{x \in X} \lim_{n \to \infty} f_n(x).$$  Can somebody explain to me why is that? Again, this might be really simple...
74329
Stuck with this problem from Zgymund's book.   Suppose that $f_{n} \rightarrow f$ almost everywhere and that $f_{n}, f \in L^{p}$ where $1&lt;p&lt;\infty$. Assume that $\|f_{n}\|_{p} \leq M &lt; \infty$. Prove that:  $\int f_{n}g \rightarrow \int fg$ as $n \rightarrow \infty$ for all $g \in L^{q}$ such that $\dfrac{1}{p} + \dfrac{1}{q} = 1$.  Right, so I estimate the difference of the integrals and using Hölder end up with:  $$\left|\int f_{n} g - \int fg\right| \leq \|g\|_{q} \|f_{n} - f\|_{p}$$  From here I'm stuck because we are not assuming convergence in the seminorm but just pointwise convergence almost everywhere. How to proceed?
23505
Let $-\infty\lt a\lt b \lt \infty$. Let $C[a,b]$ is space of continuous functions and a function $\|.\|:C[a,b]\to R$ is given by $\|f\|:=\int_a^b |f(x)|d(x)$  a). Show that $\|\cdot\|$ is a norm in $C[a,b]$  b). Show that $C[a,b]$ with this norm is not a Banach space.  Well the 1st question is kind of easy. Non-negativity, Homogenity, and Triangle inequality is easily proven. At the 2nd question I have a problem.   I know the $$C[0,1]$$ space with Norm $$\|f\| = \int_0^1|f(x)|d(x)$$ is not complete that means not Banach-Space. But how do I prove the generality?   Does anyone has a suggestion?  Thanks
22963
If we are in a sequence space, then the $ l^{p} $-norm of the sequence $ \mathbf{x} = (x_{i})_{i \in \mathbb{N}} $ is $ \displaystyle \left( \sum_{i=1}^{\infty} |x_{i}|^{p} \right)^{1/p} $.  The $ l^{\infty} $-norm of $ \mathbf{x} $ is $ \displaystyle \sup_{i \in \mathbb{N}} |x_{i}| $.  Prove that the limit of the $ l^{p} $-norms is the $ l^{\infty} $-norm.  I saw an answer for $ L^{p} $-spaces, but I need one for $ l^{p} $-spaces. Besides, I didn’t really understand the $ L^{p} $-answer either.  Thanks for your help!
43389
I found  the following exercise and I'd like to know if my answer is correct.      Let $(X, \mathscr A, \mu)$ a finite measure space. Let $\{f_n\}$ a sequence of measurable functions such that $\|f_n\|_p\le M$ for a real constant $M$ $(1&lt;p&lt;\infty)$ and suppose that $f_n \xrightarrow{\text{a.e.}} f$. Then $$\int_X f_n g \,d\mu \to \int_X fg \,d\mu$$ for all $g$ in $\mathscr{L}^q (X)$, where $q^{-1}=1- p^{-1}$.   Since $\int |f_n|^p \,d\mu\le M^p $ by Fatou's lemma it follows that $\int |f|^p \,d\mu\le M^p&lt;\infty$, so $f$ belongs to $\mathscr{L}^p (X)$. Let $g$ in $\mathscr{L}^q$ arbitrary but fixed.   Then by Hölder inequality it follows that $$\|f_n g\|_1 \le \|f_n\|_p\|g\|_q\le M \|g\|_q&lt;\infty$$ and $$\|f g\|_1 \le \|f\|_p\|g\|_q \le M \|g\|_q&lt;\infty$$ thus $f_n g$ and $fg$ belongs to $\mathscr L ^1(X)$. Now since $g$ is in $\mathscr L ^q (X)$, then $g$ is finite a.e.,without loss of generality we may assume that $g$ is real-valued so $f_n g \xrightarrow{ a.e} fg$.   Let $\nu (A) =  \int_A |g|^q \,d\mu $, so $\nu $ is absolutely continuous with respect to $\mu$, also $\nu$ is a finite measure on $(X, \mathscr A )$. We can use the $\epsilon-\delta$ definition of absolutely continuous, so given $\epsilon&gt;0$ there is a $\delta&gt;0$ such that for $\mu(A)&lt;\delta$ then $\nu (A) &lt;\epsilon^q$ for any $A$ in $\mathscr A$.   Now by Egoroff's thm exists a $B$ in $\mathscr A$ such that $(f_ng) (x)\xrightarrow{\text{uniformly}} (fg) (x)$ for $x$ in $B$ and $\mu(X\setminus B) &lt;\delta$. Let $N$ such that for all $n\ge N$, $|(f_ng)(x)-(fg)(x)|&lt;\epsilon$ for all $x\in B$. Thus   \begin{align*}\left |\int_X (f_n-f) g \, d\mu \right|&amp;\le \int_B |f_ng-fg| \, d\mu +\int_{X-B}|f_n g|\, d\mu +\int_{X-B}|fg|\,d\mu \\[6pt] &amp;\le \int_B |f_ng-fg| d\mu + 2M \left( \int_{X-B}|g|^qd\mu \right)^{1/q}\\[6pt] &amp;\le\epsilon \mu(X)+2M  (\nu(X\setminus B))^{1/q}\\[6pt] &amp;\le \epsilon (\mu(X)+2M)\end{align*}  Since $\mu(x)&lt;\infty$ and $M&lt;\infty$ the result follows.