There is some sense in saying that the limit of $\ell^p$ norms  $\displaystyle \lim_{p \to \infty} \lVert x \rVert _p = \lVert x \rVert _{\infty}$  is uniform in $x$?  Maybe with this definition  $(\forall \varepsilon &gt; 0)(\exists M&gt;0)(\forall x \in \ell^q) \big( p&gt;M \ \Longrightarrow \ | \ \lVert x \rVert_p - \lVert x \rVert_{\infty} \ | &lt; \varepsilon \big)$  where $q \in [1, \infty)$ is fixed, and $p$ take values in $(q, \infty)$.  This is true? If not, someone have a counterexample?  Thanks
48623
The $L^p$ norm in $\mathbb{R}^n$ is \begin{align} \|x\|_p = \left(\sum_{j=1}^{n} |x_j|^p\right)^{1/p}. \end{align} Playing around with WolframAlpha, I noticed that, if we define the "scaled" $L^p$ norm in $\mathbb{R}^n$ to be  \begin{align} \| x \|_p = \left(\frac{1}{n}\sum_{j=1}^{n} |x_j|^p\right)^{1/p} \end{align} then \begin{align} \lim_{p \to 0} \|x\|_p &amp;= \left( \prod_{j=1}^{n} |x_j| \right)^{1/n}, \end{align} which is the geometric mean of the coordinates' absolute values. This is interesting maybe because the $L^p$ norm doesn't have a nice limit at zero.  My questions:   How do I prove this? Is this definition of "scaled $L^p$ norm" interesting, or known by another name, or used anywhere? Is there any interesting reason to define the $L^0$ norm as the geometric mean, as above? Further reading?   Thanks!
120404
Let $ c_0 = \{ x = (x_n)_{n \in \mathbb N} \in l^\infty : \lim_{n \to \infty} x_n = 0\}$. Show that $c_0$ is a Banach space with the norm $\rVert \cdot \lVert_\infty$  I am capable of showing the space where the limit of $x_n$ exists is normed linear space but am having trouble with showing that the limit of Cauchy sequences must converge to 0.   Let $(x^{(n)})_{n \in \mathbb N}$ be a Cauchy sequence in $c_0$ such that $x^{n} = (x^n_1, x^n_2,...)$. Fix $k \in \mathbb N$ consider the sequence $(x^n_k)_{n \in \mathbb N}$ in $\mathbb F$. For any $n,m \in \mathbb N$  $\lvert x^n_k - x^m_k \rvert \le \sup_{k \in \mathbb N} \lvert x^n_k - x^m_k \rvert = \lVert x^n - x^m \rVert_\infty \lt \epsilon $   (1)  Thus $x^n_k$ is Cauchy in $\mathbb F$ and so has limit $y_k$ such that $y = (y_1,y_2,...)$ and y is the limit of $x^n$  To show that such a y exists we look at the value of $\lvert y_n - y_m \rvert \le \lvert y_n - x^N_n \rvert + \lvert x^N_n - x^N_m \rvert + \lvert x^N_m - y_m \rvert \lt \epsilon$ for all $n,m \ge N$     (2)  The middle expression on RHS of (2)  is $\lt \epsilon/3$ by (1)  The other two are also $\lt \epsilon/3$ follow from $x^N_k$ being Cauchy and converging to $y_k$  This shows that $\lim_{n \to \infty} y_n$ exists but we still have not shown that $y \in c_0$.  I know that to show y tends to 0 i should show that $\lvert y_k \rvert \lt \epsilon$ for $k \ge N$  This is where I am stuck. Perhaps $\lvert y_k \rvert = \lvert \lim_{n \to \infty} x^n_k \rvert$ and then we can take the limit function outside the absolute value sign by continuity? Then we might say due to it being a Cauchy sequence $x^n_k \lt \epsilon$. I know this last bit isn't at all convincing so I could do with some help.
47565
I get that the limit is  $$\lim\limits_{n\to\infty}f(x)=1$$ because $$\lim\limits_{n\to\infty}(1+x^{2n})=1,\,\,\lvert x\rvert&lt;1.$$  Is this right?
75374
Prove: $\displaystyle\liminf_{n\to\infty}(-a_n)=-\limsup_{n\to\infty}(a_n)$   My general idea was: if $a$ is partial limit (PL) of $a_n$, then $-a$ is a PL of $-a_n$ so it follows that $s$ is the maximal PL iff $-s$ is the minimal PL.   But how do you show it rigorously ?
98160
This is a small exercise that I just can't seem to figure out. When I see it I'll probably go 'ahhh!', but so far I haven't made any progress.  I'd like to prove that any linear functional $\phi$ on $\ell^\infty$ that is positive, i.e. for any $(x_n) \in \ell^\infty$ such that $x_n \geq 0$ for all $n \in \mathbb{N}$ we have $\phi((x_n)) \geq 0$, is a bounded linear operator and has $\|\phi \| = \phi((1,1,...))$.  Nothing I have tried worked so far, so any hints would be appreciated, thanks!
107404
I am trying to prove $$X_n \xrightarrow{d} X, Y_n \xrightarrow{d} a \implies Y_n X_n \xrightarrow{d} aX$$ where $a$ is a constant. What I tried: Let $g:\mathbb R\to \mathbb R$ an arbitrary uniformly continuous, bounded function. It suffices to show $\mathbb E[g(Y_n X_n)] \to \mathbb E[g(aX)]$. We have  $$\left \lvert \int g(Y_nX_n) - g(aX) \,dP \right \rvert \leq \left \lvert \int g(Y_n X_n) - g(aX_n) \, dP \right \rvert +\left \lvert \int g(aX_n) - g(aX) \, dP \right \rvert,$$ where the right summand goes to $0$ by assumption as $a$ is constant. Now I want to use uniform continuity of $g$ to estimate the left summand: Choose $\delta &gt; 0 $ such that $$\left \lvert g(Y_n X_n) - g(aX_n) \right \rvert &lt; \epsilon,$$ whenever $|Y_n X_n - aX_n| &lt; \delta.$ Since convergence in distribution to a constant implies convergence in probability, we have can control $P(|Y_n - a | &gt; \delta)$. But is it possible to get a bound on $|X_n|$? I assume that probability in distribution does not imply some form of boundedness... Any help is appreciated.
117646
In this case, $k,l,p,q\geq0$ and are integers. I have attempted the substitution $u=x^p$ and $v=y^q$ $$\lim_{(x,y)\to(0,0)} \frac{x^ky^l}{x^{2p}+y^{2q}}=\lim_{(u,v)\to(0,0)} \frac{u^{k/p}v^{l/q}}{u^2+v^2}$$  Then, note that $|u|&lt;\sqrt{|u|^2+|v|^2}$ and $|v|&lt;\sqrt{|u|^2+|v|^2}$. So we have  $$\bigg|\frac{u^{k/p}v^{l/q}}{u^2+v^2}\bigg|=\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}&lt;\frac{(|u|^2+|v|^2)^{k/(2p)+l/(2q)}}{|u|^2+|v|^2}$$ Thus, the limit equals $0$ when $\frac{k}{p}+\frac{l}{q}&gt;2$. We need to show that otherwise, the limit does not exist.   If $\frac{k}{p}+\frac{l}{q}=2$, taking the limit along the axes yields $0$ but letting $x^p=y^q$ gives  $$\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}=\frac{|u|^{k/p+l/q}}{2|u|^2}\to\frac{1}{2}$$ so the limit does not exist.   Similarly, if $\frac{k}{p}+\frac{l}{q}&lt;2$, taking the limit along the axes still yields $0$, but letting $x^p=y^q$ instead gives  $$\frac{|u|^{k/p}|v|^{l/q}}{|u|^2+|v|^2}&gt;\frac{|u|^{2}}{2|u|^2}\to\frac{1}{2}$$ showing that the limit does not exist.  I'm not sure if all of the steps I've taken are correct - especially some of the inequalities as $(u,v)\to(0,0)$. Also, for the last two cases, I'm not exactly sure if the absolute values around $u$ and $v$ should be there or whether I should have let $|x|^p=|y|^q$.
66783
Suppose we have a Hilbert space $\mathcal H$, with projections $P,Q$, and let $R$ be the projection onto $P\mathcal H\cap Q\mathcal H$, then I would like to show that $(PQ)^n\to R$ strongly.   We can clearly reduce to the case where $P\mathcal H\cap Q\mathcal H=0$ and $P\mathcal H+Q\mathcal H=\mathcal H$, in which case we must show that $(PQ)^n\to0$ strongly. Now, if the limit $y=\lim_{n\to\infty}(PQ)^n x$ exists, then if it is nonzero, then $$\|y\| = \|PQ y\|&lt;\|Q y\|\le\| y\|$$ which is a contradiction, so once we have convergence, we're done. However, I don't see quite how to show convergence in a general infinite dimensional case.
19432
Let $x,y \in L_p$   such that  $\|x\|_p=\|y\|_p=1$ , $1&lt; p&lt;\infty$ and $x\neq y.$ Why is $\|x+y\|_p&lt;2$  ?  I'm not sure how to start the proof.. I don't know how to handle integral of $(x+y)^p$ and it seems that using the binomial theorem won't be a great success.
70674
I am trying to show that for nonnegative $f$ on $\mathbb{R}$ if $||f||_1&lt;\infty$, we have  $$ \lim_{p\uparrow \infty}||f||_p=||f||_\infty $$ I have tried to fiddle around with  $$ \left(\int_\mathbb{R}|f|^p|f|^{1-p} \right)^{1/p}&lt;\infty $$ but haven't found a nice way to get an estimate on $\int_E|f|^p$ from this. Honestly I am pretty uncomfortable with non finite measure spaces.   Any hints would be awesome! Any references for problems similar to this would also be awesome.  edit: this is not a duplicate of the linked question as 1) my measure space is not finite, 2) my function is in $L^1$ and not necessarily in $L^\infty$
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.
115817
I am curious about whether $|x|^p$ with $0&lt;p&lt;1$ satisfy $|x+y|^p\leq|x|^p+|y|^p$ for $x,y\in\mathbb{R}$.   So far my trials show that this seems to be right... So can anybody confirm whether this is right or wrong and give a proof or counterexample?
34986
I not solve the follow limit $$\lim_{p\rightarrow 0} \bigg[\int_{\Omega} |f|^p d\mu \bigg]^{1/p} = \exp\bigg[ \int_{\Omega} \log|f|d\mu \bigg],$$  where $(\Omega, \mathcal{F}, \mu)$ is a probability space and $f,\log |f| \in L^1(\Omega).$  Can someone help me?  Thank you!
21562
If $1\leq p &lt; \infty$ then show that $L^p([0,1])$ and $\ell_p$ are not topologically isomorphic unless $p=2$.  Maybe I would have to use the Rademacher's functions.
121151
Are all the norms of $L^p$ space equivalent? That is, for any $p,~q \in R^+$, there exist two positive number $C_1,~C_2$ such that $$ C_1\|u\|_{L^q} \le \|u\|_{L^p} \le C_2\|u\|_{L^q}. $$
95036
I know that $\infty-\infty$ is indeterminate, the how about $\lim_{x\to \infty}(x-x)$ and $\lim_{x\to \infty}x-\lim_{x\to \infty}x$?   Thank you.
54056
Let $p \in (0,2)$ and let $\xi_n, n \geq 1$, be iid random variables. Show that the following two conditions are equivalent:   With probability one, the limit $$ \lim_{n \rightarrow \infty} \frac{1}{n^{1/p}} \sum_{k=1}^n \xi_k$$ exists and is finite. $\mathbb{E}\lvert \xi_i \rvert^p &lt; \infty$ AND either $\mathbb{E} \xi = 0$ or $p \leq 1$.   I have tried using Holder's inequality but didn't really get anywhere. I don't really have any idea how to approach the problem for either direction...