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If $r\le s$ then $l^r\subseteq l^s$. How can I prove there is a continuous injection $l^r\hookrightarrow l^s$? The suggestion was to use the fact that $\Vert x\Vert_r\le\Vert x\Vert_s$. Also how can I prove that $l^1$ is dense in $l^2$?

Thank you!!

Armando j18eos
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Rock
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1 Answers1

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Consider the map $$\phi: l_r \to l_s : f \mapsto f.$$ This map is well defined, that is $$f \in l_r \Rightarrow f \in l_s;$$ to see this just observe the following inequalities for $f \in l_r$: $$||f||_s < || f ||_r < \infty.$$ This shows that $f \in l_s$. Clearly this map is injective. Suppose $\phi(f) = \phi(g)$, then $f = \phi(f) = \phi(g) = g$, which proves injectivity.

For continuity: just use the epsilon definition of continuity in metric spaces for the map $\phi$.

For density: Pick any $\epsilon$ and any $f \in l_2$. Construct a function $g \in l_1$ such that $||f - g||_2 < \epsilon$. Since you picked everything arbitrarily you've proven that $l_1$ is dense in $l_2$.

HINT FOR DENSITY: Take any $f = (a_i)_{i \in \mathbb{N} }\in l_2$. There exists some $N \in \mathbb{N}$ such that $\sum_{n= N}^\infty a_n^2 < \epsilon$. Can you use this information to construct some $g \in l_1$ such that $||f-g||_2< \epsilon$? So $||f ||_2^2= \sum_{n=0}^{N-1}a_n^2 + \sum_{n= N}^\infty a_n^2$.

Here is a link worth looking at: $L^p$ and $L^q$ space inclusion

It is a bit different but I think it will help you.

Community
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Leo
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  • I'm not sure on the injection part. Injection means that the map $\phi$ you defined exists, how can I prove this? Is it suficient to prove that the norm of x in $l_s$ is not bigger than the norm in $l_r$? As for the density part, can I pick y=x / ∥x∥ (where x is a sequence of $x_n$)? This way I prove that if y and x are in $l_1$ then y is also in $l_2$. Thank you very much for your help! – Rock Jun 20 '16 at 11:13
  • I've included some edits that show injectivity. Try to do continuity yourself. As for density, write down the formula for || f ||_2. For some $f$ can you write $f $ as the sum of two elements $g$ and $e$ one of which is an element $l_1$ and the second of which has $||.||_2$ norm smaller than $\epsilon$? – Leo Jun 20 '16 at 12:07
  • Thanks. My idea would be to take $g_i$ = $f_i$$^2$ for all i, but then ||f-g||$_2$ would be zero? This isn't possible because the inclusion is not proper... Can you give me another hint? Or flat out the solution, althoufg thinking about it is actually useful. – Rock Jun 20 '16 at 17:11
  • You have a pretty clear hint for density now :) – Leo Jun 20 '16 at 18:35
  • @user1952009 I was hoping that he/she would figure that out by himself ;) – Leo Jun 20 '16 at 18:57
  • Do I? ;) I can't think of anything better than this and this seems wrong because the difference is exactly zero and so $l_1$=$l_2$. I have now thought this might not be true, I have only found an example of equality, this doesn't mean there do not exsist functions which are in $l_2$ but not in $l_1$. But I did say for all f... so this doesn't seem totally right. Oh dear, help, please!

    Oh I have a question: in the first answer you wrote for every $f$ in $l_2$, in the comments you said for some $f$. I would say it has to be for every $f$, correct?

     – Rock Jun 20 '16 at 19:01
  • Oh no, I missed the comment! ;) (Not French, but a lady here, btw) So this popped into my mind. What if I add an $\epsilon$ to $g$? Definitely not sure about this though, it seems pretty useless really – Rock Jun 20 '16 at 19:07
  • Here are a couple more hints: Do you you think $f := \sum_{n=0}^Na_n$ is in $l_1$? What happens when you subtract it from an element in $g \in l_2$ for which it holds that $f(i)=g(i)$ for $0 \leq i \leq N$? – Leo Jun 20 '16 at 19:11
  • It is, if the $a_n$s are not $\infty$, and surely if the $a_n$ are "my" $d_i$s ). If I subtract it, the element is still in $l_2$ and it's a tail of a converging series so its limit is zero?! – Rock Jun 20 '16 at 19:20
  • if you subtract everything but the tail, the tail itself has ||.||_2 norm less then $\epsilon$, which concludes the proof. (This of course only holds if you pick N large enough, take another look at the hint ;) ) – Leo Jun 20 '16 at 19:25
  • But why would I need to do that? Where is the mistake with the $g_i$ = $f_i$$^2$idea? (the $g_i$ are the elements of the sequence) Is there a problem if the difference is exactly zero? Thank you so much! – Rock Jun 20 '16 at 19:35
  • I'm sorry, I have one last question. I was discussing this with a friend and he told me to use the properties of ortonormal spaces. If $V$ = $l_1$ and $V_o_r_t_o_g_o_n_a_l$ = $V^o$= (closure of $V^o$) = {0}, then the closure of $l_1$ = $l_2$. By proving <$x$,$y$>$_2$ = 0 for all $y$ in $l_1$, $x$ = 0. Is this correct? Thank you for your help!! – Rock Jun 21 '16 at 11:43