The problem consists of a linear space $(ℓ^2 (N, R)$ of sequences $x = x_1, x_2, ...$ with defined norms for $||*||_2$ and $||*||$.
I want to show that the identity mapping $I(x) = x$ from $(ℓ^2 (N, R), ||*||_2)$ to $(ℓ^2 (N, R), ||*||)$ is continuous.
How am I supposed to go about this? Should I use the norms and from them work with the normal $\delta$ and $\epsilon$ definition of continuity?
Any explanation is greatly appreciated.
The identity map is continuous iff there is $c \ge 0$ such that
$||I(x)|| \le c||x||_2$ for all $x \in l^2$
But you have not told us, how are the norms $||∗||_2$ and $||∗||$ are defined !
Without this information a further investigation is not possible !
A linear operator $T : (X, \|\cdot\|_X) \rightarrow (Y, \|\cdot\|_Y)$ is continuous if and only if it is bounded, i.e. there exists $M > 0$ such that $\forall x \in X$ $$ \|T(x)\|_Y \leq M\|x\|_X.$$
Now apply this result to your operator $I$ from $(\ell_2, \|\cdot\|_2)$ to $(\ell_2, \|\cdot\|)$. You get that $I$ is continuous if and only if there exists $M > 0$ such that $\forall x \in \ell_2$ $$ \|I(x)\| \leq M\|x\|_2.$$ But $I(x) = x$ so that $\|I(x)\| = \|x\|$. Now for arbitrary norms one cannot claim that such a $M>0$ would exists. But depending on what you know about the norms $\|\cdot\|$ and $\|\cdot\|_2$ (how they are defined), you may be able to prove the existence of a $M>0$ such that for all $x \in \ell_2$ $$ \|x\| \leq M\|x\|_2.$$
The norms are defined as $$\|x\|_2 = (\sum_{n=1}^{\infty} |x_n|^2)^{1/2} < \infty$$ and $$\|x\| = \sum_{n=1}^{\infty} \frac{|x_n|}{n}$$
Using your comments I need to show that for every $x ∈ l^2$ :
$$\|x\| ≤ M \|x\|_2$$
Which means that:
$$\sum_{n=1}^{\infty} \frac{|x_n|}{n} < M \left(\sum_{n=1}^{\infty} |x_n|^2\right)^{1/2}$$
I am thinking that I have to make an assumption about $x_n \rightarrow x$ and show that it holds? And if so, the continuity is confirmed.
Now we are in a position to answer your question. We have to show that there is $M \ge 0$ such that
$\sum_{n=1}^{\infty} \frac{|x_n|}{n} \le M \left(\sum_{n=1}^{\infty} |x_n|^2\right)^{1/2}$ for all $(x_n) \in l^2$.
To this end let $(x_n) \in l^2$ and $y=(1/n)$. Then $y \in l^2$ and with $x:=(|x_n|)$, by Cauchy - Schwarz:
$\sum_{n=1}^{\infty} \frac{|x_n|}{n}=(x,y) \le ||x||_2*||y||_2=M (\sum_{n=1}^{\infty} |x_n|^2)^{1/2}$
with $M=\sqrt{\frac{\pi^2}{6}}$
Also, your function $y$ is a nifty trick since you know that the sum $\frac{1}{n^2}$ is $\frac{\pi^2}{6}$ before hand? I mean, I could use an arbitrary function that I know would converge to an M that belongs to $l2$?
Thank you.
– J. Smolinski Mar 16 '17 at 22:45