Is the image of this operator on $\ell^2$ closed? Suppose that $\ell^2 = \biggl\{(x_n)_n \in \mathbb{K}^{\mathbb{N}_0} \biggm|
\sum_{n=1}^{\infty}|{x_n}^2| < +\infty \biggr\}$ is a Hilbert-space with the inproduct $\langle\cdot,\cdot\rangle_2: \ell^2 \to \ell^2: (x,y) \mapsto \sum_{n=1}^\infty \overline{x_n}y_n$.
Consider the operator $f: \ell^2 \to \ell^2: (x_0, x_1, \ldots) \mapsto (x_0, 0, x_1, 0, \ldots)$.
I'm supposed to check whether the image of $f$ is closed or not. First of all, I don't think the image can be the whole $\ell^2$-space, because there's no element that's mapped on $\left(\frac{1}{n}\right)_n$. I tried to reason on like this:
If one should show that $\operatorname{Im}(f)$ is closed, it's sufficient that $\overline{\operatorname{Im}(f)} \subset \operatorname{Im}(f)$ (I'm having the habit to use $\overline{\cdot}$ for closure). Therefore, choose an element $(x_n)_n$ in $\overline{\operatorname{Im}(f)}$. As the element lies in the closure of $\operatorname{Im}(f)$, there should be a sequence $(x_n^k)_n$ that converges to $(x_n)_n$. That means that $\forall n: \lim_{k\to\infty}x_n^k=x_n$. I'm stuck at this point, because I don't know what the image of $f$ is.
Is this a good start or is there an easier way?
 A: Your notation seems off, but it seems you have the basic idea down. Try this: 
Take $x$ in the closure of the image of $f$ and choose a sequence $(x_n)$ from the image of $f$ such that $x_n\rightarrow x$ in norm. Note the even coordinates of any $x_n$ are all zero. 
If one denotes the $i$'th coordinate of a vector $y$ by $y(i)$, then for each $i$, since convergence in $\ell_2$ implies coordinatewise convergence, we have $\lim\limits_{n\rightarrow\infty} x_n(i)=x(i)$. 
In particular, $x(i)=0$ if $i$ is even.  Let $z=(x_1,x_3,\ldots)$. Then $f(z)=x$; which shows $x$ is in the image of $f$.
Thus the image of $f$ is closed.
Alternatively, you could use the fact that $f$ is an isomorphism (isometry in fact): 
Take $(y_n)$  in $f(\ell_2)$ with $y_n\rightarrow y$ for some $y\in\ell_2$. Then $(f^{-1}(y_n))$ is Cauchy in $\ell_2$ and thus converges to some $x\in\ell_2$. Then since $f(f^{-1}(y_n))=y_n$ and since $f$ is continuous, we have $y_n\rightarrow f(x)$. It follows that $f(x)=y$. It follows that $f(\ell_2)$ is closed.
You could also note (prove!) that the image of $f$ is simply the set of all vectors $x$ in $\ell_2$ such that $x(i)=0$ whenever $i $ is even. This set is closed (argue as in the first part of this answer).
A: I hope this is correct, 
We will show that the image of $f$ is complete and consequently closed as $\ell^2$ is a Banach-Space.
So let 
$$f:\ell^2\rightarrow \ell^2$$
be as above.
Then consider some arbitrary Cauchy sequence $(fx_n)_{n\in\mathbb N}\subseteq f(\ell^2)$ i.e 
$$\forall \epsilon>0 \quad\exists N\in \mathbb N \;\text{s.t} \;\forall n,m\geq N:||fx_n-fx_m||_2<\epsilon$$
Moreover notice that the operator $f$ is isometric i.e for all $x\in\ell^2$ we have
$$||fx||_2=||x||_2$$
Hence for our sequence we have,
$$||fx_n-fx_m||_2=||f(x_n-x_m)||_2=||x_n-x_m||_2<\epsilon$$
and consequently $(x_n)\subseteq\ell^2$ is also a Cauchy sequence what means that there exist some $x_0\in\ell^2$ such that,
$$||x_n-x_0||_2\rightarrow 0\quad n\rightarrow \infty$$
Finally since 
$$||fx_n-fx_0||_2=||x_n-x_0||_2\rightarrow 0\quad n\rightarrow \infty$$
it follows that $\ell^2-\lim\;fx_n$ is $fx_o\in f(\ell^2)$ implying that the image is closed.
I am not sure if  I missed something so do not hesitate to correct me...
