Let $T\in L(X)$, where $T^2=0$, such that $X_0$ is an algebraic complement of $\ker T$ that is $X=\ker T\bigoplus X_0$, and $X$ is a Banach space for the norm $||.||$, now if we consider the norm $|y+x_0|=||y||+||x_0+\ker T||_{X/\ker T}$, $y\in \ker T, x_0\in X_0$ , we will have that $X$ is also complete for this new norm. Now I am trying to see that we will have that these norms are equivalent iff $X_0$ is closed in $X$ in terms of $||.||$

I still haven't managed to figure out why $(X,|.|)$ will be complete for this new norm, let's suppose we have a cauchy sequence in this new norm $y_n+x_n$ where $y_n\in \ker T$ and $x_n\in X_0$ we will have that $|y_n+x_n-(y_m+x_m)|=||y_n-y_m||+||x_n-x_m+\ker T||_{X/\ker T}$, and so we can conclude that $y_n$ will be a cauchy sequence and since $\ker T$ is closed we will have that $y_n\rightarrow y\in \ker T$, and we will also have that $x_n$ is a cauchy sequence in $X/\ker T$ and so it converges to some $x+\ker T$, since this is a Banach space also because $\ker T$ is closed, and so we will have that there exists $k_n\in \ker T$ such that $||x_n+k_n-x||\rightarrow 0$, now I cannot get beyond this point if I could prove that the $k_n's$ are a cauchy sequence I could finish the argument but without it, I dont see how to do it, any helps here is also aprecciated.

After a lot of failed attempts, my final idea was consider the function $I:(X,||.||)\rightarrow (X,|.|)$ and to show that is continuous, by showing that it closed. If we have that is continuous we get that $||x||\leq a|x|, a>0,\forall x\in X$, and then since we are working with Banach spaces we can use the Open mapping theorem to conclude that its inverse is also bounded and so we get that they are equivalent. To prove that the function was indeed closed we need to use sequences and the fact that $X_0$ is closed.

Now my question is assuming that my proof works, which I haven't figured out all the details yet, is there a simpler way to proving this result, because all we need to see is that there is some $c>0$ such that either $||.||\leq c|.|$ or $|.|\leq c||.||$, but I was not able to find out without using the fact that $I$ would be a continuous function between the $2$ topologies. Does anyone have any ideas on how to try and do this differently? Thanks in advance!


To finish your completeness argument for $(X,|\cdot|)$:

Directly from the definition of an algebraic complement it follows that the quotient map

$$q:X_0\to X_{/\ker T}\;,\;\;x_0\mapsto [x_0]$$

is bijective. Now $[x_n]\to [x]$ in $X_{/\ker T}$ and since the above quotient map is surjective, we can assume $x\in X_0$. Then from the definition of $|\cdot|$ it follows that $(y_n+x_n)\to (y+x)$ in $(X,|\cdot|)$.

To your second question: I also would use the open mapping theorem and cannot see a way of how it can be avoided. I would approach it like this:

A classical application of the open mapping theorem is, that if a banach space $(X,\|\cdot\|$) is the algebraic direct sum of two closed subspaces $V,W$, then the norm $|v+w|_1=\|v\|+\|w\|$ $v\in V, w\in W$ is equivalent to $\|\cdot\|$. Also, if $X_0$ is closed, again by the open mapping theorem the above defined quotient map $q$ is an isomorphism, so in that case the two norms $\|\cdot\|$ and $\|[\,\cdot\,]\|_{X/\ker T}$ on $X_0$ are equivalent and now one can combine these to facts to show that $\|\cdot\|$ is equivalent to $|\cdot|$:

$$X_{\|\cdot\|}\approx \ker T_{\|\cdot\|}\oplus_1{X_0}_{\|\cdot\|}\approx \ker T_{\|\cdot\|}\oplus_1{X_0}_{\|\cdot\|_{X/\ker T}}= X_{|\cdot|}$$

On the other hand, $X_0$ is always a closed subspace of $(X,|\cdot |)$, since it is isomorphic (in fact isometric) to the banach space $ X_{/\ker T}$, so if $\|\cdot\|$ and $|\cdot|$ are equivalent, they induce the same topology, so $X_0$ is also closed in $(X,|\cdot|)$.

Notice: We only used that $\ker T$ is closed, and not that $T^2=0$. Hope this helps.

  • $\begingroup$ Yes that helps alot and I did not know about that fact , just one question , yes it will be true that $|.|$ will be equivalent to $||.||$ in $X_0$ but I dont quite see how that gives us the equivalence in the middle because I can only show that $||y||+||x||\leq ||y||+a|x|$, I cant really find a constant such that $(||y||+||x||)\leq c(||y||+|x|)$ @lulu $\endgroup$
    – Something
    Jun 11 '20 at 20:00
  • 1
    $\begingroup$ @Something : $c=\max\{1,a\}$ will work. $\endgroup$ Jun 11 '20 at 22:58

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