Apparently the ability to give counter-examples, particularly in analysis, is a sign of good understanding of the statements of theorems. These, sometimes natural, sometimes pathologic constructions reveal the significance of each individual assumption, disproving the hypothesis when one of the former is dropped.

Consider the following test of the requirement of completeness of a linear space arising in the formulation of the Banach-Steinhaus theorem. It is the proof of incompleteness of the given space that is interesting. The following construction is suggested:

$$X=\left\{ \left. x=(x_k) \right| \sum \left| x_k\right| <\infty \right\}$$

endowed with the norm

$$\Vert x\Vert=\sup_k\left|x_k\right|$$

where \((x_k)\) stands for a sequence of complex numbers.

There is some sense in which the selected norm is "unnatural" for the given space and that sense is precisely that normed space \(\left(X,\Vert .\Vert\right)\) is incomplete.

I began thinking in the direction of the harmonic series which seems to be the source of a great number of various counter-examples. After some juggling I considered the sequence

$$x_n=(x_k)_n=\frac{1}{k+n}-\frac{1}{1+k+n}$$

which is really a sequence of infinite sequences. Indeed the partial sum, after "telescoping" gives

$$\sum_{k=1}^n x_n=1-\frac{1}{1+k+n}$$

which tends to 1 and so \(x\in X\). It turns out that \(x_n\) is Cauchy in \(\left(X,\Vert .\Vert\right)\). Now \(x_n\) is Cauchy in \(\left(X,\Vert .\Vert\right)\), since

$$\Vert x_n - x_m \Vert = \sup_k\left|\frac{1}{k+n}-\frac{1}{k+m}+\frac{1}{1+k+m}-\frac{1}{1+k+n}\right|\to 0$$

Taking \(y=\left(\frac{1}{k}\right)\) we note that

$$\lim _{n\to 0}\Vert x_n - y \Vert=0$$

Another example using some "off the shelf" sequences demonstrates the non-triviality of the result of Hahn-Banach extension theorem. The main lesson here is that it is easy to construct an extension for a functional, but giving a

$$x=(1,1,1,...)$$

$$y=(0,-1,0,-1,...)$$

\(x\in c\), hence \(g(x)=f(x)=1\). \(y\in l_{\infty}\backslash c\), hence \(g(y)=0\). So \(g(x)+g(y)=1\). On the other hand

$$x+y=(1,0,1,0,...)$$

Clearly \((x+y)\in l_{\infty}\backslash c\), so \(g(x+y)=0 \ne g(x)+g(y)\), so \(g\) is non-linear.

Consider the following test of the requirement of completeness of a linear space arising in the formulation of the Banach-Steinhaus theorem. It is the proof of incompleteness of the given space that is interesting. The following construction is suggested:

$$X=\left\{ \left. x=(x_k) \right| \sum \left| x_k\right| <\infty \right\}$$

endowed with the norm

$$\Vert x\Vert=\sup_k\left|x_k\right|$$

where \((x_k)\) stands for a sequence of complex numbers.

There is some sense in which the selected norm is "unnatural" for the given space and that sense is precisely that normed space \(\left(X,\Vert .\Vert\right)\) is incomplete.

I began thinking in the direction of the harmonic series which seems to be the source of a great number of various counter-examples. After some juggling I considered the sequence

$$x_n=(x_k)_n=\frac{1}{k+n}-\frac{1}{1+k+n}$$

which is really a sequence of infinite sequences. Indeed the partial sum, after "telescoping" gives

$$\sum_{k=1}^n x_n=1-\frac{1}{1+k+n}$$

which tends to 1 and so \(x\in X\). It turns out that \(x_n\) is Cauchy in \(\left(X,\Vert .\Vert\right)\). Now \(x_n\) is Cauchy in \(\left(X,\Vert .\Vert\right)\), since

$$\Vert x_n - x_m \Vert = \sup_k\left|\frac{1}{k+n}-\frac{1}{k+m}+\frac{1}{1+k+m}-\frac{1}{1+k+n}\right|\to 0$$

Taking \(y=\left(\frac{1}{k}\right)\) we note that

$$\lim _{n\to 0}\Vert x_n - y \Vert=0$$

Another example using some "off the shelf" sequences demonstrates the non-triviality of the result of Hahn-Banach extension theorem. The main lesson here is that it is easy to construct an extension for a functional, but giving a

*linear*extension is far from obvious. Take \(f(x)=\lim_n x_n\) where \(x\) is a convergent sequence. Consider the space of convergent sequences \(c\) as a subspace of bounded sequences \(l_{\infty}\). Define \(g\) on \(l_{\infty}\) by \(g(x)=f(x)\) for \(x\in c\) and \(g(x)=0\) for \(x\in l_{\infty}\backslash c\). It can be checked that \(g\) is a non-linear extension of \(f\) to \(l_{\infty}\). Indeed, consider$$x=(1,1,1,...)$$

$$y=(0,-1,0,-1,...)$$

\(x\in c\), hence \(g(x)=f(x)=1\). \(y\in l_{\infty}\backslash c\), hence \(g(y)=0\). So \(g(x)+g(y)=1\). On the other hand

$$x+y=(1,0,1,0,...)$$

Clearly \((x+y)\in l_{\infty}\backslash c\), so \(g(x+y)=0 \ne g(x)+g(y)\), so \(g\) is non-linear.

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