Given the normed space $\ell^\infty$ of all bounded sequences of (real or complex) numbers with the norm given by $$||x||:= \sup_{j\in Z^+} |\xi_j|,$$ for each $x:=(\xi_j)_{j=1}^\infty$ in $\ell^\infty$, and given the linear operator $T \colon \ell^\infty \to \ell^\infty$ defined as $$T(\xi_j)_{j=1}^\infty := (\frac{\xi_j}{j})_{j=1}^\infty,$$ while $T$ is bounded, how to compute the norm of $T$, and how to find its range?

What is the range of $T$? And what is its norm? Is the range a closed set in $\ell^\infty$?

This operator being injective admits an inverse. (This inverse is also a linear operator from the range to the domain space.) Is the inverse a bounded operator too?

  • $\begingroup$ To find the norm is not difficult, so I'd like to give a couple of hints which will allow you to find the answer yourself. It is easy to find an estimate of $\lVert Tx\rVert_{\infty}$ in terms of $\lVert x \rVert_{\infty}$, and this will give you an upper bound $\lVert T\rVert\le C$ (I'm not revealing the exact value of $C$, you can easily find it yourself). To prove that this upper bound cannot be strict, just test the inequality $\lVert Tx\rVert_{\infty}\le C\lVert x \rVert_\infty$ with the sequence $x=(1, 1, 1,\ldots)$. $\endgroup$ – Giuseppe Negro Feb 16 '13 at 13:27
  • $\begingroup$ The value of $C$ should be at least $1$, and since in this particular case $||Tx|| = ||x||$, can we conclude that the norm is $1$? $\endgroup$ – Saaqib Mahmood Feb 16 '13 at 13:48
  • $\begingroup$ Yes, that's what I thought. $\endgroup$ – Giuseppe Negro Feb 16 '13 at 15:41

It is obvious that $\Vert Tx\Vert_\infty\leq\Vert x\Vert_\infty$, hence $\Vert T\Vert\leq 1$. Since $\Vert T(1,0,0,\ldots)\Vert_\infty=\Vert (1,0,\ldots)\Vert_\infty$, we conclude that $\Vert T\Vert=1$.

One can easily verify that $$T(l^\infty)=\left\{(\xi_j):\exists C>0 \text{ such that } |\xi_j|\leq C/j\text{ for all j}\right\}$$

$T(l^\infty)$ is not closed in $l^\infty$: Let $x_n=(1,\dfrac{1}{\sqrt{2}},\ldots,\dfrac{1}{\sqrt{n}},0,0,\ldots)=T(1,\sqrt{2},\ldots,\sqrt{n},0,\ldots)$. The sequence $(x_n)$ obviously converges to $x=\left(1/\sqrt{j}\right)_{j=1}^\infty\in l^\infty$, which does not lie in $T(l^\infty)$ (if it did, what would be it's pre-image?).

The inverse operator $T^{-1}$ is not bounded: Consider the sequence $(x_n)\subseteq T(l^\infty)$ as above. This sequence is bounded but the image $\left\{T^{-1}x_n\right\}$ is not, since $\Vert T^{-1}x_n\Vert_\infty=\sqrt{n}$.

You could also argue in this way: If $T^{-1}$ were bounded, then for every Cauchy sequence $(y_n)\in T(l^\infty)$, the sequence $(T^{-1} y_n)$ would also be Cauchy, and hence would converge to some $x\in l^\infty$ since $l^\infty$ is complete. But then, since $T$ is bounded, $y_n$ would converge to $Tx$. Then $T(l^\infty)$ would be complete, contradicting the fact that it is not closed in $l^\infty$ (this argument can actually be used to show that if $T:X\rightarrow Y$ is an isomorphism between a Banach space $X$ and some normed space $Y$ such that $T$ and $T^{-1}$ are bounded, then $Y$ is Banach).

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  • $\begingroup$ Thank you so much! This answers my question satisfactorily enough. $\endgroup$ – Saaqib Mahmood Feb 22 '13 at 7:43

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