I was watching a video lecture on bounded linear operators from one normed linear space to another. It was stated that if $T$ sends bounded sets in $X$ to bounded sets in $Y$ then $T$ is a bounded operator. I found this to be very hard to prove. The complication, as I understand it, arises for sequences $x_n$ converging to zero in $X$ where we might need an arbitrary large constant $K$ to get $\|Tx_n\|\leq K\|x_n\|$ even though there exists an constant $C$ such that $\|Tx_n\|\leq C$. Another approach would be to prove that mapping bounded sets to bounded implies continuity at some point, for example zero, but I have not succeeding in proving this either. Also in the video lecture, it had not yet been proven that linear bounded and linear continuous operators are the same, whereas I suspect this is not the standard approach.

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    $\begingroup$ If $\Vert Tx\Vert\le C$ for all $x$ of norm $1$, then for any non-zero $x$, $\Vert T(x/\Vert x\Vert)\Vert \le C$. But this implies $\Vert Tx\Vert\le C\Vert x\Vert$. $\endgroup$ – David Mitra Jan 13 '14 at 19:30
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    $\begingroup$ @DavidMitra: that seems like it should be an answer, to me $\endgroup$ – Ben Millwood Jan 13 '14 at 20:12

It follows from the homogeneity of the norm (that is $\Vert \alpha x\Vert=|\alpha|\Vert x\Vert$):

If $\Vert Tx\Vert≤C$ for all $x$ of norm $1$, then for any non-zero $x$,

$$\biggl\lVert T\left(\frac{x}{\lVert x\rVert}\right)\biggr\rVert≤C.$$

But this implies $\Vert Tx\Vert≤C\Vert x\Vert$ for all $x$.


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