# linear map of bounded sets into bounded sets implies a bounded operator

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.

• 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 $\Vert Tx\Vert\le C\Vert x\Vert$. – David Mitra Jan 13 '14 at 19:30
• @DavidMitra: that seems like it should be an answer, to me – Ben Millwood Jan 13 '14 at 20:12

## 1 Answer

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 $$x$$.