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The definition of a bounded linear operator is a linear transformation $T$ between two normed vectors spaces $X$ and $Y$ such that the ratio of the norm of $T(v)$ to that of $v$ is bounded by the same number, over all non-zero vectors in $X$.

What is this definition saying, is it saying that the norm of vectors is preserved under the transformation?

Why does ratio come into it?, the ratio of what?

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  • $\begingroup$ That a linear operator is bounded means simply that it is Lipschitz. It turns out to be equivalent to continuity in case of normed spaces. $\endgroup$ – tomasz Feb 27 '16 at 15:24
  • $\begingroup$ Details are covered in other posts. For context, the interest arises because a bounded linear operator is continuous. It's easy to show that a linear operator on a finite dimensional space is bounded (and therefore continuous). Things become more interesting in infinite dimensional spaces. $\endgroup$ – Tom Collinge Feb 27 '16 at 15:40
  • $\begingroup$ It might also be instructive to see what a not-bounded operator looks like. For example, the derivative operator $C^1(\mathbb{R}) \to C(\mathbb{R})$, where $C^1(\mathbb{R})$ is equipped with the $\sup$ norm inherited from $C(\mathbb{R})$. You can make the operator bounded by equipping $C^1(\mathbb{R})$ with the norm $u \mapsto \| u \|_\infty + \| u' \|_\infty$. $\endgroup$ – Eric Thoma Mar 1 '16 at 4:52
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Let $B_r(x)$ denote the open ball of radius $r$ centered at $x$: $$ B_{r}(x) = \{ y : \|x-y\| < r \}. $$ A linear transformation $T$ is continuous at $x$ iff, for every $\epsilon > 0$ there exists $\delta > 0$ such that $$ T(B_{\delta}(x)) \subseteq B_{\epsilon}(Tx) $$ Because of linearity, the above is equivalent to $$ T(B_{\delta}(0))\subseteq B_{\epsilon}(0). $$ So $T$ is continuous at $0$ iff it is continuous at every $x$. Furthermore, because of how linear transformations commute with scalar multiplication, continuity at $0$ is equivalent to the existence of $\delta > 0$ such that $$ T(B_{\delta}(0)) \subseteq B_{1}(0). $$ Indeed the above holds for some $\delta > 0$ iff $$ T(B_{\delta\epsilon}(0))\subseteq B_{\epsilon}(0). $$ Another equivalent is that $T(B_{1}(0))$ is contained in some finite ball $B_{r}(0)$. That is, $T$ is continuous at every $x$ iff there exists $r > 0$ such that $$ T(B_{1}(0)) \subseteq B_{r}(0). $$ Equivalently, there exists $r > 0$ such that $$ \|Tx\|_{Y} \le r\|x\|_{X},\;\;\; x\in X. $$ This last condition is the definition of a bounded transformation.

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What the definition is saying is that there exists a constant $C>0$ such that, for every $v\in X$: $$\|T(v)\|\le C\|v\|.$$ In other words, the ratio $\frac{\|T(v)\|}{\|v\|}$ is bounded by the same constant $C$, regardless $v\in X$. Note that the latter requires $v\ne 0$ for the ratio to be defined.

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    $\begingroup$ It seems that the OP know this already. $\endgroup$ – user99914 Feb 27 '16 at 12:56
  • $\begingroup$ @AugSB where $v$ is any vector in $x$ right? $\endgroup$ – Al jabra Feb 27 '16 at 13:30
  • $\begingroup$ Of course, it can be any $v\in X$. Additionally, if you use the ratio definition, $v\ne 0$. $\endgroup$ – AugSB Feb 27 '16 at 14:04
  • $\begingroup$ See the answers below. Example: For $X=Y=R^2$ each with the norm $\|(x,y)\|=\sqrt {x^2+y^2},$ and $T(x,y)=(2 x,3 y)$ we have $ \|T\|=3.$ $\endgroup$ – DanielWainfleet Mar 1 '16 at 7:05

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