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operator norm
Definition
Let $A\colon{\mathsf{V}}\to{\mathsf{W}}$ be a linear map between normed vector spaces ${\mathsf{V}}$ and ${\mathsf{W}}$. To each such map (operator) $A$ we can assign a nonnegative number $\A\_{{{\rm op}}}$ defined by
$\A\_{{{\rm op}}}:=\mathop{\sup_{{{\mathbf{v}}\in{\mathsf{V}}}}}_{{{\mathbf{v% }}\neq{\bf 0}}}\frac{\A{\mathbf{v}}\}{\{\mathbf{v}}\},$ 
where the supremum $\A\_{{{\rm op}}}$ could be finite or infinite. Equivalently, the above definition can be written as
$\A\_{{\rm op}}:=\mathop{\sup_{{{\mathbf{v}}\in{\mathsf{V}}}}}_{{\{\mathbf{v% }}\=1}}\A{\mathbf{v}}\=\mathop{\sup_{{{\mathbf{v}}\in{\mathsf{V}}}}}_{{0<\% {\mathbf{v}}\\leq 1}}\A{\mathbf{v}}\.$ 
By convention, if ${\mathsf{V}}$ is the zero vector space, any operator from ${\mathsf{V}}$ to ${\mathsf{W}}$ must be the zero operator and is assigned zero norm.
$\A\_{{{\rm op}}}$ is called the the operator norm (or the induced norm) of $A$, for reasons that will be clear in the next section.
Operator norm is in fact a norm
Definition  If $\A\_{{{\rm op}}}$ is finite, we say that $A$ is a bounded. Otherwise, we say that $A$ is unbounded.
It turns out that, for bounded operators, $\\cdot\_{{{\rm op}}}$ satisfies all the properties of a norm (hence the name operator norm). The proof follows immediately from the definition:
 Positivity:

Since $\A{\mathbf{v}}\\geq 0$, by definition $\A\_{{\rm op}}\geq 0$. Also, $\A{\mathbf{v}}\=0$ identically only if $A=0$. Hence $\A\_{{\rm op}}=0$ only if $A=0$.
 Absolute homogeneity:

Since $\\lambda A{\mathbf{v}}\=\lambda\A{\mathbf{v}}\$, by definition $\\lambda A\_{{\rm op}}=\lambda\A\_{{\rm op}}$.
 Triangle inequality:

Since $\(A+B){\mathbf{v}}\=\A{\mathbf{v}}+B{\mathbf{v}}\\leq\A{\mathbf{v}}\+\B% {\mathbf{v}}\$, by definition $\A+B\_{{\rm op}}\leq\A\_{{\rm op}}+\B\_{{\rm op}}$.
The set $L({\mathsf{V}},{\mathsf{W}})$ of bounded linear maps from ${\mathsf{V}}$ to ${\mathsf{W}}$ forms a vector space and $\\cdot\_{{{\rm op}}}$ defines a norm in it.
Example
Mathematics Subject Classification
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Comments
help: a proof about uniformly partiallly differentiaable
I faced a problem:
f:U>R is uniformly partiallly differentiaable at a , i.e.,
for any epsilon, there exists a delta so that:
[f(a+tu)f(a)]/tDfu(a)<=epsilon
for all u belongs to Rn and u=1 and all t with 0<t<delta.
show that f is diffentiable at a
*Dfu(a) is the patial dirivative of f in the direction of u.
I got a proof but I am not confident in it because this kind of basic concept problem need to be very accurate..but I am quite confused about some symbol and concept .