Proof. Write \(||L||=\sup _{x\neq 0}\dfrac {||L(x)||}{||x||}\) and check the three axioms. Throughout, the definition gives the working inequality \begin {equation} ||L(x)||\leq ||L||\,||x||\qquad \text {for all }x\in X, \tag {$\ast $} \end {equation} which is the form in which the operator norm is nearly always used.

Positivity and definiteness

Each quotient is non-negative, so \(||L||\geq 0\). If \(||L||=0\) then \((\ast )\) gives \(||L(x)||\leq 0\), hence \(L(x)=0\), for every \(x\); so \(L\) is the zero operator. Conversely the zero operator plainly has norm \(0\).

Homogeneity

For a scalar \(\alpha \), \[||\alpha L||=\sup _{x\neq 0}\frac {||\alpha L(x)||}{||x||} =\sup _{x\neq 0}\frac {|\alpha |\,||L(x)||}{||x||} =|\alpha |\sup _{x\neq 0}\frac {||L(x)||}{||x||}=|\alpha |\,||L||,\] the constant \(|\alpha |\geq 0\) passing through the supremum.

Triangle inequality

For any \(x\neq 0\), using the triangle inequality in \(Y\) and then \((\ast )\) for each operator, \[\frac {||(L+T)(x)||}{||x||}\leq \frac {||L(x)||+||T(x)||}{||x||} \leq ||L||+||T|| .\] The right-hand side does not depend on \(x\), so it is an upper bound for the set whose supremum is \(||L+T||\), giving \(||L+T||\leq ||L||+||T||\).

Only bounded operators are in \(\mathcal {L}(X,Y)\), which is what makes the supremum finite and the definition meaningful in the first place. □

Proof. Write \(Q=\sup _{x\neq 0}\dfrac {||L(x)||}{||x||}\) for the quantity defined to be \(||L||\). We show \(N=P=S=Q\) by circular inequalities.

\(Q=S\)

For \(x\neq 0\) put \(\hat {x}=x/||x||\), a unit vector. By homogeneity of \(L\) and of the norm, \[\frac {||L(x)||}{||x||}=\left \|L\!\left (\frac {x}{||x||}\right )\right \| =||L(\hat {x})|| ,\] so the set of quotients over all \(x\neq 0\) is exactly the set of values \(||L(u)||\) over unit vectors \(u\). The two suprema are therefore equal.

\(S\leq P\)

The unit sphere \(||x||=1\) is contained in the unit ball \(||x||\leq 1\), and a supremum over a larger set is at least as big.

\(P\leq Q\)

Every \(x\) with \(0<||x||\leq 1\) is in particular a non-zero vector, so its quotient is one of those over which \(Q\) is taken.

\(Q\leq N\) and \(N\leq Q\)

If \(||L(x)||\leq M||x||\) for all \(x\), then dividing by \(||x||\) for \(x\neq 0\) shows \(M\) is an upper bound for the quotients, so \(Q\leq M\); taking the infimum over such \(M\) gives \(Q\leq N\). Conversely \(||L(x)||\leq Q||x||\) holds for all \(x\) — for \(x\neq 0\) by definition of the supremum, and for \(x=0\) because both sides vanish — so \(Q\) is one of the admissible \(M\) and \(N\leq Q\).

Chaining, \(Q\leq N\leq Q\) and \(Q=S\leq P\leq Q\), so all four agree. □

Proof. \((1)\implies (2)\) Since \(X\neq \emptyset \), let \(x_0\in X\), then \(T\) is continuous at \(x_0\).

\((2)\implies (3)\) Let \(\varepsilon >0\) be given. Since \(T\) is continuous at \(x_0\), \(\exists \delta >0\ni \forall x\in X\), \[||x-x_0||<\delta \implies ||T(x)-T(x_0)||<\varepsilon \] Let \(x\in X\), \(||x-0|| =||x||=||x-x_0+x_0||=||(x+x_0)-x_0||\leq \delta \)

So that \begin {align*} ||T(x)-T(0)|| &=||T(x)|| =||T(x+x_0-x_0)||\\ &=||T(x+x_0)-T(x_0)||<\varepsilon \hspace {0.3cm} \text {by linearity} \end {align*}

\(\implies T\) is continuous at \(0\in X\).

\((3)\implies (4)\)

Since \(T\) is continuous at \(0\), \(\exists \delta >0\ni \forall x\in X\), \(||x||=||x-0||<\delta \)
\(\implies ||T(x)||=||T(x)-T(0)||<1\). Therefore \[\Bigg |\Bigg |\frac {\delta x}{2||x||}\Bigg |\Bigg |=\frac {\delta }{2}<\delta \]
\[\implies 1>\Bigg |\Bigg |T\Bigg (\frac {\delta (x)}{2||x||}\Bigg )\Bigg |\Bigg |=\frac {\delta ||T(x)||}{2||x||}\] so, \(\displaystyle {\frac {||T(x)||}{||x||}<\frac {2}{\delta }\implies ||T||\leq \frac {2}{\delta }<\infty }\)

\((4)\implies (1)\)

Let \(\varepsilon >0\) be given and choose \(\delta =\dfrac {\varepsilon }{||T||+1}\).

Let \(x,y\in X\ni ||x-y||<\delta \). Then \begin {align*} ||T(x)-T(y)|| &=||T(x-y)||\\ &=\frac {||T(x-y)||}{||x-y||}||x-y||\\ &\leq ||T||||x-y||\\ &\leq ||T||.\frac {\varepsilon }{||T||+1}\\ &<\varepsilon . \end {align*}

and so \(T\) is continuous at \(x\). Since \(x\) is arbitrary in \(X\), \(T\) is continuous on X. □

Proof. Let \(\{T_n\}^{\infty }_{n=1}\) be Cauchy in \(\mathcal {L}(X,Y)\), \(x\in X\), \(\varepsilon >0\) be given. \(\exists N\in \mathbb {N}\ni m,n>N\) \[\implies ||T_n-T_m||<\frac {\varepsilon }{||x||}\] so if \(m,n>N\), then \(||T_n(x)-T_m(x)||\leq ||T_n-T_m||||x||<\varepsilon \)
\(\implies \{T_n(x)\}^{\infty }_{n=1}\) is Cauchy in \(Y\), and since \(Y\) is complete, \(T_n(x)\rightarrow T(x)\in Y\). Since this limit is unique \(\forall x\in X\), it is a mapping from \(X\) to \(Y\).

To show linearity, let \(\alpha , \beta \in \mathbb {F}\) and \(x,y\in X\). Then \begin {align*} T(\alpha x+\beta y) &=\lim _{n\rightarrow \infty }T_n(\alpha x+\beta y)\\ &=\lim _{n\rightarrow \infty }(\alpha T_n(x)+\beta T_n(y))\\ &=\alpha \lim _{n\rightarrow \infty }T_n(x)+\beta \lim _{n\rightarrow \infty }T_n(y)\\ &=\alpha T(x)+\beta T(y) \end {align*}

Proving that \(T\) is linear.

Let \(\varepsilon >0\) be given. \(\exists N\in \mathbb {N}\ni m,n>N\implies ||T_m-T_n||<\varepsilon \). Then \(\forall x\in X\) \begin {align*} ||T(x)-T_m(x)|| &=\lim _{n\rightarrow \infty }||T_n(x)-T_m(x)||\\ &\leq \lim _{n\rightarrow \infty }||T_n-T_m||||x||\\ &<\varepsilon ||x|| \end {align*}

That is \(||T-T_m||<\varepsilon \). Thus \begin {align*} ||T|| &\leq ||T-T_m||+||T_m||\\ &<\varepsilon +||T_m||\\ &<\infty \hspace {0.5cm} \text {(since sum of finite is finite)}. \end {align*}

Showing that \(T\) is bounded.

Hence \(T\in \mathcal {L}(X,Y)\) and since \(T_n(x)\rightarrow T(x)\).

\(\mathcal {L}(X,Y)\) is complete. \(\implies \mathcal {L}(X,Y)\) is a Banach space, being a normed linear space. □