Proof It is enough to show that $(ii)$ and $(iii)$ are equivalent, since $(i)\Leftrightarrow (iii)$ follows from [Reference Contreras, Payá and Werner4, Theorem 1, proof of $(ii)\Rightarrow \left( ii i\right)$ ].

Assume that $(iii)$ holds and write $M=\overline {\operatorname *{\mathrm {span}}}M_T$ , recall that

$$ \begin{align*}M_T=\{x\in S_H: \|Tx\|=\|T\|=1\}.\end{align*} $$

Let $\{x_n\}$ in $S_H$ be a maximizing sequence for T, that is, $\|Tx_n\|\rightarrow 1$ . We can write

$$ \begin{align*}x_n = z_n+w_n,\end{align*} $$

where $z_n\in M$ and $w_n\in M^\perp $ with

(2.2) $$ \begin{align} 1=\|x_n\|^2=\|z_n\|^2+\|w_n\|^2. \end{align} $$

We also have $\langle P_Tz_n, P_Tw_n\rangle = 0$ . Therefore,

$$ \begin{align*}\|P_Tx_n\|^2=\|P_Tz_n\|^2+\|P_Tw_n\|^2.\end{align*} $$

Since $P_T$ is a positive operator and $z_n\in \overline {\operatorname *{\mathrm {span}}}M_T = \overline {\operatorname *{\mathrm {span}}}M_{P_T}$ , we have

(2.3) $$ \begin{align} P_Tz_n =z_n. \end{align} $$

Together with Equation (2.2) and the fact that $\|Tx\| =\|P_Tx\|$ for all $x\in H$ , we get

(2.4) $$ \begin{align} \lim \|P_Tw_n\|^2 & =1- \lim\|P_Tz_n\|^2\\\nonumber & = 1-\lim\|z_n\|^2\\\nonumber & =\|w_n\|^2. \end{align} $$

Now, if $\lim \|w_n\| \neq 0$ , then we have $\lim \frac { \|P_Tw_n\|^2}{\|w_n\|^2} = 1$ . Since $w_n\in M^\bot $ , we will then have $\|P_T|_{M^\bot }\|=1$ .

Observe that, $M^\bot =M_T^\bot $ . Therefore, we will have $\|P_T|_{M_T^\bot }\|=1$ . But, by the assumption, $\|P_T|_{M^\perp }\|<1$ . Therefore, $\lim \|w_n\| = 0$ . That is, $\lim \|x_n -z_n\| =0$ and $\|z_n\|\rightarrow 1$ .

Take the sequence $\{y_n\}$ to be $y_n =\frac {z_n}{\|z_n\|}$ . Then, by Equation (2.3), $y_n\in M_T$ . Also, since $\|z_n\|\rightarrow 1$ , we get

$$ \begin{align*}\lim\|x_n-y_n\|= \lim\|x_n-z_n\| = 0.\end{align*} $$

This implies $(ii)$ .

Conversely assume that $(ii)$ holds. If possible, assume that $\|P_T|_{{M_T}^\perp }\|=\|T\|.$ Then there exists a sequence $\{w_n\} $ in $S_{M_T^\perp }$ such that $\lim \|Tw_n\|=\lim \|P_Tw_n\| = 1$ . Since ${w_n\in M_T^\perp }$ , we have $d(w_n, M_T)\geq \|w_n\|=1$ for all n, which contradicts $(ii)$ .

Proof The proof of this lemma utilizes a similar idea employed in the proof of [Reference Ruess and Stegall24, Theorem 1].

Suppose $B\in {B}({X\times Y})$ with $\|B\|=1$ and the given condition holds. Assume that the norm of $B(X\times Y)$ is not strongly subdifferentiable at B. Then, by the definition of strong subdifferentiability, there exists $\varepsilon>0$ and bounded bilinear functionals $C_n$ on $X\times Y$ with $\|C_n\|< \frac {1}{n}$ such that

(2.5) $$ \begin{align} \left|\left\|B+C_n\right\|-\left\|B\right\|-\operatorname{Re}\phi\left(C_n\right)\right| \geq \varepsilon\left\|C_n\right\| \text { for all } \phi\in J_{{B}(X\times Y)^*}(B). \end{align} $$

Using the assumed condition, choose a number $\delta>0$ which satisfies the following.

If $\operatorname {Re} B(x, y)> 1-\delta $ for some $(x, y)\in S_X\times S_{Y}$ , then there exists $(x_0, y_0)\in S_X\times S_{Y}$ satisfying

1. (i) $B(x_0, y_0) =1$ ,

2. (ii) $\|x-x_0\|<\frac {\varepsilon }{4}$ , $\|y-y_0\|<\frac {\varepsilon }{4}.$

We can assume, w. l. o. g, that $\delta <\frac {\varepsilon }{2}$ . From Equation (2.5), we can find $C_n$ such that $\|C_n\|<\frac {\delta }{2+\delta }$ which satisfies the Inequality (2.5). Let $C = C_n$ . Choose an element $\left (x, y\right ) \in B_X \times B_{Y}$ such that

$$ \begin{align*}\left(B+C\right)\left(x, y\right)>\left\|B+C\right\|-\delta\left\|C\right\|.\end{align*} $$

Then we have:

$$ \begin{align*} \left\|B\right\| \geq \operatorname{Re} B\left(x, y\right) & =\left(B+C\right)\left(x, y\right)-\operatorname{Re} C\left(x, y\right) \geq\left(B+C\right)\left(x, y\right)-\left\|C\right\| \\ & \geq\left\|B+C\right\|-(1+\delta)\left\|C\right\|\\ &\geq\|B\|-(2+\delta)\left\|C\right\|\\ &> 1-\delta. \end{align*} $$

Hence, $\operatorname {Re} B\left (x, y\right )>1-\delta $ . By choice of the number $\delta $ , there exists $(x_0, y_0)\in S_X\times S_{Y}$ such that $B(x_0, y_0) =1$ and $\|x-x_0\|<\frac {\varepsilon }{4}$ and $\|y-y_0\|<\frac {\varepsilon }{4}$ .

Recall the isometric isomorphism $B(X\times Y) \cong (X \tilde {\otimes }_\pi Y)^*$ . Let c be the bounded linear functional on $X \tilde {\otimes }_\pi Y$ corresponding to the bilinear functional C. Then

(2.6) $$ \begin{align} \left|C\left(x, y\right)-C\left(x_0, y_0\right)\right| & =\left|c\left(x \otimes y-x_0 \otimes y_0\right)\right| \\\nonumber & \leq\left\|C\right\|\left\|x \otimes y-x_0 \otimes y_0\right\|_\pi\\\nonumber & \leq \left\|C\right\|\left\|(x-x_0)\otimes y + x_0\otimes (y-y_0)\right\|_\pi\\\nonumber & \leq \|C\| \frac{\varepsilon}{2}. \end{align} $$

We also have:

(2.7) $$ \begin{align} \left\|B+C\right\| \geq\operatorname{Re} \left(B+C\right)\left(x_0, y_0\right)=\left\|B\right\|+\operatorname{Re} C\left(x_0, y_0\right), \end{align} $$

From the Inequalities (2.6) and (2.7), we have

$$ \begin{align*}\begin{aligned} 0 & \leq\left\|B+C\right\|-\left\|B\right\|-\operatorname{Re} C\left(x_0, y_0\right) \\ & <\left(B+C\right)\left(x, y\right)+\delta\left\|C\right\|-\left\|B\right\|-\operatorname{Re} C\left(x_{0}, y_0\right) \\ & <\delta\left\|C\right\|+\operatorname{Re} C(x, y)-\operatorname{Re} C(x_0, y_0) \\ &\leq \delta\left\|C\right\|+ \left|C\left(x, y\right)-C\left(x_0, y_0\right)\right| \\ & < \left\|C\right\|(\delta+\frac{\varepsilon}{2})\\ &< \|C\|\varepsilon, \end{aligned} \end{align*} $$

which is a contradiction to the Inequality (2.5), since $(x_0, y_0)\in J_{{B}(X\times Y)^*}(B)$ . Therefore, the norm of $B(X\times Y)$ is strongly subdifferentiable at B.

Proof Assume that an operator $T\in S_{{B}(X, Y)}$ and the conditions $(i), (ii)$ hold. Since Y is uniformly smooth, we have the isometric isomorphism between the following Banach spaces.

$$ \begin{align*}{B}(X, Y)\cong{B}(X, Y^{**}) \cong(X\hat{\otimes}_\pi Y^*)^* \cong {B}(X\times Y^*).\end{align*} $$

So it is enough to prove that the norm of ${B}(X\times Y^*)$ is strongly subdifferentiable at T when viewed as a bounded bilinear functional on $(X\times Y^*)$ . Using Lemma 2.5, it is enough to prove the following property (P).

(P): For each $\varepsilon>0$ , there exists $\delta>0$ such that, whenever $(x, y^*)\in S_{X}\times S_{Y^*}$ satisfies $T(x, y^*)>1-\delta $ , there exists $x_0\otimes y_0^*\in J_{B(X\times Y^*)^*}(T)$ satisfying $\|x-x_0\|<{\varepsilon }$ and $\|y-y_0^*\|<{\varepsilon }$ .

To see this, let $\varepsilon> 0$ . Since Y is uniformly smooth, by Theorem 2.6, we have the following property.

(P1): We can find $\delta _1> 0$ such that whenever an element $(y, y^*)\in S_Y\times S_{Y^*}$ satisfies $|y^*(y)|> 1-\delta _1$ , there exists an element $y_0^*\in J_{Y^*}{(y)}$ with $\|y^*-y_0^*\|<\frac {\varepsilon }{2}$ .

From $(i)$ , we also have the following.

(P2): There exists $\delta _2>0$ such that whenever an element $x\in B_X$ satisfies $\|Tx\|>1-\delta _2$ , we have $d(x, M_T)<\frac {\delta _1}{2}$ .

Define

$$ \begin{align*}\delta = \min\left\{\frac{\delta_1}{2},\delta_2,\varepsilon\right\}.\end{align*} $$

Choose an element $(x, y^*)\in S_X\times S_{Y^*}$ such that $T(x, y^*)>1-\delta $ . Then

$$ \begin{align*}\|Tx\|\geq T(x, y^*)> 1-\delta.\end{align*} $$

Then, by property (P2), there exists $x_0\in M_T$ such that

$$ \begin{align*} \|x-x_0\|<\frac{\delta_1}{2}. \end{align*} $$

Therefore,

$$ \begin{align*} |y^*(Tx_0)|= |y^*(Tx_0-Tx+Tx)|> \operatorname{Re}y^*(Tx_0-Tx+Tx)>\frac{-\delta_1}{2}+1-\delta>1-\delta_1. \end{align*} $$

Since $Tx_0\in S_Y$ , by property (P1), there exists $y_0^*\in J_{Y^*}{(Tx_0)}$ such that $\|y^*-y_0^*\|<{\varepsilon }.$

Thus, we have $(x_0, y_0^*)\in J_{{B}(X\times Y^*)^*}(T)$ which satisfies the requirements of property (P) above. Therefore, the norm of ${B}(X\times Y^*) $ is strongly subdifferentiable at T.

Proof The implication $(ii)\Rightarrow (i)$ follows from Proposition 2.7.

We will now prove the implication $(i)\Rightarrow (ii)$ . For, it is enough to prove that, for each $\varepsilon>0$ , there exists $\delta>0$ such that

(2.8) $$ \begin{align} d(x, M_T)\leq \varepsilon \text{ whenever } \|T(x)\|>1-\delta \text{ for some } x\in B_{\ell_p}. \end{align} $$

To begin, assume that the norm of ${B}(\ell _p, \ell _q)$ is strongly subdifferentiable at an operator $T\in S_{{B}(\ell _p, \ell _q)}$ . Observe that, we have the following isometric identification,

$$ \begin{align*}{B}(\ell_p, \ell_q) \cong (\ell_p\hat{\otimes}_\pi \ell_q^*)^*.\end{align*} $$

under the map

(2.9) $$ \begin{align} {B}(\ell_p, \ell_q)\ni T \rightarrow \tilde{T}\in (\ell_p\hat{\otimes}_\pi \ell_q^*)^*, \end{align} $$

where the action of $\tilde {T}$ on $\ell _p\hat {\otimes }_\pi \ell _q^*$ is given by

$$ \begin{align*}\tilde{T}\left(\sum_{i=1}^nx_i\otimes y_i^*\right) = \sum_{i=1}^ny_i^*(Tx_i).\end{align*} $$

Therefore, the norm of the dual space $(\ell _p\hat {\otimes }_\pi \ell _q^*)^*$ is strongly subdifferentiable at $\tilde {T}$ .

Fix $\varepsilon>0$ and let $\delta _1>0$ be the modulus convexity of the space $\ell _p$ corresponding to $\varepsilon $ . Apply Remark 2.2 to obtain $\delta>0 $ such that

(2.10) $$ \begin{align} d(v,J_{\ell_p\hat{\otimes}_\pi \ell_q^*}(\tilde{T}))< \delta_1 \text{ whenever } \tilde{T}(v)>1-\delta \text{ for some } v\in B_{\ell_p\hat{\otimes}_\pi \ell_q^*}. \end{align} $$

We will now prove that T satisfies the requirements of (2.8) corresponding to the $\delta>0$ found in (2.10). For, let $x\in S_{\ell _p}$ such that $\|Tx\|>1-\delta $ . Then there exists a bounded linear functional $y^*\in S_{\ell _q^*}$ such that $ y^*(Tx) = \|Tx\|$ . But

$$ \begin{align*}\tilde{T}(x\otimes y^*) = y^*(Tx) = \|Tx\|>1-\delta.\end{align*} $$

Therefore, from the expression (2.10), it follows that there exists an element ${u\in J_{\ell _p\hat {\otimes }_\pi \ell _q^*}(\tilde {T})}$ such that

$$ \begin{align*}\|x\otimes y^*-u\|_\pi<\delta_1.\end{align*} $$

At this stage, we will try to get a representation of u. For, we will look at the set $J_{{B}(\ell _p, \ell _q)^*}(T)$ . It is clear that

$$ \begin{align*}u\in J_{\ell_p\hat{\otimes}_\pi \ell_q^*}(\tilde{T})\subseteq J_{{B}(\ell_p, \ell_q)^*}(T).\end{align*} $$

Observe that $J_{{B}(\ell _p, \ell _q)^*}(T)$ is a $w^*$ -closed extremal subset of the closed unit ball of $ {{B}(\ell _p, \ell _q)^*}$ , where a subset E of a convex set C is called extremal if $x,y\in E$ whenever $x,y\in C$ and $tx+(1-t)y\in E$ , $t\in (0,1)$ .

Hence, $J_{{B}(\ell _p, \ell _q)^*}(T)$ is the $w^*$ -closed convex hull of its extreme points. In particular,

$$ \begin{align*}u\in J_{{B}(\ell_p, \ell_q)^*}(T) = \overline{\operatorname*{\mathrm{conv}}}^{w^*}(\operatorname{ext}J_{{B}(\ell_p, \ell_q)^*}(T)).\end{align*} $$

That is,

$$ \begin{align*}u=w^*-\lim_{k \rightarrow \infty} \sum_n^{m_k} t_n^k \varphi_n^k,\end{align*} $$

where $ \sum _n^{m_k} t_n^k =1$ for each k, and $\varphi _n^k\in \operatorname {ext}J_{{B}(\ell _p, \ell _q)^*}(T)$ for all $n,k$ , consequently, $(\varphi _n^k)(T) =1$ .

Since $J_{{B}(\ell _p, \ell _q)^*}(T)$ is extremal, we have

(2.11) $$ \begin{align} \operatorname{ext}J_{{B}(\ell_p, \ell_q)^*}(T)\subseteq \operatorname{ext}B_{{B}(\ell_p, \ell_q)^*}. \end{align} $$

It is well known that ${K}(\ell _p, \ell _q)$ is an M-ideal in ${B}(\ell _p, \ell _q)$ [Reference Harmand, Werner and Werner14]. Therefore, by Lemma 2.9, we have

$$ \begin{align*}\operatorname{ext}B_{{B}(\ell_p, \ell_q)^*} = \operatorname{ext}B_{{K}(\ell_p, \ell_q)^\perp}\cup \operatorname{ext}B_{{K}(\ell_p, \ell_q)^*}.\end{align*} $$

From the elementary theory of projective tensor products, we know that

$$ \begin{align*}{K}(\ell_p, \ell_q)^* = \ell_p\hat{\otimes}_\pi\ell_q^*.\end{align*} $$

Hence

$$ \begin{align*}\operatorname{ext}B_{{B}(\ell_p, \ell_q)^*} = \operatorname{ext}B_{{K}(\ell_p, \ell_q)^\perp}\cup \operatorname{ext}B_{\ell_p\hat{\otimes}_\pi\ell_q^*}.\end{align*} $$

From Equation (2.11), we conclude that

$$ \begin{align*}\operatorname{ext}J_{{B}(\ell_p, \ell_q)^*}(T)\subset\operatorname{ext}B_{{K}(\ell_p, \ell_q)^\perp}\cup \operatorname{ext}B_{\ell_p\hat{\otimes}_\pi\ell_q^*}.\end{align*} $$

Therefore, either $\varphi _n^k\in \operatorname {ext}B_{{K}(\ell _p, \ell _q)^\perp }$ or $\varphi _n^k\in \operatorname {ext}B_{\ell _p\hat {\otimes }_\pi \ell _q^*}$ for each n and k. Define the set,

$$ \begin{align*}L^k=\{n \in \mathbb{N}: \varphi_n^k\in \operatorname{ext}B_{{K}(\ell_p, \ell_q)^\perp}\}.\end{align*} $$

Define

$$ \begin{align*}z_n^k =\sum_{n\in L^k}t_n^k\varphi_n^k.\end{align*} $$

Then, for each k, the element $z_n^k\in {K}(\ell _p, \ell _q)^\perp $ . Since the net $\{z_n^k\}$ is bounded, passing onto a subnet if necessary, we assume that $\{z_n^k\}$ converges ( $w^*$ -topology) to some x as k goes to $\infty $ . Thus, $x\in {K}(\ell _p, \ell _q)^\perp $ , since ${K}(\ell _p, \ell _q)^\perp $ is $w^*$ -closed.

We also have,

$$ \begin{align*}\lim_{k\rightarrow\infty}z_n^k = u-\lim_{k\rightarrow\infty}\sum_{n\notin L^k}t_n^k\varphi_n^k,\end{align*} $$

where the above limit is taken in the $w^*$ -topology on ${B}(\ell _p, \ell _q)^*$ . But then, since the elements u and $\sum _{n\notin L^k}t_n^k\varphi _n^k$ are in ${\ell _p\hat {\otimes }_\pi \ell _q^*}$ , we must have $x\in {\ell _p\hat {\otimes }_\pi \ell _q^*}$ . Therefore, $x=0$ . That is, we have

(2.12) $$ \begin{align} w^*-\lim_{k\rightarrow\infty}\sum_{n\in L^k}t_n^k\varphi_n^k = 0, \text{ and } w^*-\lim_{k\rightarrow\infty}\sum_{n\notin L^k}t_n^k\varphi_n^k = u. \end{align} $$

For each k, let $t^k = \sum _{n\in L^k}t_n^k$ . Then the net $\{t^k\}$ in $\mathbb {R}$ is bounded, passing onto a subnet if necessary, we assume that $\{t^k\}$ converges to some $t\in \mathbb {R}$ as k goes to $\infty $ .

Suppose, $t\neq 0$ . Then we can find $\varepsilon _1>0$ such that $t^k>\varepsilon _1$ for all k. This would imply,

$$ \begin{align*}\sum_{n\notin L^k}t_n^k=1-t^k<1-\varepsilon_1.\end{align*} $$

But then, since the norm is $w^*$ -lower semi-continuous,

$$ \begin{align*}\|u\|\leq \liminf_{k\rightarrow\infty}\left\|\sum_{n\notin L^k}t_n^k\varphi_n^k\right\|\leq \sum_{n\notin L^k}t_n^k <1-\varepsilon_1<1.\end{align*} $$

Which is not possible, since $\|u\|=1$ . Therefore, $t= 0$ and $L^k \subsetneq \{1,2, \ldots , m_k\}$ for all but finitely many k. Therefore,

$$ \begin{align*}\operatorname{ext}B_{\ell_p\hat{\otimes}_\pi\ell_q^*}\cap J_{{B}(\ell_p, \ell_q)^*}(T)\neq \emptyset.\end{align*} $$

From Theorem 2.10, we know that

$$ \begin{align*}\operatorname{ext}B_{(\ell_p\hat{\otimes}_\pi \ell_q^*)} =\operatorname{ext}B_{{K}(\ell_p, \ell_q)^*} = \{x\otimes y: x\in S_{\ell_p}, y\in S_{\ell_q^*}\}.\end{align*} $$

Now, for each $n\notin L^k$ , we have $\varphi _n^k\in \operatorname {ext}B_{\ell _p\hat {\otimes }_\pi \ell _q^*}$ and hence

$$ \begin{align*}\varphi_n^k= u_n^k \otimes v_n^{*k}, \text{ where } u_n^k\in S_{\ell_p}, v_n^{*k}\in S_{\ell_q^*}.\end{align*} $$

Consequently, by Equation (2.12),

(2.13) $$ \begin{align} u=w^*-\lim_{k \rightarrow \infty} \sum_{n\notin L^k} t_n^k u_n^k \otimes v_n^{*k}. \end{align} $$

Now, fix $\varphi _{n_0}^{k_0}\in \operatorname {ext}B_{\ell _p\hat {\otimes }_\pi \ell _q^*}\cap J_{{B}(\ell _p, \ell _q)^*}(T)\neq \emptyset .$ Since $(\varphi _{n_0}^{k_0})(T) =1$ , we have $v_{n_0}^{*k_0}(T u_{n_0}^{k_0}) =1$ . Hence $\|Tu_{n_0}^{k_0}\|=1$ and thus $M_T\neq \emptyset $ .

Let elements $x_0^*\in S_{{\ell _p}^*}$ and $y_0\in S_{\ell _q}$ be such that $x_0^*(x) =1$ and $y^*(y_0)=1$ , recall from the beginning that $y^*(Tx)=1$ . Then $x_0^*\otimes y_0$ defines a compact operator from $\ell _p$ to $\ell _q$ or in other words $x_0^*\otimes y_0\in \ell _{p}^*\check {\otimes }_\varepsilon \ell _q$ . Therefore, we can find k such that

$$ \begin{align*}\left|x_0^*\otimes y_0\left(u-\sum_{n\notin L^k} t_n^k u_n^k \otimes v_n^{*k} \right)\right|<{\delta_1-\|x\otimes y^*-u\|_\pi}.\end{align*} $$

Thus for the above k, we have

(2.14) $$ \begin{align} \left|x_0^*\otimes y_0\left(x\otimes y^*-\sum_{n\notin L^k} t_n^k u_n^k \otimes v_n^{*k}\right)\right|<{\delta_1}. \end{align} $$

It remains to prove that $d(x, M_T)<\varepsilon $ . If possible, suppose that $d(x, M_T)>\varepsilon $ .

Recall that the elements $x_0^*\in S_{\ell _p^*}$ and $y_0\in S_{\ell _q}$ satisfies $x_0^*(x) =1$ and $y^*(y_0)=1$ . Then, by Remark 2.8, we get

$$ \begin{align*}\operatorname{diam}S(x_0^*, \delta_1, B_{\ell_p})<\varepsilon.\end{align*} $$

Since we are assuming that $d(x, M_T)>\varepsilon $ and $x\in S(x_0^*, \delta _1, B_{\ell _p})$ , we can conclude that

$$ \begin{align*}y\notin S(x_0^*, \delta_1, B_X) \quad\text{ for all } y\in M_T.\end{align*} $$

Therefore, by the definition of the slice.

$$ \begin{align*}\operatorname{Re}x_0^*(y)<1-\delta_1 \quad\text{ for all }y\in M_T.\end{align*} $$

Since $M_T = -M_T$ , we have

(2.15) $$ \begin{align} |\operatorname{Re}x_0^*(y)|<1-\delta_1 \quad\text{ for all } y\in M_T. \end{align} $$

We can assume, w. l. o. g., the elements $u_n^k$ in Equation (2.13) satisfies $x_0^*(u_n^k)\in \mathbb {R}$ . To see this, note that, if $x_0^*(u_n^k) = r_n^ke^{i\theta _n^k}$ , where $0\leq r_n^k\leq 1$ and $0\leq \theta _n^k< 2 \pi $ , then take $\varphi _n^k = e^{-i\theta _n^k}u_n^k\otimes e^{i\theta _n^k}v_n^{*k}$ . Also, note that $\sum _{n\notin L^k} t_n^k\leq 1$ .

So, for the same k above, since the elements $u_n^k$ are in $M_T$ for all $n\notin L^k$ ,

$$ \begin{align*} |\operatorname{Re}x_0^*\otimes y_0(u_n^k \otimes v_n^{*k})| & = |\operatorname{Re}x_0^*(u_n^k)v_n^{*k}(y_0)|\\ & < 1-\delta_1. \end{align*} $$

Therefore,

$$ \begin{align*} \left|\operatorname{Re}x_0^*\otimes y_0\left(x\otimes y^*-\sum_{n\notin L^k} t_n^k u_n^k \otimes v_n^{*k}\right)\right| &=\left|\operatorname{Re}x_0^*(x)y^*(y_0)- \sum_{n\notin L^k} t_n^k \operatorname{Re} x_0^*\otimes y_0(u_n^k \otimes v_n^{*k})\right|\\ &\geq 1-\left| \sum_{n\notin L^k} t_n^k\operatorname{Re}x_0^*(u_n^k)v_n^{*k}(y_0)\right|\\ &\geq 1-\sum_{n\notin L^k}t_n^k\left|\operatorname{Re}x_0^*(u_n^k)v_n^{*k}(y_0)\right|\\ &\geq 1-\sum_{n\notin L^k}t_n^k(1-\delta_1)\\ &\geq\delta_1. \end{align*} $$

Which is a contradiction to the Inequality (2.14). Hence $d(x, M_T)<\varepsilon $ .

Proof Suppose the norm of ${B}(X, Y^{**})$ is strongly subdifferentiable at T. We have $(X\hat {\otimes }_\pi Y^*)^* \cong {B}(X, Y^{**})$ . Hence the norm of $(X\hat {\otimes }_\pi Y^*)^*$ is strongly subdifferentiable at $\tilde {T}$ (see Expression (2.9) in Theorem 2.11). Let $\varepsilon>0$ and $\delta _1$ be the modulus convexity of X corresponding to $\varepsilon $ . Using Theorem 2.1, choose $\delta>0 $ such that whenever an element $u\in B_{X\hat {\otimes }_\pi Y^*}$ satisfies $\tilde {T}(u)>1-\delta $ , we have $d(u,J_{X\hat {\otimes }_\pi Y^*}(T))< \delta _1$ .

Let $x\in S_X$ such that $1-\delta <\|Tx\|$ . Then there exists a bounded linear functional $y^*\in S_{Y^*}$ such that $ y^*(Tx) = \|Tx\|$ .

Since the element $x\otimes y^*\in B_{X\hat {\otimes }_\pi Y^*} $ satisfies $\tilde {T}(x\otimes y^*)>1-\delta $ , there exists ${u\in J_{X\hat {\otimes }_\pi Y^*}(T)}$ such that $\|x\otimes y^*-u\|_\pi <\delta _1$ .

By our assumption, the pair has $(X, Y^*)$ has the property $(N)$ . Therefore, we can write

$$ \begin{align*} u =\sum_{i=1}^\infty\lambda_ix_i\otimes y_i^*, \text{ where } \sum_{i=1}^\infty\lambda_i =1, \quad x_i\in S_X,\text{ and } y_i^*\in S_{Y^*}. \end{align*} $$

Since $\tilde {T}(u) = 1$ , we get $\tilde {T}(x_i\otimes y_i^*) = 1$ . Consequently, $\|T(x_i)\|= \|y_i^*\|= 1$ for all i and thus $M_T\neq \emptyset $ . Moreover,

(2.16) $$ \begin{align} \left\|x\otimes y^*-\sum_{i=1}^\infty\lambda_ix_i\otimes y_i^*\right\|_{X\hat{\otimes}_\pi Y^*}<\delta_1. \end{align} $$

We will now prove that $d(x, M_T)<\varepsilon $ . If possible, suppose that $d(x, M_T)>\varepsilon $ . Let the element $x_0^*\in S_{X^*}$ be such that $x_0^*(x) = 1$ . An argument, as in the proof of Theorem 2.11, using the slice of $B_X$ corresponding to the above $\delta>0$ and $x_0^*$ will obtain that

(2.17) $$ \begin{align} \left\|x_0^*(x)y^*-\sum_{i=1}^\infty\lambda_{i}\operatorname{Re}x_0^*(x_i)y_i^*\right\|_{Y^*}> \delta_1. \end{align} $$

Since the projective tensor norm is larger than the injective tensor norm, we get, using Inequality (2.17)

$$ \begin{align*} \left\|x\otimes y^*-\sum_{i=1}^\infty\lambda_ix_i\otimes y_i^*\right\|_{X\hat{\otimes}_\pi Y^*}&\geq \left\|x\otimes y^*-\sum_{i=1}^\infty\lambda_ix_i\otimes y_i^*\right\|_{X\check{\otimes}_\varepsilon Y^*}\\ &=\left\| x\otimes y^*-\sum_{i=1}^\infty\lambda_ix_i\otimes y_i^*\right\|_{{B}(X^*, Y^*)}\\ &\geq \left\|x_0^*(x)y^*-\sum_{i=1}^\infty\lambda_i\operatorname{Re}x_0^*(x_i)y_i^*\right\|_{Y^*} \\ &> \delta_1. \end{align*} $$

Which is a contradiction to Inequality (2.16). Hence $d(x, M_T)<\varepsilon $ .

Proof We have ${B}({X, \ell _\infty }) = {B}(X, c_0^{**})$ . Since the pair $(X, c_0^*)=(X, \ell _1)$ has property $(N)$ , the result follows from Theorem 2.14.

Proof We have $B(\ell _1) = \left (\bigoplus _{n=1}^\infty \ell _1\right )_{\infty }$ under the isometric identification $T \mapsto (Te_1, Te_2, \ldots )$ with $\|T\| =\sup _j\{\|Te_j\|\}$ .

Therefore, T is an $SSD$ point of $B(\ell _1)$ if and only if $(Te_1, Te_2, \ldots )$ is an $SSD$ point of $\left (\bigoplus _{n=1}^\infty \ell _1\right )_{\infty }$ .

From [Reference Franchetti and Payá10, Theorem 2.5], the element $(Te_1, Te_2, \ldots )$ is an $SSD$ point of $\left (\bigoplus _{n=1}^\infty \ell _1\right )_{\infty }$ if and only if $Te_j$ is an $SSD$ point of $\ell _1$ for all j satisfying $\|Te_j\| = 1$ and the element $(\|Te_j\|)\in \ell _\infty $ is an $SSD$ point of $\ell _\infty $ .

Since $\ell _1 =c_0^*$ , it can be easily verified using Theorem 2.1 that $Te_j$ is an $SSD$ point of $\ell _1$ if and only if $Te_j\in c_{00}$ . Since $\ell _\infty =\left (\oplus _{n=1}^\infty \mathbb {F}\right )_\infty $ , using [Reference Franchetti and Payá10, Theorem 2.5] again, we get that $(\|Te_j\|)\in \ell _\infty $ is an $SSD$ point of $\ell _\infty $ if and only if $\{j\in \mathbb {N} : \|Te_j\| =1\}\neq \emptyset $ and $\sup \{\|Te_j\| : \|Te_j\| <1\}<1.$

Proof For $q< p$ , we have ${K}(X, Y) = {B}(X, Y)$ [Reference Harmand, Werner and Werner14, Chapter VI, Corollary 5.14]. Therefore, the results hold since compact operators are weak-to-norm sequentially continuous.

Therefore, we can assume that $1<p\leq q<\infty $ . Let $T\in {B}(X, Y)$ , we can take $\|T\|=1$ . Let the sequence $\{x_n\}$ be a maximizing sequence for T and $\{x_n\}$ converges weakly to $x\neq 0$ (we are denoting subsequence of $\{x_n\}$ by $\{x_n\}$ itself). Since X is $(M_p)$ and Y is $(M_q)$ , it follows from Equation (3.2) that

(3.3) $$ \begin{align} 1-\|x\|^p =\lim\|x_n-x\|^p \end{align} $$

and

(3.4) $$ \begin{align} 1=\lim\|Tx_n\|^q =\|Tx\|^q+\lim\|Tx_n-Tx\|^q. \end{align} $$

Note that $\|Tx\|\leq 1$ and $\lim \|Tx_n-Tx\|\leq 1$ , consequently, since $q\geq p\geq 1$ ,

$$ \begin{align*} 1=\lim\left\|Tx_n\right\|^{q} &=\|Tx\|^q+\lim\|Tx_n-Tx\|^q\\ &\leq \|Tx\|^p+\lim\|Tx_n-Tx\|^p\\ &\leq \|Tx\|^p+\lim\|x_n-x\|^p\\ &= \|Tx\|^p+ 1-\|x\|^p. \end{align*} $$

Therefore,

(3.5) $$ \begin{align} \|Tx\|= \|x\|. \end{align} $$

Since $\|x\|\leq 1$ , using Equations (3.3) and (3.4) together with the fact that $q\geq p> 1$ , we get

(3.6) $$ \begin{align} 1-\|x\|^q\geq 1-\|x\|^p. \end{align} $$

and

(3.7) $$ \begin{align} \lim \|x_n-x\|^q = \lim (\|x_n-x\|^p)^{q/p} = (1-\|x\|^p)^{q/p}\leq 1-\|x\|^p. \end{align} $$

If possible, assume that the sequence $\{\|x_n-x\|\}$ does not converge to $0$ . Then there exists $\varepsilon>0$ and a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $\|x_{n_k}-x\|>\varepsilon $ for all k. Then, by Equations (3.6) and (3.7),

$$ \begin{align*} \lim \left\|T\left(\frac{x_{n_k}-x}{\|x_{n_k}-x\|}\right)\right\|^q & =\frac{1-\|Tx\|^q}{\lim\|x_{n_k}-x\|^q}\\ &=\frac{1-\|x\|^q}{\lim\|x_{n_k}-x\|^q}\\ &\geq\frac{1-\|x\|^p}{1-\|x\|^p}\\ &=1. \end{align*} $$

Since the sequence $\left \{\frac {x_{n_k}-x}{\|x_{n_k}-x\|}\right \}$ converges weakly to $0$ , by Lemma 3.3,

$$ \begin{align*} \|T\|_e\geq \lim \left\|T\left(\frac{x_{n_k}-x}{\|x_{n_k}-x\|}\right)\right\| \geq 1 =\|T\|. \end{align*} $$

Which is not possible by our assumption. Therefore, the sequence $(\|x_n-x\|)$ converges to $0$ . Hence, by Equations (3.5) and (3.3), $\|Tx\| =\|x\|=1 =\|T\|$ .

Proof Proof of Theorem 3.1

Let $T \in {B}(\ell _p, \ell _q) $ such that $\|T\|_e<\|T\|$ . To see that the operator $T\in \operatorname {SSD}\{{B}(\ell _p, \ell _q)\}$ , it is enough to verify, by Theorem 2.11, $\liminf d(x_n, M_T) = 0$ for every maximizing sequence $\{x_n\}$ of T.

Let the sequence $\{x_n\}$ be a maximizing sequence for T, then there exists a subsequence of $\{x_n\}$ (we will denote the subsequence by $\{x_n\}$ itself) converges weakly to some $x\in \ell _p$ by the reflexivity of $\ell _p$ .

If $x=0$ , then by Lemma 3.3, $\|T\|=\|T\|_e$ , which is not possible.

Therefore, $x\neq 0$ , by Lemma 3.4, $\|Tx\| =\|T\|$ , and $\|x\|=1$ . Thus $x\in M_T$ .

Since the sequence $\{x_n\}$ is a maximizing sequence for T, from the proof of Lemma 3.4, the sequence $\{\|x_n-x\|\}$ converges to $0$ . Thus $d(x_n, M_T)\rightarrow 0$ . Hence, by Theorem 2.11, the operator $T\in \operatorname {SSD}\{{B}(\ell _p, \ell _q)\}$ .

The denseness of the set $\operatorname {SSD}\{{B}(\ell _p, \ell _q)\}$ is now follows from Lemma 3.2.

Proof The proof follows directly from the inclusion, $\left \{T \in {B}(\ell _p, \ell _q): \|T\|_e<\|T\|\right \}\subseteq \operatorname {SSD}\{{B}(\ell _p, \ell _q)\}.$

Proof Fréchet differentiability implies Gateaux differentiability (smoothness). Hence, by Theorem 3.8, $(i)\Rightarrow (ii)$ .

Conversely, if $(ii)$ holds, then by Theorem 3.8, T is smooth. Also, by Theorem 3.1, T is an $SSD$ point of ${B}(\ell _p, \ell _q)$ . Since Gateaux differentiability together with strong subdifferentiability implies Fréchet differentiability, the norm of ${B}(\ell _p, \ell _q)$ is Fréchet differentiable at T.