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The fact that each $T\in B(X,Y)$ is weakly compact does not imply $X$ or $Y$ reflexive. For example, every non-weakly compact operator $T:\ell_\infty\to Y$ is an isomorphism on a subspace isomorphic t …

If $X$ and $Y$ are Banach spaces, $T:D(T)\subset X\to Y$ is a closed operator and there exists a closed subspace $N$ of $Y$ such that $\operatorname{Im}(T)\cap N=\{0\}$ and $\operatorname{Im}(T)\oplus …

The answer is yes. Indeed, let us denote $Y_h=\ell_\infty(B_{Y^*})$. For every $L\in\mathcal{K}(X,Y_h)$, $\|T\|_m=\|JT\|_m=\|JT-L\|_m\leq \|JT-L\|$. Hence $\|T\|_m\leq \|JT\|_e$. Conversely, since …

If $E$ is a Banach space and $T:E\to E$ is an injective bounded operator with closed range $R(T)$, then there exists a number $\delta>0$ such that any bounded operator $S:E\to E$ with $\|S-T\|<\delta$ …

The operator $A$ on $H^q$ is similar to $C^*$ for $C$ acting on $H^p$. Let us consider the functions $e_n(t)= e^{i2\pi nt}$ ($n\in\mathbb{N}\cup \{0\}$). For $1<p<\infty$ and $1/p+1/q=1$, the map tha …

An operator $T:X\to Y$ is Banach-Saks if for every bounded sequence $(x_n)$ in $X$ there is a subsequence $(Tx_{n_k})$ such that the Cesàro means $N^{-1}(\sum_{k=1}^N Tx_k)$ from a norm-convergent sub …

Let $\mathcal{K}$, $\mathcal{W}$ and $\mathcal{C}$ denote the compact, weakly compact and completely continuous operators, respectively, and let $\mathcal{W}^{-1}\circ \mathcal{K}$ denote the operator …

[J. Lindenstrauss and L. Tzafriri. Classical Banach spaces I. Sequence spaces. Springer 1977]. In page 76, after Prop. 2.c.3, it says that the proof of 2.c.3 shows that an operator $T:\ell_p\to\ell_p$ …

If $S(x_1,x_2,x_3,\ldots)=(0,x_1,x_2,\ldots)$ is the unilateral shift, it is easy to see that $S-\lambda I$ is bounded below for $|\lambda|<1$: $\|(S-\lambda I)x\| \geq \|Sx\|-\|\lambda x\|= (1-|\lamb …

First take $x^*\in A(\mathbb{D})^*$ that is non-norm-attaining. Next pick a non-zero $g\in A(\mathbb{D})$, and define $T: A(\mathbb{D})\to A(\mathbb{D})$ by $Tf = x^*(f)\cdot g$. Thus $\|T\|=\|x^*\|\ …

There is a Banach algebra version of the result (including the Banach space version) in Theorem 10.30 (page 264) of W. Rudin's book Functional analysis, 2nd ed. McGraw-Hill 1991.

For $\lambda\neq 0$, it is not difficult to show that $A^*A-\lambda I$ bijective if and only if so is $AA^*-\lambda I$. Moreover the kernels of $A^*A-\lambda I$ and $AA^*-\lambda I$ have the same dime …

A maybe different proof can be obtained using techniques related with the single-valued extension property (SVEP for short) of an operator. A good reference for this topic is the book [Aiena]: Pietro …

Let $V$ be the Volterra operator, $(Vf)(t)=\int_0^t f(s) ds$, acting on the Hilbert space $L_2(0,1)$, and let us denote $A=(I+V)^{-1}$. Then $\|A\|=1$ and $\sigma(A)=\{1\}$ [Halmos, A Hilbert space …

The answer is no: you can even have $T_1=T_3=0$ and $T_2$ equal to the identity $id$ on an infinite dimensional Banach space. Indeed, consider the following commutative diagram with exact rows: $$\ …