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How about this: define $T : \ell^1 \to \ell^1$ by $$T(a_1, a_2, a_3, \cdots, a_n, \cdots) = (0, (1-1/2)a_2, (1-1/3)a_3, \cdots, (1-1/n)a_n, \cdots)$$ Then $$||T|| = \sup_{||x|| = 1} ||Tx|| = \sup_n (1-1/n) = 1$$ However, for all non-zero $x \in \ell^1$, $||Tx|| < ||x|| = ||T|| \cdot ||x||$.


Let $f(x):= x^{-1/p} \mathbf{1}_{(1,\infty)}$. Then $f^q$ is integrable because $\frac{q}{p} > 1$. But $f^p = \frac{1}{x}$ (on $(1,\infty)$) is not. That is, $f \in L^q(\mathbb{R})$ but $f \notin L^p(\mathbb{R}) $.


Hint: Try to prove the following result $\textbf{Result}:$ If for a sequence $(x_n)$ in a Banach space $X$, the sequence $(f(x_n))$ is bounded for all $f\in X^*$, then the sequence $(\|x_n\|)$ is bounded. Now suppose that you have proved the result. Suppose $(T_n)$ converges to $T$ in weak operator topology, i.e., \begin{align*} |f(T_nx)-f(Tx)| ...


"Naturally identified" leaves enough wiggle room to identify an infinite subset $X$ of $\mathbb N$ with the sequence in $\mathcal N$ that enumerates the elements of $X$ in increasing order. The set of all these increasing sequnces is a closed subset of $\mathcal N$.


Let $U$ be a subspace of a Banach space $V$. Suppose $x\in U$ is an interior point of $U$, which means that $$ B(x,\varepsilon)=\{y:\|x-y\|<\varepsilon\}\subseteq U $$ for some $\varepsilon>0$. Since $U$ is a subspace, also $$ B(0,\varepsilon)=-x+B(x,\varepsilon) $$ is contained in $U$. Now, for every $v\in V$, $v\ne0$, the vector $$ ...


This might seem redundant but also seems correct to me. If we take the sequence $(S_n)$ to be the constant sequence of $\textbf{Identity operators}$ then this sequence converges strongly to the Identity operator. Since $(T_n)$ converges to $T$ weakly and in this case the sequence $T_nS_n$ is nothing but the sequence $(T_n)$, which clearly fails to converge ...

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