# Tag Info

2

Consider $T_n f=f'(1/n)$. Since $f\in C^1$, $|T_n f|\leq \sup |f'(x)|<\infty$. On the other hand, $f_t(x)=e^{tx}$ is a $C^1$ function and $T_nf_t=te^{t/n}\rightarrow t$ as $n\rightarrow\infty$. Edit: I just noticed that you want the functionals to be bounded. In this case, consider instead $$T_n f=n(f(1/n)-f(0)).$$ We see that $T_nf\rightarrow f'(0)$, ...

2

Based on your statement of the problem, you don't assume that $X$ is closed. So I don't think it is clear on the front end that the limit of a sequence is an element of $X$. Instead here is a hint. Assume $(y_k)_{k=1}^\infty$ is a cauchy sequence. Express each $y_k = a_{1,k}e_1 + \cdots a_{n,k}e_n$ using your basis. Use the definition of cauchy to prove ...

2

Assuming that $S$ is hermitian we have that $$A=S+i(I-S^2)^{1/2}\,\,\Longrightarrow\,\,A^*=S-i(I-S^2)^{1/2},$$ and hence $$A^*A=\big(S-i(I-S^2)^{1/2}\big)\big(S+i(I-S^2)^{1/2}\big)=S^2+(1-S^2)=I,$$ as $S(I-S^2)^{1/2}=(I-S^2)^{1/2}S$, which means that $A$ is unitary.

1

Let $T_nx = \langle x_n , x \rangle$. We have $\sup_n |T_n x| < \infty$, hence by the Banach Steinhaus theorem we have $\sup_n \|T_n\| < \infty$. Since $\|T_n\| = \|x_n\|$, we have the desired result.

1

For any orthogonal projection $P$, a vector $x$ is in the range of $P$ iff $\|Px\|=\|x\|$. The condition $P \le Q$ for orthogonal projections is equivalent to $\|Px\|\le \|Qx\|$. Because $\|Qx\|\le \|x\|$ it follows that if $x \in\mathcal{R}(P)$, then $x\in\mathcal{R}(Q)$.

Only top voted, non community-wiki answers of a minimum length are eligible