I'm practicing for an operator theory exam, and doing exercises from previous exams. One of them is this, it is a test and it is assumed that there is only one correct solution.

Let $T:\ell_2\longrightarrow\ell_2$ be a linear bounded operator defined as $$T\Bigl(\sum_{n=1}^\infty\langle x,e_n\rangle e_n\Bigr)=\sum_{n=1}^\infty\langle x,e_{2n}\rangle e_n,$$ where $$e_n=(0,0,\ldots,\underbrace{1}_n,\ldots),\ \forall n\in\mathbb{N}.$$ Show which of the following statements is true:

(a) $T'=T$.

(b) $a_n(T)=1,\ \forall n\in\mathbb{N}$.

(c) $T\circ T(x)=\sum_{n=1}^\infty\bigl(\langle x,e_{2n}\rangle\bigr)^2e_n$.

In this case, $T'$ is the transpose operator of $T$ and $a_n$ is the $n$-th approximation number, defined as $$a_n(T):=\inf\bigl\{\|T-S\|:\ S\in\mathcal{L}(\ell_2),\ \dim S(\ell_2)<n\bigr\}.\tag{1}$$

If I'm not wrong, I can write, for simplicity $$T(x_n)=(x_2,x_4,\ldots,x_{2n},\ldots)$$ and calculate the transpose operator $$T'(y_n)=(0,y_1,0,y_2,\ldots,0,y_n,\ldots)$$ so the first option is false. The third is also trivially false, which leaves me as a good option the second one.

But how can I prove it directly? I am unable to apply the given definition $(1)$. I have an additional lemma from Diestel, Jarchow and Tonge, Absolutely Summing Operators, which says:

Let the compact operator $u:H_1\longrightarrow H_2$ be represented as $$u=\sum_{n=1}^\infty\tau_n\langle\cdot,e_n\rangle f_n,$$ where $(e_n)$ is an orthonormal sequence in $H_1$, $(f_n)$ is an orthonormal sequence in $H_2$, an $(\tau_n)$ is a null sequence of scalars which satisfy $$0\leq\tau_{n+1}\leq\tau_n,\ \forall n\in\mathbb{N}.\tag{2}$$ Then, $\tau_n=a_n(u)$, for all $n\in\mathbb{N}$.

My class notes relax this conditions a bit and do not require that $(\tau_n)$ be a null sequence, only that the condition $(2)$ be met.

Anyway, I don't think this can be applied in this case, because $T$ is not a compact operator.

Any idea how I can continue?


1 Answer 1


Take $S=0$ then $\|T-S\|=\|T\|=1$ and $a_n(T)\le1$.

Take $S$ with $\dim \ S(l^2)<n$ and $\|S-T\|= 1-\epsilon$ for $\epsilon\ge0$. Take $y\in R(S)^\perp$, $\|y\|=1$, which is possible as the range of $S$ is finite-dimensional. Then by definition of $T$ there is $x$ with $\|x\|=\|y\|=1$ and $Tx=y$. Then $$ (1-\epsilon)^2 \ge \|(S-T)x\|^2 = \|Sx\|^2+\|Tx\|^2 + 2Re(\langle Sx,Tx\rangle) =\|Sx\|^2 +1. $$ This implies $0\le\|Sx\|^2 \le \epsilon^2 -2\epsilon = \epsilon(1-2\epsilon)$, and $\epsilon\le\frac12$, so $a_n(T)\ge\frac12$. In addition, $$ 1=\|Tx\|\le \|Sx\| +\|(S-T)x\| \le \sqrt{\epsilon^2 -2\epsilon} + 1-\epsilon, $$ which implies $\epsilon \le \sqrt{\epsilon^2 -2\epsilon}$. And $\epsilon$ has to be zero, and $a_n(T)\ge1$.

  • $\begingroup$ So, if $\|T\|=1$ then $a_n(T)=1$? $\endgroup$ Feb 3, 2021 at 13:13
  • 1
    $\begingroup$ No. Only $a_n(T)\le 1$, as in the first part of my answer. The second part uses the particular structure of $T$ to conclude $a_n(T)=1$. $\endgroup$
    – daw
    Feb 3, 2021 at 13:20

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