# Computing the appproximation numbers of an operator $T:\ell_2\longrightarrow\ell_2$

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)

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?

Take $$S=0$$ then $$\|T-S\|=\|T\|=1$$ and $$a_n(T)\le1$$.
Take $$S$$ with $$\dim \ S(l^2) and $$\|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$$.
• So, if $\|T\|=1$ then $a_n(T)=1$? Feb 3, 2021 at 13:13
• 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 $a_n(T)=1$.