8.2 Diagonal operators

For $\sigma = (\sigma _0, \dots , \sigma _{m-1}) \in \mathbb {K}^m$ let $D_\sigma $ denote the diagonal operator

\[ D_\sigma : (x_i)_i \longmapsto (\sigma _i \, x_i)_i . \]

Throughout this section, $\sigma $ is non-increasing in modulus, $|\sigma _0| \ge |\sigma _1| \ge \dots \ge |\sigma _{m-1}|$, and $n {\lt} m$.

Theorem 8.4 Same exponent

✓

Let $1 \le p \le \infty $ and $D_\sigma : \ell _p^m \to \ell _p^m$. Then $\| D_\sigma \| = \max _i |\sigma _i|$, and for every strict s-number sequence $s$ (in particular for $a_n$, $b_n$, $c_n$, $d_n$),

\[ s_n(D_\sigma ) \; =\; |\sigma _n| . \]

Proof ▼

The norm is attained at the coordinate carrying $\max _i |\sigma _i| = |\sigma _0|$. Upper bound: truncating $\sigma $ to its first $n$ entries gives a rank-$\le n$ approximant whose residual is the diagonal operator with entries $(\sigma _k)_{k \ge n}$, of norm $|\sigma _n|$; hence $s_n(D_\sigma ) \le a_n(D_\sigma ) \le |\sigma _n|$ for every s-number sequence. Lower bound for strict $s$: since $|\sigma _k| \ge |\sigma _n|$ for $k \le n$, the identity of $\ell _p^{n+1}$ factors as $\mathrm{id} = B \circ D_\sigma \circ A$ with the coordinate embedding $A$ (a contraction) and $B$ the coordinate projection followed by the diagonal operator with entries $\sigma _k^{-1}$, of norm $\le |\sigma _n|^{-1}$; then (S3) and the strict normalisation (Proposition 8.3) give $1 = s_n(\mathrm{id}) \le |\sigma _n|^{-1} s_n(D_\sigma )$.

The Hilbert numbers are not strict, and satisfy only $h_n(D_\sigma ) \le |\sigma _n|$; equality can fail for $p \neq 2$.

Theorem 8.5 Mixed exponents $p {\lt} q$

✓

Let $1 \le p {\lt} q {\lt} \infty $ and define $r$ by $\tfrac 1r = \tfrac 1p - \tfrac 1q$. For $D_\sigma : \ell _q^m \to \ell _p^m$,

\[ a_n(D_\sigma ) \; =\; \Bigl( \sum _{k = n}^{m-1} |\sigma _k|^r \Bigr)^{1/r} , \qquad \text{in particular}\qquad \| D_\sigma \| = \| \sigma \| _{\ell _r} . \]

Proof ▼

Upper bound: truncate to the first $n$ coordinates; the residual diagonal has norm equal to the $\ell _r$-norm of its entries, by Hölder’s inequality applied with the exponent pair $(q/r', \dots )$ and attained at the explicitly weighted vector $x_k = |\sigma _k|^{r/q}$. Lower bound: for a rank-$\le n$ operator $L$, a weighted flatness argument on $\ker L$ (the weighted analogue of the pigeonhole step in Theorem 8.2) produces a kernel vector on which $\mathrm{id} - L$ retains the full tail $\ell _r$-norm. The case $n = 0$, $L = 0$ yields the operator norm.

By the sandwich theorem (Theorem 4.2) the upper bound $s_n(D_\sigma ) \le \bigl(\sum _{k \ge n} |\sigma _k|^r\bigr)^{1/r}$ holds for every s-number sequence, and it needs no ordering of the diagonal.

Theorem 8.6 Reverse regime $q \le p$

✓

Let $1 \le q \le p \le \infty $ (including $p = \infty $). For $D_\sigma : \ell _q^m \to \ell _p^m$,

\[ \| D_\sigma \| = \max _i |\sigma _i| , \qquad s_n(D_\sigma ) \le |\sigma _n| \quad \text{for every s-number sequence $s$.} \]

Proof ▼

Since $q \le p$, the identity $\ell _q^m \to \ell _p^m$ is a contraction, and $D_\sigma $ factors as this contraction after the same-exponent diagonal operator on $\ell _q^m$; the ideal property (S3) transfers both bounds from Theorem 8.4. The lower bound for the norm again tests a single coordinate.

For $q {\lt} p$ the upper bound $s_n(D_\sigma ) \le |\sigma _n|$ need not be sharp: the exact values have no elementary closed form in this regime.