$\|K\|=\max_{j}s_{j}(K)$ for Compact Operator

Question

Could anyone reference me to a proof of

$$\|K\|=\max_{j}s_{j}(K)$$

$K$ being a compact operator and $s_{j}(K)$ being the singular values of the operator.

I cannot find a proof in the literature.

Answer 1

Let $K:\mathcal{H}_1\to\mathcal{H}_2$ be a compact operators where $\mathcal{H}_j$ are Hilbert spaces. The singular decomposition of $K$ is given by \begin{align} K&=\sum_{j=0}^{+\infty}s_j(K)\,e_j\otimes f_j \end{align} where $\{e_j\vert j\in\mathbb{N}\}$ is the orthonormal system of $\mathcal{H}_1$ formed by the eigenvectors of the non-negative self-adjoint operator $K^*K$, $s_j(K)=\sqrt{\lambda_j(K^*K)}$ with $\lambda_j(K^*K)$ the eigenvalues of $K^*K$, and \begin{align} f_j&=\begin{cases}s_j(K)^{-1}Ke_j&\text{if }s_j(K)\neq0,\\ 0&\text{otherwise}.\end{cases} \end{align} You can check that $\langle f_j,f_k\rangle_{\mathcal{H}_2}=\delta_{j,k}$. Furthermore, \begin{align} K^*K&=\sum_{j=0}^{+\infty}s_j(K)^2\,e_j\otimes e_j.\tag{$\star$} \end{align}

Now we compute: \begin{align} \|K\|_{\mathcal{H}_1\to\mathcal{H}_2}=\sup_{\|u\|_{\mathcal{H}_1}=1}\|Ku\|_{\mathcal{H}_2}&=\sup_{\|u\|_{\mathcal{H}_1}=1}\sqrt{\langle Ku,Ku\rangle_{\mathcal{H}_2}}\\ &=\sup_{\|u\|_{\mathcal{H}_1}=1}\sqrt{\langle K^*Ku,u\rangle_{\mathcal{H}_1}}\\ &=\max\{s_j(K)\vert j\in\mathbb{N}\}. \end{align} We used the decomposition $(\star)$ for the last equality: \begin{align} \forall u=\sum_{j=0}^{+\infty}\langle e_j,u\rangle_{\mathcal{H}_1}e_j\in\mathcal{H}_1, \end{align} \begin{align} \langle K^*K u,u\rangle_{\mathcal{H}_1}&=\left\langle\sum_{j=0}^{+\infty}s_j(K)^2\,(e_j\otimes e_j)\sum_{k=0}^{+\infty}\langle e_k,u\rangle_{\mathcal{H}_1}e_k,\sum_{\ell=0}^{+\infty}\langle e_\ell,u\rangle_{\mathcal{H}_1}e_\ell\right\rangle_{\mathcal{H}_1}\\ % &=\sum_{j=0}^{+\infty}s_j(K)^2\overline{\sum_{k=0}^{+\infty}\langle e_k,u\rangle_{\mathcal{H}_1}}\sum_{\ell=0}^{+\infty}\langle e_\ell,u\rangle_{\mathcal{H}_1}\left\langle(e_j\otimes e_j)e_k,e_\ell\right\rangle_{\mathcal{H}_1}\\ % &=\sum_{j=0}^{+\infty}s_j(K)^2\overline{\sum_{k=0}^{+\infty}\langle e_k,u\rangle_{\mathcal{H}_1}}\sum_{\ell=0}^{+\infty}\langle e_\ell,u\rangle_{\mathcal{H}_1}\left\langle\delta_{j,k} e_j,e_\ell\right\rangle_{\mathcal{H}_1}\\ % &=\sum_{j=0}^{+\infty}s_j(K)^2\overline{\sum_{k=0}^{+\infty}\langle e_k,u\rangle_{\mathcal{H}_1}}\sum_{\ell=0}^{+\infty}\langle e_\ell,u\rangle_{\mathcal{H}_1}\delta_{j,k}\delta_{j,\ell}\\ % &=\sum_{j=0}^{+\infty}s_j(K)^2\left|\langle e_j,u\rangle_{\mathcal{H}_1}\right|^2\\ % &\leq\max\{s_j(K)\vert j\in\mathbb{N}\}^2\sum_{j=0}^{+\infty}\left|\langle e_j,u\rangle_{\mathcal{H}_1}\right|^2\\ % &=\max\{s_j(K)\vert j\in\mathbb{N}\}^2\|u\|^2. \end{align} It follows that $\|K\|_{\mathcal{H}_1\to\mathcal{H}_2}\leq\max\{s_j(K)\vert j\in\mathbb{N}\}$, with equality if (and only if) $u=e_{j_0}$ where $\max\{s_j(K)\vert j\in\mathbb{N}\}=s_{j_0}$ -- the maximum is attained as the eigenvalues of the compact operator $K$ remain in a compact neighborhood of $0$ in $\mathbb{C}$.