[proofplan] We show that $\sigma(T) \subseteq [m, M] \subset \mathbb{R}$ by proving that every $\lambda \notin [m, M]$ (including non-real $\lambda$) lies in the resolvent set. The argument establishes a lower bound $\|(T - \lambda I)x\|_H \geq \delta \|x\|_H$ for such $\lambda$, and uses [Kernel-Range Duality](/theorems/551) to promote injectivity plus closed range to bijectivity. To show $m, M \in \sigma(T)$, we construct approximate eigenvectors using the positive semidefinite estimate $\|Sx\|_H^2 \leq \|S\|_{\mathcal{L}(H)} (Sx, x)_H$. [/proofplan] [step:Show that non-real $\lambda$ lies in the resolvent set] Let $\lambda = \alpha + i\beta$ with $\beta \neq 0$. Since $T$ is self-adjoint, $T - \alpha I$ is also self-adjoint, so $((T - \alpha I)x, x)_H \in \mathbb{R}$ for every $x \in H$. For $x \in H$: \begin{align*} \|(T - \lambda I)x\|_H^2 &= \|(T - \alpha I)x - i\beta x\|_H^2 \\ &= \|(T - \alpha I)x\|_H^2 + \beta^2 \|x\|_H^2 - 2\operatorname{Re}\bigl(i\beta ((T - \alpha I)x, x)_H\bigr). \end{align*} The cross term satisfies $\operatorname{Re}\bigl(i\beta ((T - \alpha I)x, x)_H\bigr) = 0$ because $((T - \alpha I)x, x)_H$ is real and $i\beta$ is purely imaginary. Therefore: \begin{align*} \|(T - \lambda I)x\|_H^2 = \|(T - \alpha I)x\|_H^2 + \beta^2 \|x\|_H^2 \geq \beta^2 \|x\|_H^2. \end{align*} This gives $\|(T - \lambda I)x\|_H \geq |\beta| \|x\|_H$ for all $x \in H$, so $T - \lambda I$ is bounded below with constant $|\beta| > 0$. A bounded-below operator is injective and has closed range. By [Kernel-Range Duality](/theorems/551), $\operatorname{Range}(T - \lambda I)$ is dense if and only if $\ker((T - \lambda I)^*) = \{0\}$. Since $(T - \lambda I)^* = T^* - \bar{\lambda}I = T - \bar{\lambda}I$ and $\bar{\lambda} = \alpha - i\beta$ also has nonzero imaginary part, the same lower bound argument shows $T - \bar{\lambda}I$ is injective. Hence $\operatorname{Range}(T - \lambda I)$ is dense and closed, therefore $T - \lambda I$ is surjective. Combined with injectivity and the bounded-below estimate, $(T - \lambda I)^{-1}$ exists and is bounded, so $\lambda \in \rho(T)$. [guided] Why does the cross term vanish? Write out the expansion: \begin{align*} \|(T - \alpha I)x - i\beta x\|_H^2 &= ((T - \alpha I)x - i\beta x,\, (T - \alpha I)x - i\beta x)_H \\ &= \|(T-\alpha I)x\|_H^2 - i\beta((T-\alpha I)x, x)_H + \overline{(-i\beta)}(x, (T-\alpha I)x)_H + \beta^2\|x\|_H^2. \end{align*} The two cross terms are $-i\beta((T-\alpha I)x, x)_H$ and $i\beta\overline{((T-\alpha I)x, x)_H}$. Since $T - \alpha I$ is self-adjoint, $((T-\alpha I)x, x)_H$ is real, so these two terms are $-i\beta c$ and $i\beta c$ for $c \in \mathbb{R}$. Their sum is zero. For the surjectivity argument: the key input from [Kernel-Range Duality](/theorems/551) is that $\overline{\operatorname{Range}(T - \lambda I)} = \ker((T - \lambda I)^*)^\perp$. Since $(T - \lambda I)^* = T - \bar{\lambda}I$ is injective (its kernel is trivial by the same bounded-below estimate applied to $\bar{\lambda}$), we get $\ker((T-\lambda I)^*)^\perp = \{0\}^\perp = H$. So $\operatorname{Range}(T - \lambda I)$ is dense. But it is also closed (bounded-below operators have closed range: if $T_0 x_n \to y$ and $\|T_0 x_n - T_0 x_m\| \geq \delta \|x_n - x_m\|$, then $(x_n)$ is Cauchy, hence convergent, and $T_0 x = y$). A dense closed subspace of $H$ is all of $H$. [/guided] [/step] [step:Show that real $\lambda \notin [m, M]$ lies in the resolvent set] Suppose $\lambda > M$ (the case $\lambda < m$ is analogous). For every $x \in H$ with $\|x\|_H = 1$: \begin{align*} ((\lambda I - T)x, x)_H = \lambda - (Tx, x)_H \geq \lambda - M =: \delta > 0. \end{align*} The operator $\lambda I - T$ is self-adjoint with $((\lambda I - T)x, x)_H \geq \delta \|x\|_H^2$ for all $x$. By the Cauchy--Schwarz inequality: \begin{align*} \delta \|x\|_H^2 \leq ((\lambda I - T)x, x)_H \leq \|(\lambda I - T)x\|_H \|x\|_H, \end{align*} so $\|(\lambda I - T)x\|_H \geq \delta \|x\|_H$. This gives a bounded-below estimate for $T - \lambda I$ with constant $\delta > 0$. Since $(T - \lambda I)^* = T - \lambda I$ is self-adjoint and also bounded below (with the same constant), it is injective. By [Kernel-Range Duality](/theorems/551), $\operatorname{Range}(T - \lambda I)$ is dense, and since $T - \lambda I$ is bounded below, the range is closed. Hence $T - \lambda I$ is bijective with bounded inverse, and $\lambda \in \rho(T)$. [/step] [step:Construct approximate eigenvectors to show $M \in \sigma(T)$] By definition of $M$ as a supremum, there exists a sequence $(x_n) \subset H$ with $\|x_n\|_H = 1$ and $(Tx_n, x_n)_H \to M$. Define $S := MI - T$. This operator is self-adjoint and positive semidefinite: $(Sx, x)_H = M - (Tx, x)_H \geq 0$ for all $x \in H$. [claim:Positive semidefinite norm bound] For any positive semidefinite self-adjoint operator $S \in \mathcal{L}(H)$ and any $x \in H$: \begin{align*} \|Sx\|_H^2 \leq \|S\|_{\mathcal{L}(H)} \cdot (Sx, x)_H. \end{align*} [/claim] [proof] Since $S$ is positive semidefinite, the Cauchy--Schwarz inequality for the semi-inner product $(Su, v)_H$ gives: \begin{align*} |(Sx, Sx)_H|^2 \leq (S(Sx), Sx)_H \cdot (Sx, x)_H. \end{align*} But we use a simpler route. We have $(S^2 x, x)_H = (Sx, Sx)_H = \|Sx\|_H^2$ (since $S$ is self-adjoint). For the self-adjoint operator $S$ with $S \geq 0$, the operator inequality $S^2 \leq \|S\|_{\mathcal{L}(H)} S$ holds, meaning $(S^2 x, x)_H \leq \|S\|_{\mathcal{L}(H)} (Sx, x)_H$ for all $x$. To verify: $\|S\| S - S^2 = S(\|S\|I - S)$. Since $0 \leq S \leq \|S\| I$ (the latter from $(Sx, x)_H \leq \|S\| \|x\|_H^2$), both $S$ and $\|S\|I - S$ are positive semidefinite. For commuting positive semidefinite operators, their product is positive semidefinite. Since $S$ commutes with $\|S\|I - S$, we have $S(\|S\|I - S) \geq 0$, giving $S^2 \leq \|S\| S$. Therefore $\|Sx\|_H^2 = (S^2 x, x)_H \leq \|S\|_{\mathcal{L}(H)} (Sx, x)_H$. [/proof] Applying the claim to $S = MI - T$ and $x = x_n$: \begin{align*} \|(MI - T)x_n\|_H^2 \leq \|MI - T\|_{\mathcal{L}(H)} \cdot ((MI - T)x_n, x_n)_H = \|MI - T\|_{\mathcal{L}(H)} \cdot (M - (Tx_n, x_n)_H). \end{align*} Since $(Tx_n, x_n)_H \to M$, the right side tends to $0$. Therefore $\|(T - MI)x_n\|_H \to 0$ with $\|x_n\|_H = 1$, so $T - MI$ is not bounded below and hence not invertible. This gives $M \in \sigma(T)$. The proof that $m \in \sigma(T)$ is analogous: define $S' := T - mI \geq 0$, choose $(x_n)$ with $(Tx_n, x_n)_H \to m$, and apply the same positive semidefinite estimate to show $\|(T - mI)x_n\|_H \to 0$. [/step] [step:Establish the operator norm identity $\|T\|_{\mathcal{L}(H)} = \max\{|m|, |M|\}$] From the previous steps, $\sigma(T) \subseteq [m, M]$, so the spectral radius satisfies $r(T) = \max\{|m|, |M|\}$. For [self-adjoint operators](/page/Self-Adjoint%20Operators), the spectral radius equals the operator norm: $r(T) = \|T\|_{\mathcal{L}(H)}$. This identity follows from the norm characterisation $\|T\|_{\mathcal{L}(H)} = \sup_{\|x\|_H = 1} |(Tx, x)_H| = \max\{|m|, |M|\}$, which holds because $T$ is self-adjoint and the quadratic form $(Tx, x)_H$ determines the norm. [/step]