Bounded Operators and the Adjoint

A Hilbert space comes with a class of maps that respect its linear and metric structure, the bounded linear operators, and these maps form an algebra rich enough to carry a spectral theory. The single construction that makes the theory work is the adjoint, the operator that moves to the other side of an inner product, and it exists for every bounded operator because the Riesz representation theorem turns the functional it defines into a vector. This post builds the operator norm, the adjoint, and the self-adjoint operators the spectral theorem decomposes [@rudinFA1991; @reedSimon1980]. Throughout, $H$ is a real Hilbert space and an operator means a bounded linear map $T:H\to H$ .

#The operator norm

Definition1

A linear map $T:H\to H$ is bounded when $\norm T=\sup_{\norm x\le 1}\norm{Tx}$ is finite. The number $\norm T$ is the operator norm, and it is the smallest constant with $\norm{Tx}\le\norm T\,\norm x$ for all $x$ .

Boundedness is continuity. A bounded $T$ is Lipschitz with constant $\norm T$ , hence continuous, and a linear map continuous at $0$ has some $\delta>0$ with $\norm{Tx}\le 1$ whenever $\norm x\le\delta$ ; for $\norm x\le 1$ the vector $\delta x$ then gives $\delta\norm{Tx}=\norm{T(\delta x)}\le 1$ , so $\norm{Tx}\le 1/\delta$ and $\norm T\le 1/\delta<\infty$ . The bounded operators form an algebra $\mathcal B(H)$ under composition, and the norm controls products.

Proposition2

For operators $S,T$ and a scalar $\alpha$ , the operator norm satisfies $\norm{S+T}\le\norm S+\norm T$ , $\norm{\alpha T}=\abs\alpha\,\norm T$ , and the submultiplicative bound $\norm{ST}\le\norm S\,\norm T$ . Moreover $\mathcal B(H)$ is complete in the operator norm.

Proof

The norm and scalar identities are immediate from the supremum definition. For the product, $\norm{STx}\le\norm S\,\norm{Tx}\le\norm S\,\norm T\,\norm x$ , and taking the supremum over $\norm x\le 1$ gives $\norm{ST}\le\norm S\,\norm T$ . For completeness, let $(T_n)$ be Cauchy in operator norm. For each $x$ , $\norm{T_nx-T_mx}\le\norm{T_n-T_m}\,\norm x$ , so $(T_nx)$ is Cauchy in $H$ and converges to a vector $Tx$ by completeness of $H$ . The map $T$ is linear as a pointwise limit of linear maps. Given $\varepsilon>0$ pick $N$ with $\norm{T_n-T_m}<\varepsilon$ for $n,m\ge N$ , then let $m\to\infty$ in $\norm{T_nx-T_mx}\le\varepsilon\norm x$ to get $\norm{T_nx-Tx}\le\varepsilon\norm x$ , so $T_n-T$ is bounded, $T=T_N-(T_N-T)$ is bounded, and $\norm{T_n-T}\le\varepsilon$ for $n\ge N$ , that is $T_n\to T$ .

#The adjoint

The defining move of Hilbert space operator theory is to push an operator across the inner product. The Riesz theorem guarantees the pushed operator exists.

Theorem3

For every operator $T$ there is a unique operator $T^\ast$ , the adjoint, with

\ip{Tx}{y}=\ip{x}{T^\ast y}\qquad\text{for all }x,y\in H. \tag{1}

It satisfies $T^{\ast\ast}=T$ , $(ST)^\ast=T^\ast S^\ast$ , $\norm{T^\ast}=\norm T$ , and the C-star identity $\norm{T^\ast T}=\norm T^2$ .

Proof

Fix $y$ . The map $x\mapsto\ip{Tx}{y}$ is linear and bounded, since $\abs{\ip{Tx}{y}}\le\norm{Tx}\norm y \le\norm T\norm y\,\norm x$ , so by the Riesz theorem there is a unique vector, call it $T^\ast y$ , with $\ip{Tx}{y}=\ip{x}{T^\ast y}$ for all $x$ . Uniqueness of the representing vector makes $y\mapsto T^\ast y$ well defined, and it is linear because $\ip{x}{T^\ast(\alpha y+\beta y')}=\ip{Tx}{\alpha y+\beta y'}=\alpha\ip{x}{T^\ast y}+\beta\ip{x}{T^\ast y'}$ holds for all $x$ . It is bounded with $\norm{T^\ast}\le\norm T$ , since putting $x=T^\ast y$ in Equation (1) gives $\norm{T^\ast y}^2=\ip{T^\ast y}{T^\ast y}=\ip{T(T^\ast y)} {y}\le\norm T\,\norm{T^\ast y}\,\norm y$ , which divides to $\norm{T^\ast y}\le\norm T\,\norm y$ . Now that $T^\ast$ is bounded its own adjoint exists, and symmetry of the real inner product gives $\ip{T^\ast x} {y}=\ip{y}{T^\ast x}=\ip{Ty}{x}=\ip{x}{Ty}$ , so $T^{\ast\ast}=T$ . The product rule is $\ip{STx}{y}=\ip{Tx}{S^\ast y}=\ip{x}{T^\ast S^\ast y}$ .

For the norms, expand $\norm{Tx}^2=\ip{Tx}{Tx}=\ip{x}{T^\ast Tx}\le\norm x\,\norm{T^\ast Tx}\le \norm{T^\ast T}\,\norm x^2\le\norm{T^\ast}\,\norm T\,\norm x^2$ , every term finite because $T^\ast$ and so $T^\ast T$ are bounded. Taking the supremum over $\norm x\le 1$ gives $\norm T^2\le\norm{T^\ast T}\le \norm{T^\ast}\,\norm T$ , so $\norm T\le\norm{T^\ast}$ . With $\norm{T^\ast}\le\norm T$ already shown, $\norm{T^\ast}=\norm T$ . The chain then collapses to $\norm T^2\le\norm{T^\ast T}\le\norm{T^\ast}\norm T =\norm T^2$ , forcing $\norm{T^\ast T}=\norm T^2$ .

The C-star identity is the algebraic fingerprint of a Hilbert space operator algebra. It ties the norm, the product, and the involution together so tightly that the norm is determined by the algebra, and it is the axiom that abstracts $\mathcal B(H)$ into the theory of C-star algebras.

#Self-adjoint operators

The operators equal to their own adjoint are the Hilbert space analogue of real numbers, and they are the ones the spectral theorem decomposes.

Definition4

An operator $T$ is self-adjoint when $T^\ast=T$ , that is $\ip{Tx}{y}=\ip{x}{Ty}$ for all $x,y$ .

For a self-adjoint $T$ the quadratic form $x\mapsto\ip{Tx}{x}$ governs the whole operator, and its size is exactly the operator norm, a fact with no analogue for general operators and the lever the spectral theorem pulls.

Theorem5

If $T$ is self-adjoint then $\norm T=\sup_{\norm x\le 1}\abs{\ip{Tx}{x}}$ .

Proof

Let $M=\sup_{\norm x\le 1}\abs{\ip{Tx}{x}}$ . The bound $\abs{\ip{Tx}{x}}\le\norm{Tx}\norm x\le\norm T$ on the unit ball gives $M\le\norm T$ . For the reverse, self-adjointness makes the cross terms agree in the expansion

\ip{T(x+y)}{x+y}-\ip{T(x-y)}{x-y}=4\ip{Tx}{y}, \tag{2}

since $\ip{Ty}{x}=\ip{y}{Tx}=\ip{Tx}{y}$ . Each quadratic form is bounded by $M$ times the squared norm of its argument, so by the parallelogram law

4\ip{Tx}{y}\le M\big(\norm{x+y}^2+\norm{x-y}^2\big)=2M\big(\norm x^2+\norm y^2\big). \tag{3}

For $\norm x=\norm y=1$ this gives $\ip{Tx}{y}\le M$ . Whenever $Tx\neq 0$ , taking $y=Tx/\norm{Tx}$ yields $\norm{Tx}\le M$ , and the inequality $\norm{Tx}\le M$ holds trivially when $Tx=0$ . Taking the supremum over $\norm x\le 1$ gives $\norm T\le M$ , so $\norm T=M$ .

The quadratic form $\ip{Tx}{x}$ is signed, ranging over an interval whose endpoints are its extreme values, and Theorem 5 says the operator norm is the larger endpoint in absolute value. When $T$ is compact, the next post shows that endpoint is actually attained at a unit vector, and that maximiser is an eigenvector with the extreme value as its eigenvalue. This is the first step of the spectral decomposition. Self-adjointness is also closed under the limits and real-linear combinations that the decomposition needs, since $(\,\cdot\,)^\ast$ is an isometric involution by Theorem 3. The adjoint and the quadratic-form identity are the two tools the spectral theorem runs on, and both are gifts of the inner product through the Riesz representation.

Explore connections

see in the atlas

referenced by (1)

Compact Operators and the Spectral Theorem

cite

@misc{bounded-operators,
  author = {Zac Kienzle},
  title  = {Bounded Operators and the Adjoint},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/bounded-operators}
}

#The operator norm

Definition1

Proposition2

Proof

#The adjoint

The defining move of Hilbert space operator theory is to push an operator across the inner product. The Riesz theorem guarantees the pushed operator exists.

Theorem3

For every operator $T$ there is a unique operator $T^\ast$ , the adjoint, with

\ip{Tx}{y}=\ip{x}{T^\ast y}\qquad\text{for all }x,y\in H. \tag{1}

It satisfies $T^{\ast\ast}=T$ , $(ST)^\ast=T^\ast S^\ast$ , $\norm{T^\ast}=\norm T$ , and the C-star identity $\norm{T^\ast T}=\norm T^2$ .

Proof

#Self-adjoint operators

The operators equal to their own adjoint are the Hilbert space analogue of real numbers, and they are the ones the spectral theorem decomposes.

Definition4

An operator $T$ is self-adjoint when $T^\ast=T$ , that is $\ip{Tx}{y}=\ip{x}{Ty}$ for all $x,y$ .

Theorem5

If $T$ is self-adjoint then $\norm T=\sup_{\norm x\le 1}\abs{\ip{Tx}{x}}$ .

Proof

\ip{T(x+y)}{x+y}-\ip{T(x-y)}{x-y}=4\ip{Tx}{y}, \tag{2}

since $\ip{Ty}{x}=\ip{y}{Tx}=\ip{Tx}{y}$ . Each quadratic form is bounded by $M$ times the squared norm of its argument, so by the parallelogram law

4\ip{Tx}{y}\le M\big(\norm{x+y}^2+\norm{x-y}^2\big)=2M\big(\norm x^2+\norm y^2\big). \tag{3}

Explore connections

see in the atlas

referenced by (1)

Compact Operators and the Spectral Theorem

cite

@misc{bounded-operators,
  author = {Zac Kienzle},
  title  = {Bounded Operators and the Adjoint},
  year   = {2026},
  month  = {05},
  url    = {https://zackienzle.com/blog/bounded-operators}
}