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Observation lower bounds give Hilbert control right inverses

Statement

Assume Countable Choice. Let $E$ and $U$ be real or complex Hilbert spaces over the same field, and let $O : E \to U$ be bounded and linear. Suppose that $∥ O x ∥_{U} \geq c ∥ x ∥_{E}$ for every $x \in E$ , where $c > 0$ . Let $O^{*} : U \to E$ be the Hilbert adjoint and set $Q = O^{*} O$ . Then $Q : E \to E$ has a bounded two-sided inverse, and $R = O Q^{- 1} : E \to U$ is a bounded linear right inverse of $O^{*}$ , meaning $O^{*} R = I_{E}$ , with $∥ R ∥ \leq c^{- 1}$ .

For the stronger trajectory conclusion, let $Y$ be a normed space over the same field and $T > 0$ . Write $C ([0, T]; Y)$ for the continuous $Y$ -valued functions with supremum norm. Suppose that a bounded linear map $J : E \to C ([0, T]; Y)$ and a linear map $i : C ([0, T]; Y) \to U$ satisfy $i J = O$ . Then $R_{C} = J Q^{- 1}$ is bounded into $C ([0, T]; Y)$ , $i R_{C} = R$ , and $∥ R_{C} ∥ \leq ∥ J ∥ c^{- 2}$ . In particular, when $O$ itself has values in both spaces with the same representatives and is bounded as a map $E \to C ([0, T]; Y)$ , one may take $J = O$ and $i$ to be the inclusion or the map to the corresponding equivalence class.

Facts & Assumptions

[A1]

Countable Choice means that every countable family of nonempty sets admits a choice function; it is assumed for the adjoint and Lax–Milgram interfaces below (The Axiom of Countable Choice (AC_ω)).

[F1]

A Hilbert space is a complete inner-product space. The inner products here are linear in the first variable; the bounded Hilbert adjoint is characterized by $⟨ O x, u ⟩_{U} = ⟨ x, O^{*} u ⟩_{E}$ (Hilbert space, The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[F2]

Cauchy–Schwarz gives $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , and the norm of a bounded operator is the supremum of its output norm on the unit ball (Cauchy–Schwarz: |⟨ x,y⟩|≤‖x‖ ‖y‖, with equality exactly for dependent pairs, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F3]

Under Countable Choice, Lax–Milgram gives a unique solution $z$ to $a (z, v) = F (v)$ for all $v$ when $a$ is a bounded coercive real bilinear or complex sesquilinear form and $F$ is a bounded functional conjugate-linear in the test variable. If the coercivity constant is $α$ , then $∥ z ∥ \leq α^{- 1} ∥ F ∥$ (The Lax--Milgram theorem, The Axiom of Countable Choice (AC_ω)).

[F4]

A norm on a vector space over $R$ or $C$ is nonnegative, separates zero, is absolutely homogeneous, and satisfies the triangle inequality; its metric is $d (y, z) = ∥ y - z ∥_{Y}$ (Normed vector space over an absolutely valued field), using the usual absolute value or complex modulus.

[F5]

In a real or complex normed space, vector addition and scalar multiplication are continuous, and $∣ ∥ y ∥_{Y} - ∥ z ∥_{Y} ∣ \leq ∥ y - z ∥_{Y}$ , so the norm map is continuous (Vector addition and scalar multiplication are continuous in a normed space, The reverse triangle inequality in a normed space). For a complex space, the reverse-triangle estimate also follows on its underlying real normed space, with the same metric and norm.

[F6]

Every continuous real-valued function on a closed real interval $[a, b]$ , $a \leq b$ , is bounded and attains its maximum and minimum (Continuous functions on a closed interval attain their extrema).

[F7]

A linear map $A : X \to Z$ between normed spaces over the same real or complex field is bounded when there is a real $M \geq 0$ such that $∥ A x ∥_{Z} \leq M ∥ x ∥_{X}$ for every $x \in X$ (A bounded linear operator between normed spaces).

[F8]

A map $g : (X, d_{X}) \to (Z, d_{Z})$ is continuous at $x_{0}$ when for every $ε > 0$ there is $δ > 0$ such that $d_{X} (x, x_{0}) < δ$ implies $d_{Z} (g (x), g (x_{0})) < ε$ ; it is continuous when this holds at every point (Continuity of a map between metric spaces, at a point and globally, in the ε-δ form). For the trajectories here, $[0, T]$ carries the metric $∣ t - r ∣$ and $Y$ carries $∥ y - z ∥_{Y}$ ; real-valued norm functions use the usual absolute-value metric.

Proof

[A1] Countable Choice means that every countable family of nonempty sets admits a choice function; it is assumed for the adjoint and Lax–Milgram interfaces below (The Axiom of Countable Choice (AC_ω)).

[F1] A Hilbert space is a complete inner-product space. The inner products here are linear in the first variable; the bounded Hilbert adjoint is characterized by $⟨ O x, u ⟩_{U} = ⟨ x, O^{*} u ⟩_{E}$ (Hilbert space, The Hilbert-space adjoint of a bounded operator, Hilbert-adjoint identities).

[F2] Cauchy–Schwarz gives $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , and the norm of a bounded operator is the supremum of its output norm on the unit ball (Cauchy–Schwarz: |⟨ x,y⟩|≤‖x‖ ‖y‖, with equality exactly for dependent pairs, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F3] Under Countable Choice, Lax–Milgram gives a unique solution $z$ to $a (z, v) = F (v)$ for all $v$ when $a$ is a bounded coercive real bilinear or complex sesquilinear form and $F$ is a bounded functional conjugate-linear in the test variable. If the coercivity constant is $α$ , then $∥ z ∥ \leq α^{- 1} ∥ F ∥$ (The Lax--Milgram theorem, The Axiom of Countable Choice (AC_ω)).

Given: The spaces, maps and lower bound in the statement, with Countable Choice.

Goal: Prove the invertibility, right-inverse identity and bounds, including the conditional trajectory lift.

1.1givenA1F1F2F3F7

Define $a (z, v) = ⟨ O z, O v ⟩_{U}$ . It is linear in $z$ and conjugate-linear in $v$ (bilinear in the real case). By Cauchy–Schwarz and boundedness of $O$ , $∣ a (z, v) ∣ \leq ∥ O ∥^{2} ∥ z ∥_{E} ∥ v ∥_{E}$ , while $Re a (z, z) = ∥ O z ∥_{U}^{2} \geq c^{2} ∥ z ∥_{E}^{2}$ . For each $f \in E$ , the functional $F_{f} (v) = ⟨ f, v ⟩_{E}$ is bounded and conjugate-linear, with $∥ F_{f} ∥ \leq ∥ f ∥_{E}$ . Lax–Milgram therefore gives a unique $z \in E$ with $a (z, v) = ⟨ f, v ⟩_{E}$ for all $v$ , and $∥ z ∥_{E} \leq c^{- 2} ∥ f ∥_{E}$ .

1.2givenF4F5F8

Under the trajectory hypotheses, give $C ([0, T]; Y)$ pointwise addition and scalar multiplication. The zero function is continuous. For continuous $g, h$ and a fixed scalar $a$ , the continuity of the vector operations gives continuity of $g + h$ and $a g$ . Explicitly, at $t_{0}$ the increments satisfy $∥ (g + h) (t) - (g + h) (t_{0}) ∥_{Y} \leq ∥ g (t) - g (t_{0}) ∥_{Y} + ∥ h (t) - h (t_{0}) ∥_{Y}$ and $∥ a g (t) - a g (t_{0}) ∥_{Y} = ∣ a ∣ ∥ g (t) - g (t_{0}) ∥_{Y}$ , which tend to zero as $t \to t_{0}$ within $[0, T]$ . All vector-space identities hold pointwise, so this is a vector space over the same field.

2.1step 1.1F1F2F7

The adjoint identity and conjugate symmetry give $a (z, v) = ⟨ O^{*} O z, v ⟩_{E} = ⟨ Q z, v ⟩_{E}$ . Thus the solution in step 1.1 satisfies $Q z = f$ , since an inner product with every $v$ vanishes only for the zero vector. Uniqueness also gives injectivity of $Q$ . Hence $Q$ is bijective; its inverse is linear because $Q$ is linear, and the estimate in step 1.1 gives $∥ Q^{- 1} ∥ \leq c^{- 2}$ .

2.2givenF4F5F6step 1.2F8

For $g \in C ([0, T]; Y)$ , the reverse triangle inequality gives $∣ ∥ g (t) ∥_{Y} - ∥ g (t_{0}) ∥_{Y} ∣ \leq ∥ g (t) - g (t_{0}) ∥_{Y}$ , so $t \mapsto ∥ g (t) ∥_{Y}$ is continuous. The extreme value theorem on $[0, T]$ makes $∥ g ∥_{\infty} = \sup_{0 \leq t \leq T} ∥ g (t) ∥_{Y}$ a finite nonnegative real number, attained at some $t$ . This quantity vanishes exactly for the zero function. Pointwise absolute homogeneity gives $∥ a g ∥_{\infty} = ∣ a ∣ ∥ g ∥_{\infty}$ , including $a = 0$ ; the pointwise triangle inequality gives $∥ g + h ∥_{\infty} \leq ∥ g ∥_{\infty} + ∥ h ∥_{\infty}$ . Thus the stated supremum norm makes $C ([0, T]; Y)$ a normed space, without any completeness assumption on $Y$ .

3.1givenF1F2step 2.1F7

Set $R = O Q^{- 1}$ . It is bounded and linear, and $O^{*} R = Q Q^{- 1} = I_{E}$ . For $f \in E$ put $z = Q^{- 1} f$ . Then $∥ R f ∥_{U}^{2} = ⟨ Q z, z ⟩_{E} = ⟨ f, z ⟩_{E} \leq ∥ f ∥_{E} ∥ z ∥_{E}$ , while $c ∥ z ∥_{E} \leq ∥ R f ∥_{U}$ . If $R f = 0$ the desired estimate is immediate; otherwise combine the two inequalities and divide by $∥ R f ∥_{U}$ to obtain $∥ R f ∥_{U} \leq c^{- 1} ∥ f ∥_{E}$ . Taking the unit-ball supremum proves $∥ R ∥ \leq c^{- 1}$ .

4.1givenF2step 2.1step 3.1step 2.2F7∎

Under the additional trajectory hypotheses, define $R_{C} = J Q^{- 1}$ . For every $f \in E$ , $i R_{C} f = i J Q^{- 1} f = O Q^{- 1} f = R f$ . Also $∥ R_{C} f ∥_{C ([0, T]; Y)} \leq ∥ J ∥ ∥ Q^{- 1} f ∥_{E} \leq ∥ J ∥ c^{- 2} ∥ f ∥_{E}$ , which gives the asserted operator bound. No identification of $O^{*}$ with an evolution endpoint map is used.

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