Open mapping and closed graph theorems
Use Banach-space completeness to turn surjectivity or graph closedness into quantitative continuity statements.
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The question for today
What you’ll understand
- State open mapping and bounded inverse theorems for Banach spaces.
- Apply closed graph only to an everywhere-defined operator on its stated domain.
- Diagnose a bijection with unbounded inverse when one normed space is incomplete.
A continuous linear map can compress some directions so strongly that recovering the input becomes unstable. Open mapping shows that surjectivity between Banach spaces prevents a global version of this failure, while closed graph gives another route to boundedness when continuity is not initially known. Both theorems rely on completeness of the spaces actually used as domain and target. A closed unbounded differential operator can still exist on a proper domain, so the word closed must never be treated as a universal synonym for bounded.
Surjectivity makes bounded maps open
A bounded linear surjection between Banach spaces maps open sets to open sets. By translation and scaling, this is equivalent to the image of a unit ball containing some ball around zero. The proof starts with Baire applied to countably many dilated images, initially obtaining information about an image closure. A successive-correction argument then constructs an actual preimage as a convergent series, removing the closure. That last step is where completeness of the domain prevents an approximate preimage from disappearing outside the space.
If the surjection is also injective, its inverse is therefore bounded: a bounded linear bijection between Banach spaces is a topological isomorphism. The theorem guarantees a finite inverse bound, not a useful numerical value for it. For a noninjective surjection, the induced map from the Banach quotient X/ker T to Y is a bounded bijection. Consequently distance to the kernel is bounded by a constant times the output norm, even though a specific minimum-norm representative need not exist in a general Banach space.
A closed graph is a complete graph space
For a linear map T defined on all of Banach X and taking values in Banach Y, its graph is the set of pairs (x,Tx). If the graph is closed in the product norm topology, it is a Banach subspace of X×Y. Projection from the graph to X is a bounded linear bijection, so the bounded inverse theorem controls its inverse x↦(x,Tx). This proves T bounded. Conversely, a bounded operator has closed graph by taking limits in both coordinates.
The sequential test for graph closedness says that xₙ→x and Txₙ→y imply Tx=y. It does not say that Txₙ converges whenever xₙ converges; demanding that would already assert continuity. The distinction is vital for unbounded operators. A derivative defined on continuously differentiable functions, viewed as a proper subspace of continuous functions with the supremum norm, can have a closed graph in the ambient product yet be unbounded relative to that domain norm. Its domain is not the whole ambient Banach space, so closed graph does not supply a contradictory conclusion.
Graph norms and closed ranges resolve different issues
For a closed operator with proper domain D(T), the graph norm ∥x∥+∥Tx∥ makes its domain Banach when the ambient source and target are Banach. The operator is then bounded from this stronger domain norm into Y. This changes the normed space in the theorem rather than proving the original unbounded operator continuous in the ambient source norm. In differential equations, graph norms encode derivatives that a weaker function norm does not control, making the domain and its regularity part of the analytic data.
Closed range is another separate property. A bounded injective operator can have dense nonclosed range, with an inverse defined on that range that is unbounded in the inherited target norm. If the range were closed, it would be Banach and bounded inverse would apply. Compact diagonal operators on ℓ² offer explicit examples of this distinction. Their small diagonal entries compress coordinate directions and explain inverse instability without violating injectivity, showing why algebraic uniqueness of a solution does not imply a stable solution map.
Work it through
The same differentiable functions with two norms
Consider the identity from C¹[0,1] with norm ∥f∥∞+∥f′∥∞ to C¹[0,1] with norm ∥f∥∞. Explain why it is a bounded bijection with unbounded inverse and why bounded inverse does not apply.
The identity is linear and bijective on the common set of functions, and its output norm is at most its input norm. The source is Banach in the derivative-controlling norm.
Take fₙ(t)=sin(nt)/n. Its supremum norm is at most 1/n, but its derivative is cos(nt), whose supremum norm is one. No fixed constant can bound the stronger norm by the weaker one.
The target is incomplete: for instance, the smooth functions √((t−1/2)²+1/n) converge uniformly to |t−1/2|, which is continuous but not continuously differentiable. Thus a required Banach hypothesis fails.
The inverse is unbounded in the weaker target norm. This example illustrates the necessity of completeness, rather than contradicting bounded inverse.
The living notebook
Explore the model
Interactive lab · matrix
Invertibility and inverse size are separate
Pause and Reset, then set scale a=0.25,shear b=0. Predict singular values 1 and 0.25, so the inverse Euclidean operator norm is four. For a fresh comparison Reset and set a=1,b=0; predict inverse norm one. Every available matrix is invertible because the scale never reaches zero. The instrument shows finite inverse bounds for selected matrices, not a singular limit or a proof of the Banach open mapping theorem; readouts always concern the final selected matrix.
- Area scale · det A
- 1.4
- Singular values · σmax, σmin
- 1.597, 0.876
- Condition number · κ₂(A)
- 1.823
The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forAn inverse can exist while its norm is large, and a theorem guaranteeing boundedness need not give a practical stability constant.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let D:ℓ²→ℓ² be (Dx)ₙ=xₙ/n. Show it is bounded and injective with dense nonclosed range, and explain the inverse instability.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Finite-support vectors lie in the range. Test the candidate output yₙ=1/n and the images of coordinate unit vectors.
Compare with the solution
The operator norm is one, attained on e₁, and every diagonal entry is nonzero, so D is injective. Any finite-support output has a finite-support preimage, giving dense range. But y=(1/n) belongs to ℓ² and its only possible preimage is the constant-one sequence, which does not belong to ℓ². Hence the range is proper and, being dense, is not closed. Also ∥Deₙ∥₂=1/n while ∥eₙ∥₂=1, so the inverse on the range has no finite bound in the inherited ℓ² norm.
Keep the idea
Take it with you
- A bounded surjection between Banach spaces is open.
- Closed graph implies boundedness for an everywhere-defined linear operator between Banach spaces.
- Changing to a graph norm or a closed range changes which spaces meet the hypotheses.
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