Functional Analysis 12 | Continuity [dark version]

TL;DR

In metric spaces, continuity via open sets (preimages of open sets are open) is equivalent to sequential continuity (x_n → x implies f(x_n) → f(x)).

Briefing Cornell Notes

Briefing

Continuity in metric spaces is pinned down by a simple rule: a function is continuous exactly when it preserves limits of sequences. That equivalence matters because it turns a topological definition—preimages of open sets being open—into a practical test using sequences, which is often easier to compute with in analysis.

Formally, given metric spaces (X, d_X) and (Y, d_Y) and a map f: X → Y, continuity means that for every open set B ⊆ Y, the preimage f^{-1}[B] is open in X. The intuition is geometric: if an input varies slightly inside an open region, the output cannot “jump” outside the corresponding open region. The transcript then introduces an alternative criterion—sequential continuity—saying that whenever a sequence x_n → x in X, the images satisfy f(x_n) → f(x) in Y. In metric spaces, these two notions coincide, so continuity can be checked using sequences; outside metric spaces, they may differ.

Several examples anchor the definitions. With the discrete metric on X (where every subset is open), every function f: X → Y is automatically continuous, regardless of the metric on Y. Constant functions are also always continuous: if f(x) = y_0 for all x, then any convergent sequence x_n → x forces f(x_n) = y_0 → y_0 = f(x).

The next examples focus on maps built from norms and inner products, showing why these core analytic objects behave well under limits. In a normed space X (with the induced metric) mapping into ℝ with the absolute value metric, the norm map f(x) = ||x|| is continuous. The proof uses the triangle inequality twice: bounding ||x_n|| relative to ||x|| by writing ||x_n|| ≤ ||x_n − x|| + ||x|| and also ||x|| ≤ ||x_n − x|| + ||x_n||. Since ||x_n − x|| → 0 when x_n → x, both inequalities force ||x_n|| → ||x||.

For inner product spaces, continuity extends to the inner product in each argument. Fixing x_0 and defining f(x) = ⟨x_0, x⟩ (viewed as a complex-valued function with metric |·|), the proof again follows the limit of differences. Using linearity, |f(x_n) − f(x)| = |⟨x_0, x_n − x⟩|, and then applying Cauchy–Schwarz gives |⟨x_0, x_n − x⟩| ≤ ||x_0||·||x_n − x||. As x_n → x, the right-hand side goes to 0, so ⟨x_0, x_n⟩ → ⟨x_0, x⟩. By symmetry (repeating the argument with roles swapped), the inner product is continuous in both variables.

That continuity is then used to show a structural result: for a subset U of an inner product space, the orthogonal complement U^⊥ is closed. The argument uses sequences: if x_n ∈ U^⊥ and x_n → x, then ⟨x_n, u⟩ = 0 for every u ∈ U. Continuity of the inner product lets the limit pass inside, yielding ⟨x, u⟩ = 0 for all u ∈ U, so x ∈ U^⊥. The closure follows directly from the sequential characterization of closed sets in metric spaces.

Cornell Notes

In metric spaces, continuity can be defined either by open sets (preimages of open sets are open) or by sequences (convergent sequences map to convergent sequences). These two definitions are equivalent for metric spaces, making sequential continuity a powerful tool. The transcript demonstrates this equivalence and then proves continuity for key analytic maps: constant functions and all functions from a discrete metric space, the norm map x ↦ ||x|| on a normed space, and the inner product maps x ↦ ⟨x_0, x⟩ using Cauchy–Schwarz. Finally, continuity of the inner product is used to show that orthogonal complements U^⊥ are closed: if x_n ∈ U^⊥ and x_n → x, then ⟨x, u⟩ = 0 for every u ∈ U, so x ∈ U^⊥.

Why does sequential continuity match the open-set definition of continuity in metric spaces?

In metric spaces, the topology is determined by the metric, and open-set continuity is equivalent to preserving limits of sequences. Concretely, if f^{-1}[B] is open for every open B ⊆ Y, then whenever x_n → x in X, the images cannot “leave” the neighborhoods of f(x), forcing f(x_n) → f(x). Conversely, if every convergent sequence’s image converges to the image of the limit, then preimages of open sets must be open. The transcript emphasizes that this equivalence is special to metric spaces; for general topological spaces, sequential continuity and open-set continuity can differ.

What makes every function continuous when the domain uses the discrete metric?

With the discrete metric on X, every subset of X is open. Since continuity requires that f^{-1}[B] be open in X for every open B ⊆ Y, and f^{-1}[B] is always a subset of X, it is automatically open. So no matter how f behaves, the preimage of any open set is open, guaranteeing continuity.

How does the triangle inequality prove that the norm map x ↦ ||x|| is continuous?

Let x_n → x in a normed space. The proof bounds ||x_n|| relative to ||x|| in two directions. First, ||x_n|| = ||(x_n − x) + x|| ≤ ||x_n − x|| + ||x||, so limsup ||x_n|| ≤ ||x||. Second, ||x|| = ||(x − x_n) + x_n|| ≤ ||x − x_n|| + ||x_n||, giving ||x|| ≤ liminf ||x_n||. Since ||x_n − x|| → 0, both inequalities force ||x_n|| → ||x||.

How does Cauchy–Schwarz establish continuity of x ↦ ⟨x_0, x⟩?

Fix x_0 and define f(x) = ⟨x_0, x⟩. For x_n → x, consider |f(x_n) − f(x)| = |⟨x_0, x_n⟩ − ⟨x_0, x⟩| = |⟨x_0, x_n − x⟩|. Cauchy–Schwarz gives |⟨x_0, x_n − x⟩| ≤ ||x_0||·||x_n − x||. The factor ||x_n − x|| → 0, so the whole expression goes to 0, meaning ⟨x_0, x_n⟩ → ⟨x_0, x⟩. The same reasoning with swapped roles yields continuity in the other argument too.

Why does continuity of the inner product imply that U^⊥ is closed?

Take a sequence x_n ∈ U^⊥ with x_n → x. By definition of orthogonal complement, ⟨x_n, u⟩ = 0 for every u ∈ U and for all n. Continuity of the inner product lets limits pass inside: ⟨x_n, u⟩ → ⟨x, u⟩. Since the left side is always 0, the limit is also 0, so ⟨x, u⟩ = 0 for all u ∈ U. Therefore x ∈ U^⊥, which shows U^⊥ contains limits of convergent sequences from itself and is closed.

Review Questions

In a metric space, how would you use sequences to verify continuity of a function f: X → Y?
What two inequalities are needed to prove continuity of the norm map x ↦ ||x||, and where does the triangle inequality enter?
How does Cauchy–Schwarz turn convergence x_n → x into convergence ⟨x_0, x_n⟩ → ⟨x_0, x⟩?

Key Points

1
In metric spaces, continuity via open sets (preimages of open sets are open) is equivalent to sequential continuity (x_n → x implies f(x_n) → f(x)).
2
With the discrete metric on the domain, every function is continuous because every subset of the domain is open.
3
Constant functions are always continuous since their outputs never change, so limits are preserved automatically.
4
The norm map x ↦ ||x|| is continuous in any normed space, with the proof relying on the triangle inequality in both directions.
5
For inner product spaces, the map x ↦ ⟨x_0, x⟩ is continuous, and Cauchy–Schwarz provides the key estimate |⟨x_0, x_n − x⟩| ≤ ||x_0||·||x_n − x||.
6
Continuity of the inner product implies orthogonal complements U^⊥ are closed: limits of sequences in U^⊥ stay in U^⊥.

Highlights

Continuity in metric spaces can be checked using sequences: preserving limits is enough.

The norm map’s continuity follows from two triangle-inequality bounds that squeeze ||x_n|| toward ||x||.

Cauchy–Schwarz converts convergence of x_n to convergence of inner products ⟨x_0, x_n⟩.

Orthogonal complements U^⊥ are closed because ⟨x_n, u⟩ = 0 for all n forces ⟨x, u⟩ = 0 in the limit.

Topics

Continuity in Metric Spaces
Sequential Continuity
Norm Continuity
Inner Product Continuity
Orthogonal Complements