§ Closed graph theorem

• the graph of a function from a banach space to another banach space is a closed subset iff the function is continuous.
• Formally, given $f:X \to Y$, the set $G \equiv { (x, f(x)) }$ is closed in $X \times Y$ iff $f$ is continuous.

§ Proof: Continuous implies closed

• We must show that every limit point of the graph G is in G.

• Let $(p, q)$ be a limit point. Since everything in metric spaces is equivalent to the sequential definition, this means that $(p, q) = \lim (x_i, f(x_i))$.

• Limits in product spaces are computed pointwise, so $(p, q) = (\lim x_i, \lim f(x_i))$

• Thus, $p = lim xi$ from above. Now we calculate:

• $q = \lim f(x_i) = f (\lim xi) = f(p)$ where we use the continuity of $f$ to push the limit inside.
• Thus, $(p, q) = (p, f(p))$ which is an element of $G$.
• So an arbitrary limit point $(p, q) \in G$ is an element of $G$, and thus G is closed. Qed.

§ Proof: closed implies continuous

• Suppose G as defined above is a closed set. We must show that f is continuous, ie, $f$ preserves limits.

• Let $x_i$ be a sequence. We must show that $f(\lim x_i) = \lim f (x_i)$.

• Consider $(x_i, f(x_i))$ as a sequence in $G$. Let the limit of this sequence be $(p, q)$. Since G is closed, $(p, q)$ in G. By defn of $G$, $q = f(p)$.

• Now we compute that

• But since $q = f(p)$ (by defn of $G$), we have that $lim f(x_i) = q = f(p) = f(\lim x_i)$ which proves continuity.