The Closed Graph Theorem

# The Closed Graph Theorem

Definition: Let $f : X \to Y$ be a function. Then the Graph of $f$ is defined to be $\mathrm{Gr} (f) = \{ (x, f(x)) : x \in X \}$. |

Theorem 1 (The Closed Graph Theorem): Let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a linear operator. Then $T$ is a bounded linear operator if and only if the graph of $T$, $\mathrm{Gr} (T) = \{ (x, T(x)) : x \in X \}$ is a closed set in the product space $X \times Y$, that is, if $(x_n)_{n=1}^{\infty}$ is a sequence of points in $X$ that converges to $x \in X$ and $(T(x_n))_{n=1}^{\infty}$ converges to $y \in Y$ then $T(x) = y$. |

**Proof:**$\Rightarrow$ Suppose that $T$ is bounded (continuous). Let $(x_n, T(x_n))$ be a sequence in $\mathrm{Gr}(T)$ that converges to $(x, y)$. Then $(x_n)$ converges to $x$ and $(T(x_n))$ converges to $y$. Since $T$ is continuous, $(T(x_n))$ converges to $T(x)$. So $T(x) = y$. So $\mathrm{Gr}(T)$ is closed.

- $\Leftarrow$ Suppose that $\mathrm{Gr}(T)$ is a closed set in the product space $X \times Y$. Define a new norm $\| \cdot \|_T$ on $X$ for all $x \in X$ by:

\begin{align} \quad \| x \|_T = \| x \| + \| T(x) \| \end{align}

- We must first verify that $\| \cdot \|_T$ is indeed a norm.

- Suppose that $\| x \|_T = 0$. Then $\| x \| + \| T(x) \| = 0$. Since $\| x \| \geq 0$ and $\| T(x) \| \geq 0$ we must have that $\| x \| = 0$. But $\| \cdot \|$ is a norm so $x = 0$. Now we also have that $\| 0 \|_T = \| 0 \| + \| T(0) \| = 0 + 0 = 0$. So $\| x \|_T = 0$ if and only if $x = 0$.

- Let $\lambda \in \mathbb{C}$. Then:

\begin{align} \quad \| \lambda x \|_T = \| \lambda x \| + \| T(\lambda x) \| = | \lambda | \| x \| + | \lambda | \| T(x) \| = | \lambda |(\| x \| + \| T(x) \|) \end{align}

- Lastly, let $x, y \in X$. Then:

\begin{align} \quad \| x + y \|_T = \| x + y \| + \| T(x + y) \| \leq \| x \| + \| y \| + \| T(x) \| + \| T(y) \| = (\| x \| + \| T(x) \|) + (\| y \| + \| T(y) \|) = \| x \|_T + \| y \|_T \end{align}

- So indeed, $\| \cdot \|_T$ is a norm on $X$. Furthermore, since $\| x \| \geq 0$ and $\| T(x) \| \geq 0$ we have for all $x \in X$ that:

\begin{align} \quad \| x \| \leq \| x \| + \| T(x) \| = 1 \cdot \| x \|_T \quad (*) \end{align}

(5)
\begin{align} \quad \| T(x) \| \leq \| x \| + \| T(x) \| = 1 \| x \|_T \quad (**) \end{align}

- We aim to show that $X$ with the norm $\| \cdot \|_T$ is a Banach space. Let $(x_n)_{n=1}^{\infty}$ be a Cauchy sequence in $X$ with respect to the norm $\| \cdot \|_T$. Let $m, n \in \mathbb{N}$. Then from $(*)$ and $(**)$ we have that:

\begin{align} \quad \| x_m - x_n \| \leq \| x_m - x_n \|_T \quad (***) \end{align}

(7)
\begin{align} \quad \| T(x_m) - T(x_n) \| \leq \| x_m - x_n \|_T \quad (****) \end{align}

- From $(***)$ we have that $(x_n)_{n=1}^{\infty}$ must be a Cauchy sequence in $X$ with respect to the original norm on $X$. And from $(****)$ we have that $(T(x_n))_{n=1}^{\infty}$ must be a Cauchy sequence in $Y$ with respect to the norm on $Y$.

- Since $X$ is a Banach space, $(x_n)_{n=1}^{\infty}$ converges to some $x \in X$ and since $Y$ is a Banach space, $(T(x_n))_{n=1}^{\infty}$ converges to some $y \in Y$.

- So $(x_n)_{n=1}^{\infty}$ is a sequence of points in $X$ that converges to $x \in X$ and $(T(x_n)_{n=1}^{\infty}$ converges to $y \in Y$, so $T(x) = y$.

- Now all that remains to show is that the Cauchy sequence $(x_n)_{n=1}^{\infty}$ in $X$ converges to $x \in X$ with respect to the $\| \cdot \|_T$] norm. We have that:

\begin{align} \quad \lim_{n \to \infty} \| x_n - x \|_T = \lim_{n \to \infty} [\| x_n - x \| + \| T(x_n) - T(x) \|] = \lim_{n \to \infty} \| x_n - x \| + \lim_{n \to \infty} \| T(x_n) - y \| = 0 + 0 = 0 \end{align}

- So indeed every Cauchy sequence $(x_n)_{n=1}^{\infty}$ in $X$ converges to some $x \in X$ with respect to the $\| \cdot \|_T$ norm. So $X$ is a Banach space with respect to the $\| \cdot \|_T$ norm.

- From $(*)$ we have that $\| \cdot \|$ and $\| \cdot \|_T$ are equivalent norms by the theorem on the Equivalence of Norms on Banach Spaces page. So there exists a $C > 0$ such that for all $x \in X$:

\begin{align} \quad \| x \|_T \leq C \| x \| \end{align}

- But from $(**)$ this means that for all $x \in X$:

\begin{align} \quad \| T(x) \| \leq C \| x \| \end{align}

- So $T$ is a bounded linear operator. $\blacksquare$

Corollary 2 (An Equivalent Formulation of the Closed Graph Theorem): Let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a linear operator. Then $T$ is a bounded linear operator if and only if whenever $(x_n)_{n=1}^{\infty}$ is a sequence of points in $X$ that converges to $0 \in X$ and $(T(x_n))_{n=1}^{\infty}$ converges to $y \in Y$ then $y = 0$. |

**Proof:**$\Rightarrow$ If $T$ is a bounded linear operator then by the Closed Graph theorem above, if $(x_n)_{n=1}^{\infty}$ converges to $0$ and $(T(x_n))_{n=1}^{\infty}$ converges to $y$ then we must have that $T(0) = y$. But $T(0) = 0$, i.e., $y = 0$..

- $\Leftarrow$ Suppose that whenever $(x_n)_{n=1}^{\infty}$ converges to $0$ and $(T(x_n))_{n=1}^{\infty}$ converges to $y$ we have that $y = 0$.

- Let $(z_n)_{n=1}^{\infty}$ be a sequence in $X$ that converges to $z$, and let $(T(z_n))_{n=1}^{\infty}$ converge to $w \in Y$. Then $(z_n - z)_{n=1}^{\infty}$ converges to $0$, and $(T(z_n) - w)_{n=1}^{\infty}$ converges to $T(z) - w$. So by the assumption we must have that $T(z) - w = 0$, that is, $T(z) = w$. By the Closed Graph theorem, $T$ is bounded. $\blacksquare$