# proof of closed graph theorem

Let $T:X\to Y$ be a linear mapping. Denote its graph by $G(T)$, and let ${p}_{1}:X\times Y\to X$ and ${p}_{2}:X\times Y\to Y$ be the projections onto $X$ and $Y$, respectively. We remark that these projections are continuous^{}, by definition of the product^{} of Banach spaces^{}.

If $T$ is bounded^{}, then given a sequence
$\{({x}_{i},T{x}_{i})\}$ in $G(T)$ which converges^{} to $(x,y)\in X\times Y$, we have that

$${x}_{i}={p}_{1}({x}_{i},T{x}_{i})\underset{i\to \mathrm{\infty}}{\overset{}{\to}}{p}_{1}(x,y)=x$$ |

and

$$T{x}_{i}={p}_{2}({x}_{i},T{x}_{i})\underset{i\to \mathrm{\infty}}{\overset{}{\to}}{p}_{2}(x,y)=y,$$ |

by continuity of the projections. But then, since $T$ is continuous,

$$Tx=\underset{i\to \mathrm{\infty}}{lim}T{x}_{i}=y.$$ |

Thus $(x,y)=(x,Tx)\in G(T)$, proving that $G(T)$ is closed.

Now suppose $G(T)$ is closed. We remark that $G(T)$ is a vector subspace of $X\times Y$, and being closed, it is a Banach space. Consider the operator
$\stackrel{~}{T}:X\to G(T)$ defined by $\stackrel{~}{T}x=(x,Tx)$. It is clear that $\stackrel{~}{T}$ is a bijection, its inverse^{} being ${{p}_{1}|}_{G(T)}$, the restriction^{} of ${p}_{1}$ to $G(T)$. Since ${p}_{1}$ is continuous on $X\times Y$, the restriction is continuous as well; and since it is also surjective, the open mapping theorem^{} implies that ${{p}_{1}|}_{G(T)}$ is an open mapping, so its inverse must be continuous. That is, $\stackrel{~}{T}$ is continuous, and consequently $T={p}_{2}\circ \stackrel{~}{T}$ is continuous.

Title | proof of closed graph theorem |
---|---|

Canonical name | ProofOfClosedGraphTheorem |

Date of creation | 2013-03-22 14:48:47 |

Last modified on | 2013-03-22 14:48:47 |

Owner | Koro (127) |

Last modified by | Koro (127) |

Numerical id | 5 |

Author | Koro (127) |

Entry type | Proof |

Classification | msc 46A30 |