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Homeproof of closed graph theorem

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# proof of closed graph theorem

Let $T\colon X\to Y$ be a linear mapping. Denote its graph by $G(T)$, and let $p_{1}\colon X\times Y\to X$ and $p_{2}\colon X\times Y\to Y$ 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 $\{(x_{i},Tx_{i})\}$ in $G(T)$ which converges to $(x,y)\in X\times Y$, we have that

$x_{i}=p_{1}(x_{i},Tx_{i})\xrightarrow[i\to\infty]{}p_{1}(x,y)=x$ |

and

$Tx_{i}=p_{2}(x_{i},Tx_{i})\xrightarrow[i\to\infty]{}p_{2}(x,y)=y,$ |

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

$Tx=\lim_{{i\to\infty}}Tx_{i}=y.$ |

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 $\tilde{T}:X\to G(T)$ defined by $\tilde{T}x=(x,Tx)$. It is clear that $\tilde{T}$ 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, $\tilde{T}$ is continuous, and consequently $T=p_{2}\circ\tilde{T}$ is continuous.

## Mathematics Subject Classification

46A30*no label found*

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