PlanetMath: proof of closed graph theorem

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proof of closed graph theorem

(Proof)

Let $T\colon X\to Y$ be a linear mapping. Denote its graph by , and let $p_1\colon X\times Y\to X$ and $p_2\colon X\times Y\to Y$ be the projections onto and , respectively. We remark that these projections are continuous, by definition of the product of Banach spaces.

If is bounded, then given a sequence $\{(x_i, Tx_i)\}$ in which converges to $(x,y)\in X\times Y$ , we have that

$\displaystyle x_i = p_1(x_i,Tx_i) \xrightarrow[i\to\infty]{} p_1(x,y) = x$

and

$\displaystyle Tx_i = p_2(x_i,Tx_i) \xrightarrow[i\to\infty]{} p_2(x,y) = y,$

by continuity of the projections. But then, since

is continuous,

$\displaystyle Tx = \lim_{i\to\infty} Tx_i = y.$

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

is closed.

Now suppose is closed. We remark that 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 $\tilde Tx = (x,Tx)$ . It is clear that $\tilde T$ is a bijection, its inverse being $p_1\vert _{G(T)}$ , the restriction of to . Since is continuous on $X\times Y$ , the restriction is continuous as well; and since it is also surjective, the open mapping theorem implies that $p_1\vert _{G(T)}$ 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.