## Closed Graph Theorem implies Open Mapping Theorem

Today Wongting raised the question in the real analysis prelims at UW about proving that the Closed Graph Theorem (CGT) implies the Open Mapping Theorem (OMT). This is a standard exercise in the first course of functional analysis.

Let us recall the three basic theorems. We assume $X,Y$ are Banach spaces. Recall that a linear operator $T:X\rightarrow Y$ is bounded if and only if it is continuous on $X$.

OMT: If $T:X\rightarrow Y$ be a bounded surjective linear operator, then $T$ is an open mapping.

Inverse Mapping Theorem (IMT): If $T:X\rightarrow Y$ is a bounded bijective linear operator, then the inverse $T^{-1}$ is bounded.

CGT: Suppose $T:X\rightarrow Y$ is a linear operator. Then $T$ is bounded if and only if its graph is closed in $X\times Y$.

Standard texts tell us that OMT implies IMT and IMT implies CGT. Once we proved CGT implies OMT, we see that the above three theorems are indeed equivalent.

Theorem. CGT implies OMT.

Proof. Assume CGT is true. Let $T:X\rightarrow Y$ be a bounded surjective linear operator. We want to show (wlog, by linearity) $B(0,\delta)\subset T(B(0,1))$ for some $\delta>0$ (Notation: $B(x_0,r)$ is the open ball centered at $x_0$ with radius $r$).

Intuitively (if you have learnt some general topology), that a map is open is related to the continuity of its inverse. Therefore we want to construct the inverse of $T$, but $T$ may not be invertible. One way to do is to consider equivalence classes, or more precisely, quotient spaces.

Define an operator $G: Y\rightarrow X/N(T)$ (Notation: $N(T)$ is the null space of $T$; Note now $N(T)$ is a closed subspace of $X$) by $y\mapsto [x]=x+N(T)$ where $Tx=y$. It is easy to see this map is well defined and linear.  Then:

Claim 1: The graph of $G$ is closed.

Indeed, if $(y_n,[x_n])\in \mathrm{graph}(G)\rightarrow (y,[x])$, then $Tx_n = y_n$, $y_n \rightarrow y$ and $\displaystyle\|[x_n]-[x]\|=\inf_{z\in N(T)}\|x_n-x-z\|\rightarrow 0$. Thus by passing to subsequence, $\|x_n-x-z_n\|\rightarrow 0$ for some $\{z_n\}\subset N(T)$, and so $x_n-z_n\rightarrow x$, also $T(x_n-z_n)\rightarrow Tx$. Since $T(x_n-z_n)=Tx_n =y_n$, $y_n$ has a limit and thus Cauchy, $T(x_n)$ has a convergent subsequence, so $Tx_n \rightarrow Tx$ and $Tx=y$. We get $(y,[x])\in \mathrm{graph}(G)$.

By CGT, $G$ is bounded. Then, by linearity, we may assume there exists $\delta>0$ such that $G(B(0,\delta))\subset B(0,1)$. After proving the following second claim, we are done.

Claim 2: $B(0,\delta)\subset T(B(0,1))$.

Indeed, let $y\in B(0,\delta)$.  Then $[x]\triangleq G(y)\in B(0,1)$ and thus $\displaystyle \inf_{z\in N(T)}\|x-z\|<1$. Therefore there exists $z_0\in N(T)$ such that $\|x-z_0\|<1$. It follows that $y=Tx=T(x-z_0)\in T(B(0,1))$.

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### 3 Responses to Closed Graph Theorem implies Open Mapping Theorem

1. Ifeyinwa says:

The work is very helpful . Thanks. I need some help in homology group

2. Anonymous says:

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3. Ji says:

Thanks