Definition — Open Map
A map \(T\colon X\to Y\) is open if \(T(U)\) is open in \(Y\) for every open set \(U\subseteq X\).


Open Mapping Theorem
Let \(X\) and \(Y\) be Banach spaces. If \(T\colon X\to Y\) is a bounded, surjective linear operator, then \(T\) is an open map.

Recall that a map is continuous if the preimage of every open set is open. Suppose \(T\) is bijective and write

$$ S=T^{-1}\colon Y\to X. $$

For every open set \(U\subseteq X\), the preimage of \(U\) under \(S\) is

$$ S^{-1}(U)=T(U). $$

Thus \(S\) is continuous precisely when \(T\) is open. For a bounded linear bijection between Banach spaces, the Open Mapping Theorem therefore gives continuity of the inverse.

Proof of the Open Mapping Theorem - Step 1

The goal of this step is to show that the image of the unit ball contains a ball around 0. Let \(X\) and \(Y\) be Banach spaces, and let \(T\colon X\to Y\) be a bounded, surjective linear operator. Let

$$ B_0=B_X(0,1)=\{x\in X:\|x\|_X\lt 1\}. $$

The image of the open unit ball is

$$ T(B_0)=\{Tx:x\in X,\ \|x\|_X\lt 1\}. $$
Claim
There exists \(\delta>0\) such that \(B_Y(0,\delta)\subseteq T(B_0)\).

First recall, for any \(y\in Y\), its distance from the origin is \(d_Y(y,0)=\|y-0\|_Y=\|y\|_Y\). Thus \(y\in B_Y(0,\delta)\) means \(\|y\|_Y\lt\delta\). Also, \(y\in T(B_0)\) means that there exists \(x\in X\) with \(\|x\|_X\lt 1\) and \(Tx=y\).

The claim therefore says that every vector sufficiently close to zero in \(Y\) has a preimage of norm less than \(1\).

Let

$$ B_1=B_X\left(0,\frac12\right) =\left\{x\in X:\|x\|_X\lt\frac12\right\}. $$

For each positive integer \(k\), the dilated set \(kB_1\) is

$$ kB_1=\{kx:x\in B_1\} =B_X\left(0,\frac{k}{2}\right). $$

Indeed, if \(x=ku\) with \(\|u\|_X\lt 1/2\), then \(\|x\|_X=k\|u\|_X\lt k/2\). Conversely, if \(\|x\|_X\lt k/2\), then \(\|x/k\|_X\lt 1/2\), so \(x=k(x/k)\in kB_1\).

Every \(x\in X\) has finite norm. Choose a positive integer \(k\) large enough that \(\|x\|_X\lt k/2\). Then \(x\in kB_1\), proving

$$ X=\bigcup_{k=1}^{\infty}kB_1. $$

2. Use surjectivity to cover the codomain.

Since \(T\) is surjective, every \(y\in Y\) has the form \(y=Tx\) for some \(x\in X\). The preceding step places this \(x\) in some \(kB_1\). Hence \(y\in T(kB_1)\), and therefore

$$ Y=T(X)=\bigcup_{k=1}^{\infty}T(kB_1). $$

Now take the closure of each image in \(Y\). Each closure contains the original image and is a subset of \(Y\), so

$$ Y=\bigcup_{k=1}^{\infty}\overline{T(kB_1)}. $$

Each set \(\overline{T(kB_1)}\) is closed in \(Y\).

3. Apply the Baire Category Theorem.

The space \(Y\) is Banach, so it is complete. We have expressed it as a countable union of closed sets. By the Baire Category Theorem, at least one of the sets \(\overline{T(kB_1)}\) has nonempty interior.

Consequently, there exist a positive integer \(k\), a point \(y_0\in Y\), and a radius \(r>0\) such that

$$ B_Y(y_0,r)\subseteq\overline{T(kB_1)}. $$

Fix this choice of \(k\), \(y_0\), and \(r\) for the rest of the argument.

4. Show that scaling commutes with taking the closure.

By linearity,

$$ T(kB_1) =\{T(kx):x\in B_1\} =\{kTx:x\in B_1\} =kT(B_1). $$

We also need

$$ \overline{kT(B_1)}=k\,\overline{T(B_1)}. $$

To prove this carefully, write \(A=T(B_1)\) and show both inclusions in \(\overline{kA}=k\overline A\).

First, suppose \(y\in\overline{kA}\). Since \(Y\) is a metric space, there exists a sequence of points \(ka_j\in kA\), with \(a_j\in A\), such that \(ka_j\to y\). Dividing by the positive number \(k\) gives

$$ a_j\to\frac{y}{k}. $$

Every \(a_j\) belongs to \(A\), so \(y/k\in\overline A\). Therefore \(y=k(y/k)\in k\overline A\). This proves \(\overline{kA}\subseteq k\overline A\).

Conversely, suppose \(y\in k\overline A\). Then \(y=ka\) for some \(a\in\overline A\). Choose a sequence \(a_j\in A\) with \(a_j\to a\). Multiplication by \(k\) gives

$$ ka_j\to ka=y. $$

Since every \(ka_j\) belongs to \(kA\), we have \(y\in\overline{kA}\). Thus \(k\overline A\subseteq\overline{kA}\), proving the equality.

Applying this to \(A=T(B_1)\) gives

$$ \overline{T(kB_1)} =\overline{kT(B_1)} =k\,\overline{T(B_1)}. $$

5. Rescale the ball obtained from Baire’s theorem.

We now have

$$ B_Y(y_0,r)\subseteq k\,\overline{T(B_1)}. $$

Dividing a ball by \(k>0\) divides both its center and its radius by \(k\). Indeed,

$$ \|y-y_0\|_Y\lt r \quad\Longleftrightarrow\quad \left\|\frac{y}{k}-\frac{y_0}{k}\right\|_Y\lt\frac{r}{k}. $$

Therefore,

$$ B_Y\left(\frac{y_0}{k},\frac{r}{k}\right) \subseteq\overline{T(B_1)}. $$

Set \(a=y_0/k\) and \(\rho=r/k\). Since \(\rho>0\), we have found an open ball

$$ B_Y(a,\rho)\subseteq\overline{T(B_1)}. $$

Our next goal is to obtain a ball centered at zero.

6. Show that the image and its closure are symmetric about zero.

Suppose \(y\in T(B_1)\). Then \(y=Tx\) for some \(x\in B_1\). Since

$$ \|-x\|_X=\|x\|_X\lt\frac12, $$

we also have \(-x\in B_1\). By linearity, \(T(-x)=-Tx=-y\), so \(-y\in T(B_1)\). Thus \(T(B_1)\) is symmetric about zero.

This symmetry also holds for its closure. If \(y\in\overline{T(B_1)}\), choose \(y_j\in T(B_1)\) with \(y_j\to y\). By symmetry, \(-y_j\in T(B_1)\), and \(-y_j\to-y\). Hence \(-y\in\overline{T(B_1)}\).

Reflecting the ball \(B_Y(a,\rho)\) through zero therefore gives

$$ B_Y(-a,\rho)\subseteq\overline{T(B_1)}. $$

More explicitly, if \(z\in B_Y(-a,\rho)\), then \(\|-z-a\|_Y=\|z+a\|_Y\lt\rho\). Thus \(-z\in B_Y(a,\rho)\), and symmetry places \(z\) in the closure as well.

7. Show that midpoints remain in the image and its closure.

Let \(u,v\in T(B_1)\). Choose \(x,z\in B_1\) such that \(u=Tx\) and \(v=Tz\). By linearity,

$$ \frac{u+v}{2} =\frac{Tx+Tz}{2} =T\left(\frac{x+z}{2}\right). $$

The triangle inequality gives

$$ \left\|\frac{x+z}{2}\right\|_X \le\frac{\|x\|_X+\|z\|_X}{2} \lt\frac{\frac12+\frac12}{2} =\frac12. $$

Thus \((x+z)/2\in B_1\), so \((u+v)/2\in T(B_1)\).

The same property holds in the closure. Suppose \(u,v\in\overline{T(B_1)}\). Choose sequences \(u_j,v_j\in T(B_1)\) with \(u_j\to u\) and \(v_j\to v\). Every midpoint \((u_j+v_j)/2\) belongs to \(T(B_1)\), and

$$ \frac{u_j+v_j}{2}\to\frac{u+v}{2}. $$

Therefore \((u+v)/2\in\overline{T(B_1)}\).

8. Obtain a ball centered at zero in the closure.

Take any \(y\in Y\) with \(\|y\|_Y\lt\rho\). Then

$$ \|(a+y)-a\|_Y=\|y\|_Y\lt\rho, \qquad \|(-a+y)-(-a)\|_Y=\|y\|_Y\lt\rho. $$

Thus

$$ a+y\in B_Y(a,\rho)\subseteq\overline{T(B_1)}, \qquad -a+y\in B_Y(-a,\rho)\subseteq\overline{T(B_1)}. $$

By the midpoint property, their midpoint also belongs to the closure. That midpoint is

$$ \frac{(a+y)+(-a+y)}{2}=y. $$

Since this holds for every \(y\) with \(\|y\|_Y\lt\rho\), we conclude

$$ B_Y(0,\rho)\subseteq\overline{T(B_1)}. $$

Current State of the Proof

We have proved that every \(y\) with \(\|y\|_Y\lt\rho\) can be approximated arbitrarily closely by images of vectors in \(B_1\). Explicitly, for every \(\epsilon>0\) there exists \(x\in X\) such that

$$ \|x\|_X\lt\frac12, \qquad \|Tx-y\|_Y\lt\epsilon. $$

The remaining step is to obtain an exact preimage of each sufficiently small \(y\) with norm less than \(1\), establishing the original claim \(B_Y(0,\delta)\subseteq T(B_0)\) for some \(\delta>0\).

References

  • SFSU Fourier Analysis by Professor Chun-Kit Lai
  • Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi