The Open Mapping Theorem

Every continuous linear surjection $T : \mathbb{R}^n \to \mathbb{R}^m$ is automatically an open map. The argument is almost purely geometric:

1. Surjectivity gives absorbing. The image $T(B)$ of the open unit ball is convex, balanced (symmetric about 0), and absorbing: for every $y \in \mathbb{R}^m$, some scalar multiple $\lambda y$ lies in $T(B)$. This follows immediately from surjectivity: given $y$, pick $x$ with $Tx = y$, then $y/\|x\| = T(x/\|x\|) \in T(B)$.

2. Absorbing + convex + balanced $\implies$ nonempty interior (in finite dimensions). Pick a basis $e_1, \dots, e_m$ of $\mathbb{R}^m$. Since $T(B)$ is absorbing, for each $i$ there exists $\lambda_i > 0$ with $\lambda_i e_i \in T(B)$. By convexity and symmetry, $T(B)$ contains the convex hull of $\{\pm \lambda_1 e_1, \dots, \pm \lambda_m e_m\}$. This is a cross-polytope (a “diamond”), a solid body in $\mathbb{R}^m$ with nonempty interior.

3. Nonempty interior at the origin $\implies$ openness. Once $T(B)$ contains an open ball around 0, linearity propagates this to all open sets: $T$ maps every open set to an open set.

The deeper reason this works is compactness of the unit ball (Heine–Borel). Compact $\to$ continuous images are compact $\to$ compact convex sets with the right spanning properties have interior. Everything closes cleanly because there are only finitely many directions to account for.

By Riesz’s lemma, the closed unit ball in an infinite-dimensional Banach space is never compact, destroying the finite-dimensional argument at its root. In finite dimensions, compressing finitely many coordinate directions by positive factors always leaves a solid body with nonempty interior; in infinite dimensions, compressions $\lambda_n \to 0$ can crush the image to a set with empty interior, even though every finite-dimensional projection still looks fine.

A concrete example¶

Let $c_{00}$ be the space of real sequences with only finitely many nonzero terms, equipped with the sup-norm $\|x\|_\infty = \sup_k |x_k|$. As we saw in the Banach-Steinhaus discussion, $c_{00}$ is not complete.

Define $T : c_{00} \to c_{00}$ by

This is bounded ($\|Tx\|_\infty \leq \|x\|_\infty$). It compresses each coordinate direction by a different factor: the $n$-th direction is scaled by $\lambda_n = 1/n$, so the compressions tend to zero as $n \to \infty$.

Bijective on $c_{00}$: Injectivity is immediate: if $Tx = 0$ then $x_n/n = 0$ for all $n$, so $x = 0$. For surjectivity, given $y \in c_{00}$, set $x_n = ny_n$. Since $y$ has finite support, so does $x$, hence $x \in c_{00}$ and $Tx = y$.

Not open: For any $\varepsilon > 0$, the vector $\frac{\varepsilon}{2} e_k$ lies in $B_\varepsilon(0)$. Its unique preimage under $T$ is $\frac{k\varepsilon}{2} e_k$, which has norm $k\varepsilon/2 \to \infty$ as $k \to \infty$. So no matter how small $\varepsilon$ is, $B_\varepsilon(0)$ contains points whose preimages escape any fixed ball. Equivalently, $T(B)$ is the set $\{y \in c_{00} : |y_k| < 1/k\}$, an “infinite-dimensional box” that gets narrower in each successive direction. No sup-norm ball fits inside it.

What happens on the completion? On $c_0$, the map $T$ is no longer surjective: $y = \{1/n\} \in c_0$ requires $x_n = 1$ for all $n$, but the constant sequence $\{1, 1, 1, \ldots\} \notin c_0$. So surjectivity fails on the completion, and the OMT hypothesis is no longer met.

(1) $\iff$ (3): by the definition of surjectiveness.

(2) $\implies$ (1): If $L$ is open, then $L(B_X(0, 1))$ is an open set in $Y$ containing $L(0) = 0$. Thus there exists $r > 0$ such that $B_Y(0, r) \subset L(B_X(0, 1))$. Now pick any $y \in Y$ and any scalar $t > 0$, we can use the linearity of $L$ in the following way, $L(B_X(0, t)) = L(t B_X(0, 1)) = t L(B_X(0, 1))$.

Now since $B_Y(0, r) \subset L(B_X(0, 1))$ we also have that $B_Y(0, tr) = t B_Y(0, r) \subset L(B_X(0, t))$. Now set $t = 2 ||y|| / r$ then $y \in B_Y(0, 2 ||y||) \subset L(B_X(0, 2 ||y||/r))$, which means that there exists an $x \in B_X(0, 2||y||/r)$ such that $Lx = y$, further note that we have $||x|| < 2||y|| / r$.

(2) $\implies$ (4): If $L$ is open, then $B_Y(0, r) \subset L(B_X(0, 1))$ for some $r > 0$. Pick $y \in Y$ then $\tilde{y} := \frac{y}{||y||}\frac{r}{2} \in B_Y(0, r)$, so there exists $z \in B_X(0, 1)$ such that $Lz = \tilde{y}$. Define $x = \frac{2||y||}{r} z$, which by linearity gives us $Lx = y$. Now since $||z|| < 1$ we have $||x|| < \frac{2||y||}{r}$. Now define $C := \frac{2}{r}$ and we have $||x|| \leq C ||y||$ for any $y \in Y$.

(4) $\implies$ (2): Assume that for every $y\in Y$ there exists $x \in X$ such that $Lx = y$ and $||u||\leq C||y||$. We begin by showing that $L$ maps the unit ball to an open set. Let $y \in Y$ with $||y|| < \frac{1}{C}$, then there exists $x \in X$ such that $Lx = y$ and $||x|| < C ||y|| < 1$. This means that $y \in L(B_X(0,1))$. Thus we have shown that $B_Y(0, \frac{1}{C})\subset L(B_X(0, 1))$, making $L(B_X(0, 1))$ an open set. By linearity of $L$ this means that $L$ maps all open sets to open sets, therefore $L$ is open.

(4) $\implies$ (5): If it holds everywhere it will hold for a dense set.

(5) $\implies$ (4): Let $E \subset Y$ be the dense set, and let $y \in Y$ then want to approximate $y$ using elements in $E$.

• Step 1: Set the initial residual $r_0 = y$, then find $y_1 \in E$ so that $||y_1 - r_0|| < \varepsilon = \frac{1}{2} ||y||$.

• Step 2: Set the second residual $r_1 = r_0 - y_1 = y - y_1$, and find $y_2 \in E$ so that

• Step 3: At the $j$-the step we define $r_j = r_{j-1} - y_j$ where $y_j \in E$ with

Note that the sequence of residuals has $||r_j|| \to 0$ and thus $r_j \to 0$, and note that $y_j = r_{j-1} - r_j$ so if we sum the $y_j$s we find a telescoping series i.e.

Finally we show that we can life the inequality from the dense set to everywhere. Since $y_j \in E$ we have there exists $x_j : Lx_j = y_j$ with $||x_j|| \leq C ||y_j||$. Then

Since $y_j\in E$ we have that

Hence we can define $x:= \sum_{j=1}^{\infty} x_j$. Finally

Thus we have that $Lx = y$ and similarly find that $||x|| \leq 4C ||y||$.

(3) $\implies$ (4): The strategy is to apply the Theorem 1. Since by assumption $\forall y \in Y$ there exists $x \in X$ such that $Lx = y$ the following $E_n$ form a cover of $Y$.

Since $Y$ is complete, Baire’s theorem implies that at least one of the $E_n$’s is not nowhere dense. Say $E_m$ is the not nowhere dense set, meaning $E_m$ is dense in some ball $B(y_0, r)$ i.e. $\overline{E_m \cap B(y_0, r)} = B(y_0, r)$. We take advantage of this twice to find solution approximations in $E_m$.

Construct approximate solution: Let $\varepsilon > 0$ then for any $y\in B(y_0, r)$ there exists $\hat{y} \in E_m$ such that $||y - \hat{y}|| < \varepsilon$, and there exists $\hat{x}$ such that $L\hat{x} = \hat{y}$ with $||\hat{x}|| \leq m ||\hat{y}||$ and $||y - L\hat{x}|| < \varepsilon$ (in other words $\hat{x}$ is solution approximation of $Lx = y$), and we have

Map ball of right hand sides to the origin. Pick another $y' \in B(y_0, r)$ then we can find another solution approximation $\tilde{x}$ with the same properties above. Then note that $y - y' \in B(0, 2r)$ and that $x = \hat{x} - \tilde{x}$ is a solution approximation with $||Lx - (y - y')|| < 2\varepsilon$ and $||x|| \leq 2n(||y_0| + r + \varepsilon)$.

Scaling argument. Next pick any nonzero $y \in Y$, define $\lambda = \frac{r}{||y||}$ then $\lambda y \in B(0, 2r)$ and there is an approximate solution $u' \in X$. Define $u = \lambda u'$ and pick $\varepsilon = r / 4$, then $u$ satisfies $||Lu - y|| < \frac{1}{2} ||y||$ and $||u|| < K ||y||$ where $K:= 2n(5r/4 + ||y_0||) / r$ a constant.

Iterate. Pick $y_0 \in Y$, find $u_1 \in X$ by the previous step, then iterate by defining the residual $y_1 = y_0 - Lu_1$, and solve for $u_2 \in X$ such that $||Lu_2 - y_1|| < \frac{1}{2}||y_1||$. We again obtain a convergent series, and we can check similarly to before that its limit satisfies the required properties.