# Sufficient condition in proving open mapping theorem for Banach spaces

Statement of the open mapping theorem: Let $X, Y$ be Banach spaces, $T:X\to Y$ be a continuous linear transformation. If T is onto, then T is an open map (i.e. for any open set $O\subset X$, $T(O)\subset Y$ is open.

[EDITED] Claim: It suffices [in proving that T is an open map] to show that $T(B_X(1))$ contains an open ball centered at the origin, where $B_X(1)$ is the open ball of radius 1 centered at the origin in $X$.

I'm not seeing how this condition implies that all open sets map to open sets. Linearity of T seems to be important here, but nothing is coming to mind except the definition $T(af+bg)=aT(f)+bT(g)$ for $a,b\in\mathbb{R}$, $f,g\in X$.

Would greatly appreciate some insight. Thanks!

Suppose the condition satisfied, consider $U$ an open subset of $X$, and $x\in U$, there exists $c\in R$ such that $x+cB(0,1)\subset U$. $T(x+cB(0,1))=T(x)+cT(B(0,1))$ since $T$ is linear, this implies that $T(U)$ is open since $T(B(0,1))$ is open.

• Sorry I may have been unclear: I want to show that $T(B_X(1))$ containing an open ball centered at the origin implies that $T$ is an open map. If I am following your argument correctly, the last part presupposes that $T$ is an open map - "$T(B(0,1))$ is open." Nov 29 '15 at 17:11

I had the same question, so I will contribute the answer I found useful.

Consider the note on p. 162 of Folland's Real Analysis: Modern Techniques and Their Applications. The first statement is:

If $X$ and $Y$ are metric spaces, [the requirement that $f: X \to Y$ be open] amounts to requiring that if $B$ is a ball centered at $x \in X$ then $f(B)$ contains a ball centered at $f(x)$.

Clearly if $f$ is an open map this is satisfied. Now we want to show that this statement implies that $f$ is open. Let $U \subset X$ be any open set: we want to show this property implies $f(U)$ is open. However, by openness of $U$ we have that for any $x \in U$ we can construct a ball $B_{x}$ with $x \in B_{x} \subset U$, and by assumption we can construct a corresponding ball $B_{x}'$ such that $f(x) \in B_{x}' \subset f(B_{x}) \subset f(U)$. But then $f(x)$ is contained within an open ball contained within $U$. Since the point $x \in U$ (and corresponding $f(x)\in f(U)$) is arbitrary, we conclude that $U$ is open.

The second statement is:

Specializing still further, if $X$ and $Y$ are normed linear spaces and $f$ is linear, then $f$ commutes with translations and dilations; it follows that $f$ is open iff $f(B)$ contains a ball centered at $0$ in $Y$ when $B$ is the ball of radius $1$ about $0$ in $X$.

If $f$ is open then by the Folland's first statement above we have that $f(B)$ contains a ball centered at $f(0)$. But $f(0)=0$ (by homogeneity of $f$, $f(\mathbf{0}) = f( 0 \cdot \mathbf{v}) = 0 f( \mathbf{v}) = 0$). Conversely, let $U$ be any open set and suppose the condition in Folland's second statement is satisfied. Let $x \in U$, and let $B'$ be a ball centered at $x$ with $B' \subset U$ (by openness of $U$). Note that $B'$ can be constructed by translation/dilation of $B$; e.g. $B' = x + c B$. By Folland's first statement it suffices to show that $f(B')$ contains a ball centered at $f(x)$. By linearity of $f$ we have that $f(B') = f(x) + c f(B)$, so that actually it suffices to prove that $f(B)$ contains a ball centered at $f(0)=0$. This last part is satisfied by assumption, so we conclude $f(U)$ is open.

Yes it is sufficient, because it implies that image of any open ball has nonempty interior.

• Why couldn't $T(B_X(1))$ be a closed ball, or something else that is not itself open but does have nonempty interior? Nov 28 '15 at 21:41
• If $S\subset X$ is open then $S=\bigcup_{x\in S} B(x, r_x )$ hence $T(s) =\bigcup_{x\in S}\mbox{Int} T(B(x, r_x ))$ is open. Nov 28 '15 at 21:45