I have encountered two different versions for the First Isomorphism Theorem in the context of Normed/Banach spaces, I wanted to ask whether one implies the other.

Theorem 1: Let $X,\,Y$ be normed spaces and suppose $T:X \rightarrow Y$ is a linear map such that $T(B_X)=B_Y$ where $B_X,B_Y$ are the corresponding open unit balls. Then $\frac{X}{ker(T)} \simeq Y$ are isometrically isomorphic.

Let me note that this version can be proved directly from definitions without using some sort of heavy hammer. Also note that $T$ is automatically bounded and surjective and of norm $1$.

Theorem 2: Let $X, Y$ be Banach spaces. $T:X \twoheadrightarrow Y$ A bounded surjective operator. Then $\frac{X}{ker(T)} \simeq Y$ are isomorphic (i.e. have a linear homeomorphism between them).

I claim that Theorem 1 implies Theorem 2: Let $X, Y, T$ be as above. Define a new norm $\Vert \cdot \Vert_T$ on $X$ as follows:

$$ \Vert x \Vert_T = \Vert Tx \Vert_Y \leq \Vert T \Vert \Vert x \Vert_X$$

So this inequality must also hold in the quotient space(I use the same notation for the quotient norm):

$$\forall x \; \Vert x + ker(T) \Vert_T \leq \Vert T \Vert \Vert x + ker(T) \Vert_X$$

But by the Open Mapping Theorem: $T$ is an open map and thus There exists $r > 0$ s.t. $\Vert Tx \Vert_Y \leq r \implies \Vert x + ker(T) \Vert_X \leq 1$. So for $x \notin ker(T)$, we have $$ \Vert T\big( r \frac{x}{\Vert Tx \Vert_Y}\big) \Vert_Y \leq r \implies \Vert r \frac{x}{\Vert Tx \Vert_Y} + ker(T) \Vert_X \leq 1$$

$$\implies \Vert x + ker(T) \Vert_X \leq \frac{1}{r}\Vert x \Vert_T$$

So this gives us in the quotient space:

$$\forall x \; \Vert x + ker(T) \Vert_X \leq \frac{1}{r} \Vert x + ker(T) \Vert_T$$ From these two inequalities we conclude that $\Vert \cdot \Vert_X,\,\Vert \cdot \Vert_T$ are equivalent norms on $\frac{X}{ker(T)}$ (the identity is a homeomorphism). Note that now $T(B_{\Vert \cdot \Vert_T}) = B_Y$ as a direct consequence of the definition of $\Vert \cdot \Vert_T$. So by Theorem 1, $(\frac{X}{kerT}, \Vert \cdot \Vert_T) \simeq (Y, \Vert \cdot \Vert_Y)$ are isometrically isomorphic and thus isomorphic, but the norms are equivalent and we are done.

is the preceding proof correct?

  • $\begingroup$ I'd rather say that open mapping theorem implies theorem 2. $\endgroup$ – Eclipse Sun Jan 28 at 17:19
  • $\begingroup$ @EclipseSun It does (I have just demonstrated that here), I just wanted to make sure that Theorem 1 is connected to Theorem 2, since they are very similar, But turns out one is stronger. $\endgroup$ – pitariver Jan 28 at 17:26
  • $\begingroup$ What if $x\ne0$ but $Tx=0$? Then the second inequality in your proof is plainly wrong... $\endgroup$ – Vladimir Jan 28 at 17:26
  • $\begingroup$ @Vladimir I fixed the norm equivalence to only on the quotient. Thanks for pointing that out. $\endgroup$ – pitariver Jan 28 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.