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Oxymoron
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Im trying to prove the following proposition
Let [itex](X,\|\cdot\|_X)[/itex] and [itex](Y,\|\cdot\|_Y)[/itex] be normed vector spaces and let [itex]T:X \rightarrow Y[/itex] be a surjective linear map.
Then [itex]T[/itex] is an isomorphism if and only if there exist [itex]m,M > 0[/itex] such that
[tex]m\|x\|_X \leq \|Tx\|_Y \leq M\|x\|_X \quad \forall \, x \in X[/tex]
For the forward inclusion, I supposed that [itex]T[/itex] is an isomorphism and I want to prove that [itex]m,M > 0[/itex] exist such that
[tex]m\|x\|_X \leq \|Tx\|_Y \leq M\|x\|_X \quad \forall \, x \in X[/tex]
Now since T is an isomorphism I know three things...
1) [itex]T[/itex] is bijective
2) [itex]T[/itex] bounded
3) [itex]T^{-1}[/itex] is bounded
So I take any [itex]x \in X[/itex]. Then since [itex]T[/itex] is linear and bounded...
[tex]\|Tx\|_Y \leq M\|x\|_X[/tex]
for some [itex]M > 0[/itex]. My question is how do I show that [itex]\|Tx\|_Y \geq m\|x\|_X[/itex] at the same time? Do I use the inverse [itex]T^{-1}[/itex]?
Let [itex](X,\|\cdot\|_X)[/itex] and [itex](Y,\|\cdot\|_Y)[/itex] be normed vector spaces and let [itex]T:X \rightarrow Y[/itex] be a surjective linear map.
Then [itex]T[/itex] is an isomorphism if and only if there exist [itex]m,M > 0[/itex] such that
[tex]m\|x\|_X \leq \|Tx\|_Y \leq M\|x\|_X \quad \forall \, x \in X[/tex]
For the forward inclusion, I supposed that [itex]T[/itex] is an isomorphism and I want to prove that [itex]m,M > 0[/itex] exist such that
[tex]m\|x\|_X \leq \|Tx\|_Y \leq M\|x\|_X \quad \forall \, x \in X[/tex]
Now since T is an isomorphism I know three things...
1) [itex]T[/itex] is bijective
2) [itex]T[/itex] bounded
3) [itex]T^{-1}[/itex] is bounded
So I take any [itex]x \in X[/itex]. Then since [itex]T[/itex] is linear and bounded...
[tex]\|Tx\|_Y \leq M\|x\|_X[/tex]
for some [itex]M > 0[/itex]. My question is how do I show that [itex]\|Tx\|_Y \geq m\|x\|_X[/itex] at the same time? Do I use the inverse [itex]T^{-1}[/itex]?
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