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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