Let $(X, \| \cdot \|_X)$ and $(Y, \| \cdot \|_Y)$ be normed vector spaces and assume the map $T \in \mathcal{L}(X,Y)$ is an isomorphism. Define A scalar-valued function $\|\cdot \|_T$ on $X$ by $$\| x \|_T = \| T(x) \|_Y, x\in X$$ Prove that $\| \cdot \|_T$ is a norm on $X$ and show that it is equivalent to the original norm $\| \cdot \|_X.$
My attempt:
For all $x \in X$, $ \| x \|_T = \| T(x) \|_Y \geq 0$ because $\| \cdot \|_Y$ is a norm on $Y$.
Suppose that $\| x \|_T =0.$ By definition, we have $\| T(x) \|_Y = 0.$ Since $\| \cdot \|_Y$ is a norm on $Y$, we have $T(x) = 0.$ Since $T$ is linear and injective, we have $x = 0$.
$\|x + y\|_T = \|T(x+y) \|_Y = \|T(x) +T(y) \|_Y \leq \|T(x) \|_Y + \|T(y)\|_Y = \|x\|_T + \|y\|_T$
$\|\lambda x \|_T = \| T(\lambda x) \|_Y = |\lambda| \|T(x)\|_Y = |\lambda| \|x \|_T.$
Therefore, $\| \cdot \|_T$ is a norm.
Since $T$ is bounded, $\| T \|$ exists.
Therefore, for each $x \in X,$ we have $\| x \|_T = \| T(x) \|_Y \leq \| T \| \|x \|_X.$
As $T$ is an isomorphism, $T^{-1}$ is also bounded and hence $\| T^{-1} \|$ exists.
Therefore, $\| x \|_X = \| T^{-1}(T(x))\| \leq \| T^{-1} \| \| T(x) \|_Y = \|T^{-1} \| \| x \|_T.$ Hence, $\| \cdot \|_T$ is equivalent to $\| \cdot \|_X.$
Is my proof correct?