Proof of Transport of an Inner Product and of Hilbert Space Structure along a Linear Bijection

Each result cited is universally quantified over the data in its own statement. Differences and negatives in $X$ and in $E$ are those of Elementary Identities in a Vector Space: $x-y=x+(-y)$ and $(-1)y=-y$ by claims 2 and 5 of that lemma. By claim 1 of Inverse of a Bijection, $T^{-1}(Tx)=x$ for every $x\in X$ and $T(T^{-1}e)=e$ for every $e\in E$; and by Bijection of Sets, every $e\in E$ has exactly one preimage under $T$, namely $T^{-1}e$.

Two preliminary facts. First, $T0_{X}=0_{E}$: by claim 3 of Elementary Identities in a Vector Space, $0_{X}=0\cdot0_{X}$, so by the linearity of $T$, condition 2 of Linear Map, $T0_{X}=0\cdot T0_{X}=0_{E}$, the last step by claim 3 again, in $E$. Second, $T$ preserves differences: $T(x-y)=T(x+(-1)y)=Tx+(-1)Ty=Tx-Ty$ by the two conditions of Linear Map.

Claim 1. We verify the four conditions of Real Inner Product Spaces (i) for $x,y,z\in X$ and $\lambda\in\mathbb{R}$, using the same conditions for $\langle\cdot,\cdot\rangle_{E}$.

(a) $\langle x,y\rangle_{X}=\langle Tx,Ty\rangle_{E}=\langle Ty,Tx\rangle_{E}=\langle y,x\rangle_{X}$.

(b) $\langle x+y,z\rangle_{X}=\langle T(x+y),Tz\rangle_{E}=\langle Tx+Ty,Tz\rangle_{E}=\langle Tx,Tz\rangle_{E}+\langle Ty,Tz\rangle_{E}=\langle x,z\rangle_{X}+\langle y,z\rangle_{X}$, by condition 1 of Linear Map.

(c) $\langle\lambda x,y\rangle_{X}=\langle T(\lambda x),Ty\rangle_{E}=\langle\lambda Tx,Ty\rangle_{E}=\lambda\langle Tx,Ty\rangle_{E}=\lambda\langle x,y\rangle_{X}$, by condition 2 of Linear Map.

(d) $\langle x,x\rangle_{X}=\langle Tx,Tx\rangle_{E}$, which is nonnegative, and which vanishes only if $Tx=0_{E}$. In that case $Tx=T0_{X}$ by the first preliminary fact, so $x=0_{X}$ because $0_{E}$ has exactly one preimage under $T$.

Hence $\langle\cdot,\cdot\rangle_{X}$ is an inner product on $X$. By Real Inner Product Spaces (ii), $|x|_{X}$ is the unique nonnegative real number whose square is $\langle x,x\rangle_{X}=\langle Tx,Tx\rangle_{E}=(|Tx|_{E})^{2}$; since $|Tx|_{E}$ is nonnegative with that square, $|x|_{X}=|Tx|_{E}$. By Real Inner Product Spaces (iii) and the second preliminary fact, $d_{X}(x,y)=|x-y|_{X}=|T(x-y)|_{E}=|Tx-Ty|_{E}=d_{E}(Tx,Ty)$.

Claim 2. Let $a,b\in E$ and $\lambda\in\mathbb{R}$, and put $x=T^{-1}a$, $y=T^{-1}b$. Then $T(x+y)=Tx+Ty=a+b$ and $T(\lambda x)=\lambda Tx=\lambda a$ by Linear Map, so $x+y$ is the unique preimage of $a+b$ and $\lambda x$ that of $\lambda a$: $T^{-1}(a+b)=T^{-1}a+T^{-1}b$ and $T^{-1}(\lambda a)=\lambda T^{-1}a$, which is the linearity of $T^{-1}$. Finally $\langle T^{-1}a,T^{-1}b\rangle_{X}=\langle T(T^{-1}a),T(T^{-1}b)\rangle_{E}=\langle a,b\rangle_{E}$.

Claim 3. Suppose $E$ is a real Hilbert space. By claim 1 and The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity (iii), $d_{X}$ is a metric on $X$. Let $(x_{k})_{k\in\mathbb{N}}$ be a Cauchy sequence in $(X,d_{X})$. Since $d_{E}(Tx_{k},Tx_{\ell})=d_{X}(x_{k},x_{\ell})$ for all $k,\ell$ by claim 1, the sequence $(Tx_{k})_{k\in\mathbb{N}}$ satisfies the Cauchy condition in $(E,d_{E})$ with the same $N$ for each $\varepsilon$, so it is a Cauchy sequence in $(E,d_{E})$. Since $(E,d_{E})$ is complete by Real Hilbert Spaces (i), the sequence $(Tx_{k})$ converges to some $e\in E$. Put $x=T^{-1}e$, so that $Tx=e$. Then $d_{X}(x_{k},x)=d_{E}(Tx_{k},Tx)=d_{E}(Tx_{k},e)$ for every $k$, by claim 1, so for every real $\varepsilon>0$ the $N$ furnished by the convergence of $(Tx_{k})$ to $e$ gives $d_{X}(x_{k},x)<\varepsilon$ for all $k\ge N$: the sequence $(x_{k})$ converges to $x$ in $(X,d_{X})$. Hence $(X,d_{X})$ is complete, and $X$ with $\langle\cdot,\cdot\rangle_{X}$ is a real Hilbert space by Real Hilbert Spaces (i).

Claim 4. By claim 2, for all $i,j\in\mathbb{N}$, $\langle T^{-1}e_{i},T^{-1}e_{j}\rangle_{X}=\langle e_{i},e_{j}\rangle_{E}$, which is $0$ for $i\ne j$ since $(e_{k})$ is an orthonormal sequence; and $|T^{-1}e_{i}|_{X}=|T(T^{-1}e_{i})|_{E}=|e_{i}|_{E}=1$ by claim 1 and the preliminary identity $T(T^{-1}e)=e$. So $(T^{-1}e_{k})_{k\in\mathbb{N}}$ is an orthonormal sequence in $X$. Let $x\in X$ satisfy $\langle x,T^{-1}e_{k}\rangle_{X}=0$ for every $k\in\mathbb{N}$. Then $\langle Tx,e_{k}\rangle_{E}=\langle Tx,T(T^{-1}e_{k})\rangle_{E}=\langle x,T^{-1}e_{k}\rangle_{X}=0$ for every $k$, so $Tx=0_{E}$ by Orthonormal Basis of a Real Hilbert Space (i) applied to the orthonormal basis $(e_{k})$ of $E$, and hence $x=0_{X}$ as in condition (d) of claim 1. By Orthonormal Basis of a Real Hilbert Space (i), applied in the real Hilbert space $X$ of claim 3, $(T^{-1}e_{k})_{k\in\mathbb{N}}$ is an orthonormal basis of $X$.