# Showing that these inner product induced norms are equivalent

I am working on the following exercise:

Let $H$ be a Hilbert space, $T:H\to H$ be a self-adjoint isomorphism and suppose that $T$ is positive, that is, $\langle Tx,x\rangle\geq 0$ for all $x\in H$. Define $$[x,y]= \langle Tx,y\rangle, x,y\in H,$$ and show that this is an inner product in $H$ an that the norm induced by this inner product and the original norm of $H$ are equivalent.

Let $\|\cdot\|_0$ denote the norm induced by $[\cdot,\cdot]$ and $\|\cdot\|$ denote the original norm of H. I'm struggling to prove that $[x,x]>0$ for all $x\in H\backslash\{0\}$ and $\|\cdot\|\leq C\|\cdot \|_{0}$ for some constant $C>0$. I've done all the rest.

• $T$ is bounded so $[x,x] \leq C \langle x, x \rangle$. Once you know that $x \mapsto [x,x]^{\frac{1}{2}}$ is a norm, that shows the two norms are equivalent, by the "bounded inverse theorem." – fourierwho Jan 26 '18 at 19:31
• @fourierwho I've already shown this inequality. and you can't apply the bounded inverse theorem if you don't know if the space is still complete with the induced norm. – Filburt Jan 26 '18 at 19:36

We have

$$\lVert x\rVert^2 = \langle x,x\rangle = [T^{-1}x,x] \leqslant \lVert T^{-1}x\rVert_0\lVert x\rVert_0 \leqslant K\cdot \lVert T^{-1}x\rVert\cdot \lVert x\rVert_0 \leqslant K\cdot \lVert T^{-1}\rVert_{\text{op}} \lVert x\rVert\cdot\lVert x\rVert_0$$

by the Cauchy-Schwarz inequality for $[\,\cdot\,,\,\cdot\,]$, the inequality $\lVert y\rVert_0 \leqslant K\lVert y\rVert$, and the definition of the operator norm (with respect to $\lVert\,\cdot\,\rVert$). For $x\neq 0$, we can divide by $\lVert x\rVert$ and obtain

$$\lVert x\rVert \leqslant K\cdot \lVert T^{-1}\rVert\cdot \lVert x\rVert_0\,.$$

This inequality is of course also true for $x = 0$.

• Thanks! Now, the part I stucked first was showing that $[\cdot,\cdot]$ is indeed an inner product. Why $[x,x]>0$? – Filburt Jan 26 '18 at 20:31
• That follows from the final inequality. It is clear that $[\cdot,\cdot]$ is a positive semidefinite sesquilinear form, and that gives us Cauchy-Schwarz. Of course we can also prove it in different ways. For example the spectrum of $T$ is contained in $[\varepsilon,\lVert T\rVert_{\text{op}}]$ for some $\varepsilon > 0$, and that implies $\langle Tx,x\rangle \geqslant \varepsilon \lVert x\rVert^2$ by the spectral theory of self-adjoint operators. – Daniel Fischer Jan 26 '18 at 20:36
• But if we throw spectral theory at it, we might as well use the existence of square roots of positive self-adjoint operators and be done with writing $[x,x] = \langle Tx,x\rangle = \langle \sqrt{T}\,x,\sqrt{T}\,x\rangle$. – Daniel Fischer Jan 26 '18 at 20:46
• I was supposed to use only basic spectral theory (for compact and self-adjoint operators). Further, what is a semidefinite sesquilinear form? – Filburt Jan 26 '18 at 21:00
• A sesquilinear form on a complex vector space $E$ is a map $E\times E \to \mathbb{C}$ that is linear in its first argument, and conjugate-linear [or antilinear] in the second. A sesquilinear form $S$ is positive semidefinite if $S(x,x) \geqslant 0$ for all $x\in E$. Positive definite would be $S(x,x) > 0$ for all $x \in E\setminus \{0\}$. So, just like an inner product, except that we don't require $S(x,x) = 0 \implies x = 0$. A positive semidefinite sesquilinear form is Hermitian, i.e. $S(x,y) = \overline{S(y,x)}$ for all $x,y\in E$. – Daniel Fischer Jan 26 '18 at 21:09