# Is every norm preserved under a unitary map?

1. Feb 22, 2014

### Bipolarity

I am a bit confused, so this question may not make much sense.

A unitary operator from one vector space to another is one whose inverse and Hermitian transpose are identical.

It can be proved that unitary operators are norm preserving and inner product preserving. Which raises the question, which norm/inner product does this refer to, since the unitarity of the map was not defined with respect to an inner product?

In particular, I ask this question in application to signal processing.
Consider the space of functions that are integrable and square-integrable from -∞ to ∞. This space is a vector space, and also an inner product space under the standard function inner product. The norm induced by this inner product is the $L^{2}$ norm. It can be shown to be preserved under the Fourier transform, a result known as Parseval's theorem.

But does the Fourier transform preserve the $L^{1}$ norm, which does not appear to be induced by any inner product that I know of?

Thanks!

BiP

2. Feb 22, 2014

### mathman

Fourier transform of L1 function is bounded, with the bound ≤ norm of the original function. In other words the Fourier transform of an L1 function is L∞.

3. Feb 22, 2014

### D H

Staff Emeritus
That's one way to define a unitary operator. Note well: It doesn't make sense to talk about UU* without having some concept of an inner product. Another way to define a unitary operator is as a surjective operator that preserves the inner product. The two definitions are equivalent.

Unitary operators preserve the inner product, which means they preserve the norm induced by that inner product. They do not preserve all norms. A simple example: Consider ℝ2. Rotate $\hat x$ by 45 degrees to $\frac{\sqrt{2}}2 (\hat x + \hat y)$. The Euclidean norm (L2 norm) of that rotated unit vector is still one, but the taxicab norm (L1 norm) has changed from one to √2, and the max norm (L norm) has changed from one to √2/2.