# The Translation Action on Function Spaces

## Preliminaries

Here's something I learned from Chongchun Zeng of Georgia Tech, while chatting over some hors-d'oeuvres at the AMS Sectional Meeting.

Observation
For most Hilbert/Banach spaces that one uses in analysis of PDEs (think the Lebesgue/Sobolev/Hölder spaces over $\mathbb{R}^d$), partial differentiation is an unbounded linear operator.

Now, let $y$ be a unit vector in $\mathbb{R}^d$. Denote by $T_y(t)$ the one-parameter family of translation operators $x \mapsto x + ty$ on $\mathbb{R}^d$. Let $H$ be a translation-invariant Hilbert (Banach) space over $\mathbb{R}^d$. Then $T_y(t)$ induces by pullback a one-parameter family of actions $T_y(t)^\star: H \to H$ given by $$(T_y(t)^\star\phi)(x) = \phi\circ T_y(t).$$ By definition $T_y(t)^\star$ is an isometry.

For most reasonable $H$ (such as the Lebesgue spaces), we know that $\lim_{t\to 0} \| T_y(t)^\star f - f\| = 0$. This implies that the mapping (for fixed $y$): $\mathbb{R}\times H \to H: (t,\phi) \mapsto T_y(t)^\star \phi$ is a continuous mapping. (However, this mapping is in general not uniformly continuous even on bounded subsets.)

Example
Let $H = L^1(\mathbb{R})$. Let $\delta > 0$. Then if we consider the step function $\chi$ that equals $\delta^{-1}$ on the interval $[0,\delta]$ and $0$ otherwise, we see that $\|T(\delta)^\star \chi - \chi\| = 2$, $\|\chi\| = 1$.

The one parameter family is formally generated by a vector field $X$ on $H$, where $X$ is formally $X|_f = \lim_{t\to 0} \frac{1}{t}( T_y(t)^\star f - f)).$ Unpacking the definitions we have that $$X|_f = \partial_y f.$$ For generic functions $f\in H$, as $\partial_y$ is unbounded, this implies that the vector field $X$ is unbounded even on bounded open subsets of $X$.
In other words, we have

Proposition
For most translation-invariant Hilbert (Banach) spaces, the one-parameter family of actions $T_y(t)^\star$ is not differentiable at most points. In other words, for generic $f$, the curve $\mathbb{R} \ni t \mapsto T_y(t)^\star f$ is a nowhere differentiable curve in $H$.

On the other hand, when $f \in H$ is such that $\partial_y f \in H$, then the curve $t\mapsto T_y(t)^\star f$ is a $C^1$ curve in $H$.

## A geometric consequence

For convenience let us work with $H = L^2(\mathbb{R})$, but the same statements work for arbitrary dimensions and many other Hilbert spaces.

Fix $f\in C^\infty \cap H$. Our arguments above shows that the curve $\gamma: t \mapsto T(t)^\star f$ is a smooth curve in $H$. For dynamical problems on $H$, it is often convenient to study the dynamics near $\gamma$ by parametrizing via the normal bundle of $\gamma$. This allows us to split the problem into a finite dimensional part about motion along $\gamma$, and an infinite dimensional part orthogonal to $\gamma$.

The naive thing to do may be the following: given $t \in \mathbb{R}$, the tangent space $\mathsf{T}_{\gamma(t)} H$ (which is isomorphic to $H$ since $H$ is a linear space) can be decomposed as the direct sum $$\mathsf{T}_{\gamma(t)} H = \mathsf{T}_{\gamma(t)} \gamma \oplus P_t$$ where $P_t$ is the orthogonal complement to $\mathsf{T}_{\gamma(t)}\gamma$. This latter tangential space is given by the span of $T(t)^\star (f')$, where $f'$ denotes the derivative of $f$. And hence it is very natural and tempting to describe $$P_t = T(t)^\star( P_0) = T(t)^\star( {f'}^\perp ).$$

This description can also be interpreted as describing the normal bundle of $\gamma$ by the trivialization $\mathbb{R} \times P_0$, with the coordinate mapping $$\mathbb{R}\times P_0 \ni (t, g) \mapsto T(t)^\star f + T(t)^\star g.$$ However, we see immediately a problem: for generic $g\in P_0$, the constant $g$ curve $\mathbb{R}\times \{g\}$ is not differentiable anywhere, and hence we have a horribly irregular coordinate system.

So what is a better description? In the case of $H$ being a Hilbert space, we can take advantage of the Riemannian structure. First, consider the mapping $$\mathbb{R}\times H \ni (t,\psi) \mapsto F(t,\psi) := \langle \psi - \gamma(t), \gamma'(t) \rangle.$$ Provided $\gamma(0) \in C^\infty \cap H$ is non-trivial, we have that $\gamma'(0) \not\equiv 0$. One easily sees that $F(t, \gamma(t)) = 0$. The points $(t,\gamma(t))$ are however non-critical: the partial derivative relative to the first ($\mathbb{R}$) component of $F$ is easily computed to be $\partial_1 F(t,\psi) = - \langle \gamma'(t), \gamma'(t) \rangle + \langle \psi - \gamma(t), \gamma''(t) \rangle.$ Evaluated at $(t, \gamma(t))$ we see that $\partial_1 F(t,\gamma(t)) = - \langle \gamma'(t), \gamma'(t)\rangle \neq 0$. Hence by the implicit function theorem, there exists a real-valued smooth submersion $s$ defined on a tubular neighborhood of $\gamma$ such that $F(s(\psi),\psi) =0$.

The smoothness of our normal bundle assured, it remains to find a better coordinate system.

In general, when $\gamma$ may be higher dimensional, one approach is to treat the fibers of the normal bundle $P_t$ as finite codimensional subspaces of $H$. When performing computations, however, one should not use "partial differentiation relative to $t$". Instead, one should use the "covariant derivative". Let $\Xi$ be a smooth section of the normal bundle, which we can represent as a $H$-valued function of $t$ such that $H(t)\in P_t$. Its covariant derivative can be defined by the usual projection-based formula: $$\nabla_t \Xi(t) = \partial_t \Xi(t) - \frac{1}{|\gamma'|^2} \langle \partial_t \Xi(t), \gamma'(t) \rangle \gamma'(t).$$

In our specific example, where $\gamma$ is one dimensional, we can in fact use parallel transport with the above connection to define a coordinate system for the normal bundle. Given $g\in P_0$, we let $g(t)$ be the solution to the system $\nabla_t g = 0, \langle g(t), \gamma'(t) \rangle = 0.$ The second equation implies that $\langle g'(t), \gamma'(t) \rangle + \langle g(t), \gamma''(t) \rangle = 0$ and hence the first equation can be written as the ordinary differential equation $$\label{eq:ode:trans} g'(t) + \frac{1}{|\gamma'|^2} \langle g(t), \gamma''(t) \rangle \gamma'(t) = 0 .$$ This defines a coordinate system for the normal bundle via the assignment $$\mathbb{R}\times P_0 \ni (t,g) \mapsto (t,g(t)).$$ Notice that the coefficients of \eqref{eq:ode:trans} are smooth (in $t$) by our assumption that $\gamma(0)$ is a smooth function and hence $\gamma(t)$ is a smooth curve. This implies that solutions $g(t)$ are actually infinitely differentiable in $t$.

Let's do an explicit example. Suppose $\gamma(0)$ is some even $C^\infty_0$ function, which we denote by $f$. For convenience assume $f'$ has $L^2$ norm 1. Then any even $L^2$ function $g$ is in $P_0$, since $\gamma'(0)$ is odd. The naive parametrization will send $(t,g)$ to $f(x + t) + g(x + t)$. The parallel transport parametrization will first ask us to solve the integral-differential equation $\partial_t g(t,x) = -\int g(t,y) f''(y + t) ~dy \cdot f'(x+t)$ with initial data $g(0,x) = g(x)$. Then the parametrization sends $(t,g)$ to $f(x+t) + g(t,x)$. It is clear that if $g(x)$ is irregular, than the naive parametrization would not be smooth. On the other hand, with the parallel transport parametrization one sees directly that the parametrization is smooth, especially considering that parallel transport preserves norms so that $g(t,x)$ is bounded as an element of $H$ uniformly in $t$.

##### Willie WY Wong
###### Assistant Professor

My research interests include partial differential equations, geometric analysis, fluid dynamics, and general relativity.