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Consider the following question: Given a Lipschitz map $f$ between two Banach spaces $X$ and $Y$ with certain properties, can we construct a linear map with the same properties? In this article, we shall demonsrate that in the case of Lipschitz embeddings, the answer is positive, subject to a condition on $Y$. Most of the results here are adapted from Topics in Banach Space Theory by Albiac and Kalton, although some of the arguments have been simplified somewhat.

Definition. Let $X$ and $Y$ be Banach spaces and take $A\subset X$ open. A map $f:X\rightarrow Y$ is said to be Gâteaux differentiable at a point $x\in A$ if there exists a bounded linear map $T:X\rightarrow Y$ such that for every $u\in X$,

We say that $f$ is Fréchet differentiable at $x\in A$ if the above limit holds uniformly for $u\in B_X$.

Idea. Suppose $f:X\rightarrow Y$ is a Lipschitz embedding. Then if $x_0\in X$ and $t>0$,

so if $f$ is Gâteaux differentiable at $x_0$ then by dividing through by $t$ and taking limits we obtain that $D_f(x_0):X\rightarrow Y$ is a linear embedding, as desired.

Before we begin, we need to generalise the notion of the Lebesgue integral to functions whose range is a Banach space, rather than the real or complex numbers. This is done with the Bochner Integral. We shall not discuss the definition here as it is defined using simple functions, more or less the same as in the real case. However, one key difference is that the Radon-Nikodym theorem does not hold in general here, motivating us to define the following:

Definition. Let $X$ be any Banach space and $(E, \mathcal{A}, \mu)$ be a measure space. We say that $f:E\rightarrow X$ is strongly measurable or Bochner measurable if there exists a measurable function $g:E\rightarrow X$ such that $g$ has separable range and $g=f$ almost everywhere.

Definition. A Banach space $X$ is said to have the Radon Nikodym property (RNP) if for every bounded linear map $T:L_1[0,1]\rightarrow X$, there is a bounded and strongly measurable function $g:[0,1]\rightarrow X$ such that $T(f)=\int_{0}^{1}f(t)g(t)dt$ for $f\in L_1[0,1]$.

The following results regarding the Bochner integral are important, though the proofs do not add much to the current context so may be skipped if desired.

Theorem. A Banach space $X$ has the RNP if and only if every Lipschitz map from $[0,1]$ into $X$ is differentiable almost everywhere.

Proof. Suppose every Lipschitz map from $[0,1]$ into $X$ is differentiable almost everywhere, and let $T:L_1[0,1]\rightarrow X$ be a bounded linear map. Define $f$ from $[0,1]$ into $X$ by $f(t) = T(\chi_{[0,t]})$. By assumption, $f'$ exists almost everywhere on $[0,1]$; furthermore, since $f$ is Lipschitz, $f'$ is bounded and $f'[0,1]$ is separable, i.e. $f'$ is strongly measurable. For every $x^*\in X^*$, we can use the FTC and linearity of $x^*$ to obtain

so $T(g) = \int_0^tf'(s)g(t)ds$ for every $g$ of the form $\chi_{[0,t]}$, $t\in [0,1]$. But from these we can obtain all the indicator functions of intervals, and so by linearity and continuity we get the result for every $g\in L_1[0,1]$.

For the converse, suppose $X$ has the RNP. Let $\S$ be the linear span of all indicator functions on intervals, and let $f:[0,1]\rightarrow X$ be a Lipschitz map. Then define $T:\S\rightarrow X$ by

We may assume the intervals in this sum are disjoint, giving use well-definedness. $T$ is clearly a bounded linear operator, so by density of $S$ in $L_1[0,1]$, we may extend it to a bounded linear operator on the whole space. By assumption, we can find a strongly measurable bounded function $g:[0,1]\rightarrow X$ such that

so by the Lebesgue differentiation theorem $f'=g$ exists almost everywhere on $[0,1]$. $\square$

Lebesgue Differentiation Theorem for the Bochner Integral. Let $X$ be a Banach space and $f:\mathbb{R}^n\rightarrow X$ be a strongly measurable function such that for all compact $K\subset\mathbb{R}^n$, $\int_{K}\norm{f(\xi)}d\xi < \infty$. Then the set of Lebesgue points of $f$ has full measure, or in other words, $\lim_{\delta\rightarrow 0^+}\delta^{-n}\int_{\xi\leq\delta}\norm{f(x+\xi)-f(x)}d\xi$ for almost every $x\in\mathbb{R}^n$.

Proof. Since $f$ is strongly measurable, we may assume that $f$ has separable range. Choose $Z$ dense in $f(\R^n)$. Now by the Lebesgue differentiation theorem for real valued functions, for each fixed $z\in Z$,

except on a null set, say $A_z\subset\R^n$. But then setting $A=\bigcup_zA_z$ (which is also null), we obtain the above for every $z\in Z$. Finally, using the triangle inequality,

where $V_n$ is the volume of the $n$ dimensional unit ball. Taking the infimum of $z\in Z$ gives the result. $\square$

Now we come back to a special case of our problem.

Finite Dimensional Rademacher Theorem. Let $f:E\rightarrow Y$ be a Lipschitz map from a finite dimensional normed space $E$ to a Banach space $Y$ with the RNP. Then $f$ is Fréchet differentiable almost everywhere.

Proof. (sketch) Since $Y$ has the RNP, we can use the above characterisation of this to see that for each direction $u\in E$, $\partial_u(x)$ exists on a set of full measure $\Omega_u$. Extend this to a bounded and strongly measurable function $S(\cdot)(u)$ on $E$. We can take a sequence of mollifications $g_n$ of $f$ such that $D_{g_n}(x)(u)\rightarrow S(x)(u)$ for $u\in E$ at Lebesgue points $x\in L_u$, which is of full measure by the Lebesgue differentiation theorem. Choosing $G\subset E$ as a dense set of rational coordinates in $E$, $E'=\bigcap_{u\in G}(\Omega_u\cap L_u)$ is of full measure, and $S(x)(\cdot)|_G$ is linear for $x\in E'$, because so is $D_{g_n}(x)(\cdot)$. We can extend this to the desired derivative $T(x)(\cdot)$ of $f$ for $x\in E'$. $\square$

We would like to apply something similar to the above to obtain our result; remember that all we need for an embedding is a single point of differentiability. However, we cannot use the Lebesgue measure as our space may be infinite dimensional: Therefore we must introduce an alternative notion of a null set for arbitrary Banach spaces.

Definition. We say that a Borel set $A$ in a Banach space $X$ is Haar-null if there is a probability measure $\mu$ on $\mathcal{B}(X)$ such that $\mu(x+A)=0$ for every $x\in X$.

Notice that in this definition, $\mu(A) = 1$.

Lemma. Suppose $A$ is a subset of a separable Banach space $X$, and $E$ is a finite dimensional subspace of $X$ such that $\lambda((x+A)\cap E) = 0$ for all $x\in X$. Then $A$ is Haar-null.

Proof. Choose a probability measure $\mu$ on $X$, supported on $E$, such that for $B\subset X$, $\mu(B) = 0$ if and only if $\lambda(B\cap E) = 0$ (we could do this by, for example, setting $\mu(B) = \int_{B\cap E}\phi(\xi)d\lambda(\xi)$ for a suitable strictly positive function $\phi$). Then since $\lambda((x+A)\cap E) = 0$ for all $x\in X$, $\mu(x+A)=0$ for all $x\in X$ as required. $\square$

Lemma. If $A_n$ is a sequence of Haar-null sets in a separable Banach space $X$, then $\bigcup_nA_n$ is Haar-null.

Proof. Choose a sequence of probability measures $\mu_n$ on $A_n$ such that for each $n$, $\mu_n(x+A_n) = 0$ for all $x\in X$. Define a probability measure $\mu$ on $X$ by

Note that we obtain countable additivity by absolute convergence of the sum (so we can rearrange the terms), and clearly $\mu(X) 1$. Clearly, $\mu(x+A) = 0$ for all $x\in X$, giving the result. $\square$

Infinite Dimensional Rademacher Theorem. Let $f:X\rightarrow Y$ be a Lipschitz map from a separable Banach space $X$ to a Banach space $Y$ with the RNP. Then $f$ is Gâteaux differentiable except on a Haar-null set.

Proof. Take an increasing sequence of finite dimensional spaces $(E_n)_{n=1}^{\infty}$ whose union is dense in $X$. Denote by $D_n$ the set of points $x\in E_n$ at which $f|_{E_n}$ is differentiable with derivative $T_n(x)$, and let $A_n=X\setminus D_n$. Then if $x\in X$, $(x+A_n)\cap E_n$ has Lebesgue measure zero by Rademacher’s theorem, so by apply the preceding two lemmas, $\bigcup_{n}A_n$ is Haar-null. Choose $x_0\in \bigcap_nD_n$. Then for each $n$, $T_n(x_0):E_n\rightarrow Y$ is a linear operator bounded by $\Lip(f)$ extending $T_{n-1}(x_0)$. Hence by continuity we can define a linear operator $T(x_0):X\rightarrow Y$, again bounded by $\Lip(f)$, which we can check is the Gâteaux derivative of $f$ at $x_0$. Indeed, if $u\in X$ and $\varepsilon>0$, then choosing $v\in \bigcup_nE_n$ such that $\norm{u-v} \leq \Lip(f)^{-1}\varepsilon/3$ we obtain

provided $t$ is sufficiently small. $\square$

Note that this is no longer uniform in the direction $u$, so we lose Fréchet differentiability. Indeed, it would be too much to hope for that, since $f:\ell_2\rightarrow\ell_2$, $f(x_1, x_2, ...) = (\abs{x_1}, \abs{x_2}, ...)$ is nowhere Fréchet differentiable.

From this we deduce that since $X$ is not Haar-null, we must have at least one point of differentiability, giving the following result.

Corollary. Let $f:X\rightarrow Y$ be a Lipschitz embedding from a separable Banach space into a Banach space $Y$ with the RNP. Then $X$ is linearly isomorphic to a subspace of $Y$.