Show that $T$ is not continuous when $p = \infty.$

Question

Let $n \in \mathbb N,$ $1 \leq p \leq \infty$ and let $f \in L^p (\mathbb R^n).$ Define a function $T : \mathbb R^n \longrightarrow L^p (\mathbb R^n)$ by $$T (h) (x) = f(x + h)$$ for all $h \in \mathbb R^n$ and for all $x$ a.e. in $\mathbb R^n.$ Show that

$(1)$ $T$ is continuous when $1 \leq p \lt \infty.$

$(2)$ $T$ is not continuous when $p = \infty.$

For $(1)$ I exploited the fact that $C_c (\mathbb R^n)$ is dense in $L^p (\mathbb R^n).$ For $(2)$ I no longer have access to this result. But I think the result continues to hold for $p = \infty$ if $f \in C_c (\mathbb R^n)$ because compactly supported continuous functions are uniformly continuous. This result clearly holds for Identity function and constant functions. The result where the result might violate is the set of all continuous functions which are either not bounded or the quantity $|f(x+h) - f(x)|$ is depending upon $x.$ Any function which is in the complement of $C (\mathbb R^n)$ in $L^{\infty} (\mathbb R^n)$ will possibly also work. Could anyone please give me some more insights on $(2)\ $?

Thanks for your kind attention.

Answer 1

For a simple counterexample just let $f$ be the heaviside step function on $\mathbb R$. Then for all $h\neq h'$ we have $\|T(h)-T(h')\|_\infty=1$.

Answer 2

I think generally your remarks are spot on (except that the identity function isn’t bounded, so that doesn’t work). The original problem should more correctly have asked to show, in part 2), that $T$ need not be continuous, rather than claiming that it is not continuous.

In fact, you can argue that for the case $p=\infty$, $T$ will be continuous if and only if $f$ agrees almost everywhere with a bounded uniformly continuous function $\hat{f}$. I will sketch the argument here.

One direction is easy: If $f$ is as described, then fixing $h_1$,$h_2$ we get for a.e. $x$, $|T(h_1)(x)-T(h_2)(x)|=|\hat{f}(x+h_1)-\hat{f}(x+h_2)|<\omega(||h_1-h_2||)$ for some modulus of continuity $\omega$ depending only on $\hat{f}$, so $T$ is continuous.

Conversely, if $T$ is continuous, then let $\omega$ be a modulus of continuity for $T$ at $h=0$. Observe first that for every $x_0$ and $r>0$, $$\operatorname{esssup}_{x\in B(x_0,r)}f(x) - \operatorname{essinf}_{x\in B(x_0,r)}f(x)\leq \omega(2r) \text{.}$$

To see this, suppose by contradiction that there are sets $X$ and $Y$ of positive measure in $B(x_0,r)$ where $\inf_{x\in X,y\in Y}(f(y)-f(x))>\omega(2r)$, and then take points $x$ and $y$ of density for $X$ and $Y$. Then for small enough $\delta$, $B(x,\delta)\cap X$ and $B(y,\delta)\cap Y$ each have measure more than $1/2|B(x,\delta)|$, whereby $A:=B(y,\delta)\cap Y \cap (X+y-x)$ is of positive measure. But letting $h=x-y$, for each $z\in A$, we have $z+h \in X$, $z\in Y$, so $$T(0)(z)-T(h)(z) = f(z)-f(z+h)>\omega(2r)\geq\omega(||h||)\text{,} $$ contradicting our choice of $\omega$.

With this in mind, we define

$$\hat{f}(x_0)=\lim_{r\rightarrow 0}\operatorname{esssup}_{x\in B(x_0,r)}f(x)\text{,}$$

and note that the essential inf also gives the same function. Arguing similarly to above, one can show that

$$\left|\operatorname{esssup}_{x\in B(x_1,r)}f(x)-\operatorname{essinf}_{x\in B(x_2,r)}f(x)\right| \leq \omega(2r+||x_2-x_1||) \text{,}$$

for any two points $x_1,x_2\in\mathbb R^n$, and so passing to the limit as $r\rightarrow 0$ shows that $\hat{f}$ is uniformly continuous, indeed with the same modulus of continuity $\omega$. Lastly, argue that $f$ coincides with $\hat{f}$ at any Lebesgue point of $f$.

Remark

One final note about the above proof. Notice that in both directions, the same modulus of continuity $\omega$ is used for $f$ and for $T$. Notice also that in the converse direction, we only really needed to know $T$ was continuous at a single point, namely the origin.

Thus what we really proved was that the following are equivalent, with the same modulus of continuity $\omega$:

1. $f$ is a.e. equal to a bounded uniformly continuous function.
2. $T$ is uniformly continuous.
3. $T$ is continuous.
4. $T$ is continuous at the origin.

and if the above proof is modified slightly (which I will leave as an exercise), we can even add

5. $T$ is continuous at some arbitrary point in $\mathbb R^n$.