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Let $f\in L^1$, consider the map $t\mapsto f_t=f(x-t)$, then how can one show that $t\mapsto f_t$ is continuous? More explicitly one wants to show that $\lim_{h\to 0}|f_{t+h}-f_t|_{L^1}=0$. I tried to use approximation by $C_0(\mathbb{R})$ functions $g^n$ to approximate $f$ in $L^1$ norm. Then one has $\lim_{h\to 0}|g^n_{t+h}-g^n_t|_{L^1}=0$, but then I came across the problem: how can one show that the two limits can exchange so that one has $$\lim_{h\to 0}|f_{t+h}-f_t|_{L^1}=\lim_{h\to 0}\lim_{n\to\infty}|g^n_{t+h}-g^n_t|_{L^1}=\lim_{n\to\infty}\lim_{h\to 0}|g^n_{t+h}-g^n_t|_{L^1}=0.$$

Can someone help me with some conditions on which two limits can be exchanged, or do you have a better way of proving the continuity? Thank you!

Davide Giraudo
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Xuxu
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2 Answers2

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The translations $\tau_t$, where $\tau_t(f)(x) = f(x-t)$ are isometries of $L^1$ (more genral, any $L^p$).

Thus the family $T = \{\tau_t : t \in \mathbb{R}^n\}$ is equicontinuous.

For $f \in C_c(\mathbb{R}^n)$, we have $\lim\limits_{t \to 0} \tau_t(f) = f$ in $L^p$ by the uniform continuity of $f$. For $1 \leqslant p < \infty$, $C_c(\mathbb{R}^n)$ is dense in $L^p$.

By the theorem of Ascoli (-Bourbaki), on a family $\mathcal{F}$ of equicontinuous functions $\varphi \colon X \to Y$, the topology of pointwise convergence on $X$, the topology of pointwise convergence on a (fixed) dense subset of $X$ and the topology of uniform convergence on compact subsets of $X$ coincide.

Hence not only $\lim\limits_{t\to 0} \tau_t(f) = f$ for all $f \in L^p$ [$p < \infty$, again], the convergence is even uniform on all compact subsets of $L^p$.

If you prefer a direct argument without citing topological theorems, let $\varepsilon > 0$. Find a $g \in C_c(\mathbb{R}^n)$ with $\lVert f - g\rVert_1 < \varepsilon/3$. By the uniform continuity of $g$, you have a $\delta > 0$ such that $\lVert \tau_h(g) - g\rVert_1 < \varepsilon/3$ whenever $\lVert h\rVert < \delta$. Then, for $\lVert h \rVert < \delta$, you have

$$\lVert \tau_h(f) - f\rVert_1 \leqslant \lVert \tau_h(f) - \tau_h(g)\rVert_1 + \lVert \tau_h(g) - g\rVert_1 + \lVert g - f \rVert_1 < \varepsilon.$$

Daniel Fischer
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    Just curious, what if h goes to infinity, instead of going to zero. What's the limit of the L1 norm?? – Topoguy Jul 12 '15 at 04:43
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    @Topoguy We have $\lim\limits_{\lVert h\rVert\to\infty} \lVert \tau_h(f) - f\rVert_1 = 2\lVert f\rVert_1$. If $f$ has compact support, we have $\lVert \tau_h(f) - f\rVert_1 = 2\lVert f\rVert_1$ for large enough $\lVert h\rVert$ since the supports of $f$ and $\tau_h(f)$ are disjoint. Then use an approximation by functions with compact support to get the general limit. – Daniel Fischer Jul 12 '15 at 08:49
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    @DanielFischer Thanks! It makes sense. – Topoguy Jul 12 '15 at 16:35
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Let $g_n$ be a sequence in $C_0(\mathbb{R})$ such that $g_n\to f$ in $L^1(\Omega)$. Note that $$|f(x-t)-f(x)|\leq |f(x-t)-g_n(x-t)|+|g_n(x-t)-g_n(x)|+|g_n(x)-f(x)|$$

Can you finish?

Tomás
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