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I am asked to prove that $\tau_{h_n}f\to f$ weakly-* in $L^\infty$, where $f\in L^\infty$, $\tau_{h_n}f(x)=f(x-h_n)$ and $h_n\to0$ in $\mathbb{R}^n$.

I managed to prove it assuming $f$ is a.e. equal to a function which is continuous in almost any point. Unfortunately, that doesn't fill up the whole of $L^\infty$. So how do I go about with functions that don't have this nice property?

My try which solved the previous case was changing both $\tau_{h_n}f$ and $f$ in a null set so that $f$ would become continuous at almost any point and I could use dominated convergence on $\int|f(x-h_n)-f(x)||u(x)|dx$ by the dominant $2\|f\|_{L^\infty}\cdot|u|$ which is integrable as $u\in L^1$, but that doesn't work for functions without the above property, does it?

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MickG
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1 Answers1

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Work from the other end, using Continuity of $L^1$ functions with respect to translation.

Weak* continuity in $L^\infty$ means that for every $g\in L^1$, $$\int (\tau_{h}f)g\to \int fg,\qquad h\to 0$$ But $\int (\tau_{h}f)g= \int f(\tau_{-h}g)$, by a change of variables. Also, $$\left|\int f(\tau_{-h}g) - \int fg\right| \le \|f\|_\infty \|\tau_{-h}g -g\|_1\to 0$$ which yields the result.

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