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Let $f,g\in L^1[0,1)$ such that $fg\in L^1[0,1)$. Define the shift operator $T_tf(x) = f(x-t)$. I want to show that $$ \|f(x)g(x) - T_tf(x)T_{\tau}g(x)\|_1\to0 $$ as $t,\tau \to 0$.

I've looked at both Continuity of $L^1$ functions with respect to translation and Shifting a function is continuous and have tried to massage their arguments to work in my setting; however, the product of functions is a bit unwieldy. I know the standard method to show these things is to use the fact that continuous functions are dense in $L^1$ and then do a few triangle inequalities. I'm just not seeing the correct sequence of functions to add and subtract.

Any advice would be appreciate or a counterexample.

Doofenshmert
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2 Answers2

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Let $g(x)=x^{-1/2}$ and $$f(x)=\sum_{n=1}^\infty a_n|x-2^{-n}|^{-1/2},\quad a_n>0$$ Then $0<f<fg$ and $$ \|fg\|_1=\sum_{n=1}^\infty a_n\int\limits_0^1x^{-1/2}|x-2^{-n}|^{-1/2}\,dx$$ Let $$b_n=\int\limits_0^1x^{-1/2}|x-2^{-n}|^{-1/2}\,dx$$ Then $fg\in L^1$ for $a_n=2^{-n}b_n^{-1}.$ However for $x>2^{-2k}$ we get $$(T_{2^{-k}}f)(x)(T_{2^{-2k}}g)(x)\\ \ge a_k(x-2^{-2k})^{-1/2} \,(x-2^{-2k})^{-1/2}\\ =a_k (x-2^{-2k})^{-1}\notin L^1$$ Summarizing the product $$(T_{2^{-k}}f)(x)(T_{2^{-2k}}g)(x)$$ is not integrable for all positive integers $k.$ In particular the claim in OP is not satisfied.

Ryszard Szwarc
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  • Thanks! I had a feeling that independently adjusting the $t$ and $\tau$, things wouldn't necessarily be integrable. – Doofenshmert Mar 23 '25 at 14:16
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Define the shift operator on smooth functions with support away from the boundaries. Wrt to this operator with generator $\partial_x$ there exist different extensions according to the family of unitary boundary conditions in $L^2(0,1)$. The most simple extension uses periodic boundary condition and Fourier series in $L^2(0,1).$

Roland F
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