Group of Unitaries: Strong Continuity

Question

Let $\mathcal{L}^2(\mathbb{R})$ be the the Hilbert space of square integrable functions, shortly $\mathcal{L}^2$.

Consider the group of unitaries: $$U:\mathbb{R}\to Aut(\mathcal{L}^2):(U_vf)(x):=f(x-v)$$

Is there a "nice" way to show that it is strongly continuous: $$\|U_vf-f\|_2\to 0\text{ when }v\to 0\text{ for }f\in\mathcal{L}^2$$

Certainly, a good start is to reduce this to the dense subspace $\mathcal{C}_c^\infty$ via: $$\|U_v f+(-U_vf_0+U_vf_0-f_0+f_0)-f\|\leq \|f-f_0\|+\|U_vf_0-f_0\|+\|f_0-f\|$$ But then I argued quite technically that this indeed vanishes: $$\|U_vf_0-f_0\|\to 0\text{ when }v\to 0\text{ for }f_0\in\mathcal{C}_c^\infty$$ Is there a better way to do so?

Answer 1

If you like the Fourier transform and the Lebesgue dominated convergence theorem, then there's a trivial proof for $L^{2}(\mathbb{R})$. That should qualify as a "nice" way, but I think it's better to stick with basics using density arguments.

Answer 2

The remaining step that $\lVert U_v(f_0) - f_0\lVert_2 \to 0$ when $v \to 0$ holds for $f_0 \in C_c^{\infty}$ follows from two observations

1. $f_0$ is uniformly continuous because it has compact support

2. On a space $X$ with finite measure we have $\lVert f \rVert_2 \leq \mu(X)^{1/2} \lVert f \rVert_\infty$.

Fix $\varepsilon > 0$. Since $f_0$ is uniformly continuous, there is $\delta > 0$ such that $|v| \lt \delta$ implies $\lVert U_v (f_0) - f_0 \rVert_\infty = \sup_{x \in\mathbb{R}} | f_0(x-v) - f_0(x)| \leq \varepsilon$.

We have shown so far that $|v| \lt \delta$ implies $\lVert U_v (f_0) - f_0 \rVert_\infty \leq \varepsilon$.

Suppose the support of $f_0$ is contained in $[-R,R]$. Then $|v| \leq 1$ implies that the support of $U_v(f_0)$ is contained in $[-R-1, R+1]$. Therefore $$\lVert U_v (f_0) - f_0 \rVert_2 \leq \sqrt{2R + 2} \lVert U_v (f_0) - f_0 \rVert_\infty \leq C\varepsilon$$ whenver $|y| \lt \min\{1,\delta\}$.