3
$\begingroup$

Let $G$ be a locally compact group with unimodular Haar measure $\mu$. We consider the Hilbert space $\mathscr{H}:= L_{\mu}^2(G)$ together with the unitary representation $\pi : G \to U(\mathscr{H})$ given by $\pi(g): \mathscr{H} \to \mathscr{H}, f \mapsto f \circ r_g$, where $r_g$ is right multiplication by $g$. Hence $(\pi(g)f)(x) = f(xg)$.

We studies the following: For a continuous function $\varphi \in C_c(G)$ with compact support we define $$\pi(\varphi):= \int_G \varphi(g)\pi(g) \, d\mu(g).$$ More explicitly, for a measurable function $f: G \to \mathbb{C}$ we define $$(\pi(\varphi)f)(x) :=\int_G \varphi(g)f(xg) \, d\mu(g).$$

My question in now: Does this define a well defined representation of $C_c(G)$ on $\mathscr{H}$, where the problem is, that if $f \in L_{\mu}^2(G)$, then we need to show that $\pi(\varphi)f$ is in $L_{\mu}^2(G)$.

I doubt that this is true for a general $G$. However, I can prove the statement if $G$ is compact as follows: Note that if $G$ is compact, then $L_{\mu}^2(G) \subset L_{\mu}^1(G)$. Hence for $f \in L_{\mu}^2(G)$, \begin{align*} ||\pi(\varphi)f||_2^2 &= \int_G|(\pi(\varphi)f)(x)|^2 \, d\mu(x) \\ &= \int_G \bigg| \int_G \varphi(g)f(xg) \, d\mu (g) \bigg|^2 d\mu(x) \\ & \leq ||\varphi||_{\infty}^2 \int_G \bigg| \int_G f(xg) \, d\mu (g) \bigg|^2 d\mu(x) \\ &\leq ||\varphi||_{\infty}^2 \int_G ||f||_1^2 \,d \mu(x) \\ &= ||\varphi||_{\infty}^2 \cdot ||f||_1^2 \cdot \mu(G) < \infty. \end{align*} Where we used the invariance of the Haar measure in line 4. And the last term is less than infinity, since $\varphi$ is continuous of compact support and hence bounded, $L_{\mu}^2(G) \subset L_{\mu}^1(G)$ and hence for $f \in L_{\mu}^2(G)$ we have $||f||_1 < \infty$ and finally since $G$ is compact, the Haar measure is finite.

So the problem is can, one do a similar calculation if $G$ is not compact, or if one can find a counterexample.

$\endgroup$
0

1 Answer 1

2
$\begingroup$

It's true: $\pi(\varphi)f\in L^2(G,\mu)$.

Let $M$ be the support of $\varphi$, so $$(\pi(\varphi)f)(x) :=\int_M \varphi(g)f(xg) \, d\mu(g).$$ By the Cauchy-Schwarz inequality, we have $$|(\pi(\varphi)f)(x)|^2\le \|\varphi\|^2_2\int_M|f(xg)|^2d\mu(g)=\|\varphi\|^2_2\int_{xM}|f(g)|^2d\mu(g),$$ so $$\int_G|(\pi(\varphi)f)(x)|^2d\mu(x)\le \|\varphi\|^2_2\int_G\int_{G}\mathbb{1}_{xM}(g)|f(g)|^2d\mu(g)d\mu(x).$$ Since $g\in xM$ $\Leftrightarrow$ $x\in gM^{-1}$, this yields (I should assume $G$ $\sigma$-compact to apply Fubini, but otherwise since $f$ has $\sigma$-compact support I apply it inside a suitable $\sigma$-compact open subgroup) $$\int_G|(\pi(\varphi)f)(x)|^2d\mu(x)\le \|\varphi\|^2_2\int_G\int_{gM^{-1}}d\mu(x)|f(g)|^2d\mu(g).$$ So $$\int_G|(\pi(\varphi)f)(x)|^2d\mu(x)\le \|\varphi\|^2_2\mu(M^{-1})\int_G|f(g)|^2d\mu(g)=\|\varphi\|^2_2\mu(M^{-1})\|f\|_2^2<\infty.$$

Note that I only used (twice) left-invariance of $\mu$ (in the non-unimodular case, the right-regular representation $\pi$ is not unitary, however).

$\endgroup$
2
  • $\begingroup$ Thank you very much. To simplify a little: $$\begin{align*} ||\pi(\varphi)f||_2^2 &= \int_G |(\pi(\varphi)f)(x)|^2 \, d\mu(x) \\ &= \int_G \bigg| \int_{\mathrm{supp}(\varphi)} \varphi(g)f(xg) \, d\mu(g) \bigg|^2 \, d\mu(x) \\ & \leq ||\varphi||_2^2 \int_G\int_{\mathrm{supp}(\varphi)} |f(xg)|^2 \, d\mu(g)d\mu(x) \\ &= ||\varphi||_2^2 \int_{\mathrm{supp}(\varphi)} \int_G |f(xg)|^2 \, d\mu(x)d\mu(g) \\ &= ||\varphi||_2^2 \int_{\mathrm{supp}(\varphi)} \int_G |f(x)|^2 \, d\mu(x)d\mu(g) \\ &= \mu(\mathrm{supp}(\varphi)) \cdot ||\varphi||_2^2 \cdot ||f||_2^2 < \infty. \end{align*} $$ $\endgroup$ Commented Mar 31, 2018 at 10:09
  • $\begingroup$ Yes. Actually you're using right-invariance of $\mu$ and not left as I did. In this case the right-regular representation is unitary. Said otherwise: for $G$ possibly non-unimodular and $\mu$ left-invariant, my argument works for the right-regular representation, and your argument works for the left-regular representation (these two representations are of distinct interest when $G$ is non-unimodular; only the left one is unitary). $\endgroup$
    – YCor
    Commented Mar 31, 2018 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.