Comparison between the operator norm and the $L^1$ norm on group algebras

Question

Consider a discrete group $G$ and its group algebra over $\mathbb{C}$, $\mathbb{C}[G]$. There are four norms on it I wish to consider for this question:

1. The 2-norm given by $||\sum_{g \in G} c_gg||_2^2 = \sum_{g \in G} |c_g|^2$;
2. The 1-norm given by $||\sum_{g \in G} c_gg||_1 = \sum_{g \in G} |c_g|$;
3. The $\infty$-norm given by $||\sum_{g \in G} c_gg||_\infty = \sup_{g \in G} |c_g|$;
4. The operator norm induced by the left regular representation, i.e., if $\lambda: \mathbb{C}[G] \rightarrow \mathbb{B}(l^2(G))$ is the left regular representation, then the norm I wish to consider is $||x||_{op} = ||\lambda(x)||$.

The first question I have is for what $G$ are the operator norm and the 1-norm equivalent. We always have $||x||_{op} \leq ||x||_1$, so it's only the other direction that's interesting. Obviously, this holds whenever $G$ is finite, and my guess is that this never holds for any infinite group, but I cannot find any proof of this.

The second question is a weaker version of the first: Under what conditions would it happen that $||x||_2^2 \leq C||x||_{op}||x||_\infty, \forall x \in \mathbb{C}[G]$ for some constant $C > 0$. Or, in an even weaker form, under what conditions would it happen that whenever, for a sequence of elements $x_n$ in $\mathbb{C}[G]$ with uniformly bounded operator norms and $\infty$-norms converging to zero, we always have the 2-norms converging to zero? By Hölder's inequality, this would follow if the first question has a positive answer. Again, this is only interesting when $G$ is infinite.

Answer 1

This is an alternative, more ad hoc proof, very closely mimicking a part of section 3 of the paper https://doi.org/10.4007/annals.2013.178.1.4

The start of the argument is the same. Write $M=L(G)$. Given $\varepsilon > 0$, it suffices to prove the existence of a unitary $v \in U(M)$ such that $|(v)_g| < \varepsilon$ for all $g \in G$. We again argue by contradiction and denote by $\Delta : M \to M \overline{\otimes} M$ the comultiplication.

Note that for every $v \in U(M)$, we have

$$(\tau \otimes \tau \otimes \tau)((\Delta(v) \otimes v)(v \otimes \Delta(v))^*) = \sum_{g \in G} |(v)_g|^4 \geq \varepsilon^4 \; .$$

Denote by $K \subset \text{ball}(M \overline{\otimes} M \overline{\otimes} M)$ the $\|\cdot\|_2$-closure of the convex hull of

$$\{(\Delta(v) \otimes v)(v \otimes \Delta(v))^* \mid v \in U(M)\} \; .$$

Let $X \in K$ be the unique element of minimal $\|\cdot\|_2$. By the above estimate, $(\tau \otimes \tau \otimes \tau)(X) \geq \varepsilon^4$, so that $X \neq 0$. By uniqueness of $X$, we find that

$$(\Delta(v) \otimes v) X = X (v \otimes \Delta(v))$$

for all $v \in U(M)$. Denote $\mathcal{G} = \{v \otimes v \mid v \in U(M)\}$, which is a subgroup of $U(M \overline{\otimes} M)$. Write $P = \mathcal{G}^{\prime\prime}$. We conclude that

$$(\Delta \otimes \text{id})(a) X = X (\text{id}\otimes \Delta)(a)$$

for all $a \in P$.

Whenever $b \in M$ is self-adjoint, we have $\exp(itb) \otimes \exp(itb) \in P$ for every $t \in \mathbb{R}$. Taking the derivative at $t = 0$, it follows that $b \otimes 1 + 1 \otimes b \in P$. By linearity, $b \otimes 1 + 1 \otimes b \in P$ for all $b \in M$. In particular, $u_g \otimes 1 + 1 \otimes u_g \in P$ for all $g \in G$. We write $y_g = u_g \otimes 1 + 1 \otimes u_g$ and $S = XX^*$. It follows that

$$(\Delta \otimes \text{id})(y_g) S = X (\text{id}\otimes \Delta)(y_g) X^*$$

for all $g \in G$. Write $Q = \Delta(M) \overline{\otimes} M$. Taking left and right the conditional expectation on $Q$ and writing $E_Q(S) = (\Delta \otimes \text{id})(T)$ for some $T \in M \overline{\otimes} M$, we conclude that

$$ \| y_g T\|_2 = \|E_Q(X (\text{id}\otimes \Delta)(y_g) X^*)\|_2$$

for all $g \in G$.

Since $G$ is infinite, we can take a sequence $g_n \to \infty$ in $G$. We prove that

$$\|y_{g_n} T\|_2 \to \sqrt{2} \|T\|_2 \quad\text{and}\quad \|E_Q(X (\text{id}\otimes \Delta)(y_{g_n}) X^*)\|_2 \to 0 \; .$$

When both are proven, we conclude that $T = 0$, so that $S = 0$ and thus $X = 0$, a contradiction.

For the first limit, note that

$$\|y_{g_n} T\|_2^2 = (\tau \otimes \tau)(T^*(2 + u_{g_n}^* \otimes u_{g_n} + u_{g_n} \otimes u_{g_n}^*) T) \to 2 \|T\|_2^2 $$

because $u_{g_n} \to 0$ weakly.

For the second limit by density and uniform continuity, it suffices to prove that

$$\|E_Q(Y (\text{id}\otimes \Delta)(y_{g_n}) Z)\|_2 \to 0$$

whenever $Y = u_{h_1} \otimes u_{h_2} \otimes u_{h_3}$ and $Z = u_{k_1} \otimes u_{k_2} \otimes u_{k_3}$ with $h_i,k_i \in G$. But in that case, the expression is eventually zero.

Answer 2

This is only a partial answer, but it shows that if G has an element of infinite order then the weakest form of Q2 has a negative answer.

We use the so-called Rudin-Shapiro polynomials (really due to Shapiro), see

W. Rudin, Some theorems on Fourier coefficients. Proc. Amer. Math. Soc. 10 (1959), 855–859. MathReview | DOI 10.1090/S0002-9939-1959-0116184-5

The properties we require: there is a sequence $(\varepsilon_j)_{j\geq 1}\in \{\pm 1\}^{\mathbb N}$ such that when we set $P_n(z)=\sum_{j=1}^n \varepsilon_j z^j$ whe have $\sup_{|z|=1} |P_n(z)| \leq 5\sqrt{n}$. If I recall correctly, a construction and proof of these properties is also in Katznelson's Introduction to Harmonic Analysis.

Now pick an element $g\in G$ with infinite order and define $f_n =n^{-1/2}\sum_{j=1}^n \varepsilon_j \delta_{(g^j)}$. These are norm-1 vectors in $\ell^2(G)$ and clearly $\lVert f_n\rVert_\infty \to 0$ as $n\to\infty$. On the other hand, by general properties of ${\rm C}^\ast$-algebras, the operator norm of $f_n$ is equal to the $L^\infty$-norm of $n^{-1/2}P_n$ viewed as function on the unit circle, hence is at most $\sqrt{5}$.

Note: as remarked in Rudin's article, if we allow complex coefficients for our Fourier series then an even older example is possible; Rudin says this is due to Hardy and Littlewood. However I thought I would keep the use of the "Rudin-Shapiro polynomials" since they are more combinatorial and they were the example that came to mind first.

Answer 3

Also for an arbitrary infinite group $G$, the weakest form of Q2 has a negative answer.

Denote by $M=L(G)$ the group von Neumann algebra with its canonical trace $\tau$. For every element $a \in M$, we denote by $(a)_g = \tau(au_g^*)$ its Fourier coefficients. Given $\epsilon > 0$, I'll show below that we can find a unitary $v \in U(M)$ such that $|(v)_g| \leq \epsilon$ for all $g \in G$.

Once this is proven, the Kaplansky density theorem provides an element $a \in \mathbb{C}[G]$ such that $\|a\|_{op} \leq 1$ and $\|v-a\|_2 < \epsilon$. Then, $\|a\|_2 > 1-\epsilon$ and $|(a)_g| < 2 \epsilon$ for all $g \in G$.

I expect that one should be able to prove the existence of such a unitary $v \in U(M)$ by more elementary means, but the following argument using Sorin Popa's intertwining-by-bimodules does the job. I freely make use of some of the jargon and basic results of that theory. The necessary background can be found in Chapter 17 of the book of Claire Anantharaman and Sorin Popa, which is available here: https://www.math.ucla.edu/~popa/Books/IIunV15.pdf

The start of the argument is inspired by section 3 of the following paper: https://doi.org/10.4007/annals.2013.178.1.4

I post an alternative, more ad hoc, proof below, which is even more closely inspired by section 3 of the above mentioned paper.

In that paper, the largest Fourier coefficient of an element $a \in L(G)$ (i.e. $\|a\|_\infty$ in the notation of the question above) is called the height of $a$.

We give an argument by contradiction. Assume that we have an $\epsilon > 0$ such that for every $v \in U(M)$, there exists a $g \in G$ with $|(v)_g| \geq \epsilon$.

Denote by $\Delta : M \to M \overline{\otimes} M$ the comultiplication satisfying $\Delta(u_g) = u_g \otimes u_g$ for all $g \in G$. Denote by $E_{\Delta(M)}$ the unique trace preserving conditional expectation from $M \overline{\otimes} M$ onto $\Delta(M)$. For every $v \in U(M)$, we have that $\|E_{\Delta(M)}(v \otimes v)\|_2^2 = \sum_{g \in G} |(v)_g|^4 \geq \varepsilon^4$.

Denote $\mathcal{G} = \{v \otimes v \mid v \in U(M)\}$, which is a subgroup of $U(M \overline{\otimes} M)$. Write $P = \mathcal{G}^{\prime\prime}$. With the notation of Popa's intertwining-by-bimodules (see Definition 17.1.4 in the book cited above), we get that $P \prec_{M \overline{\otimes} M} \Delta(M)$.

Denote by $\sigma \in \operatorname{Aut}(M \overline{\otimes} M)$ the flip automorphism. I claim that $P$ equals the fixed point algebra $(M \overline{\otimes} M)^\sigma$ of $\sigma$. Clearly $P \subset (M \overline{\otimes} M)^\sigma$. Whenever $a \in M$ is self-adjoint, we have $\exp(ita) \otimes \exp(ita) \in P$ for every $t \in \mathbb{R}$. Taking the derivative at $t = 0$, it follows that $a \otimes 1 + 1 \otimes a \in P$. By linearity, $a \otimes 1 + 1 \otimes a \in P$ for all $a \in M$ and in particular for all $a \in U(M)$. Multiplying with $v \otimes v$ for $v \in U(M)$, it follows that $a \otimes u + u \otimes a \in P$ for all $a,u \in U(M)$. By linearity, we find that $a + \sigma(a) \in P$ for all $a \in M \overline{\otimes} M$ and the claim is proven.

It follows from the claim that $P$ has index $2$ in $M \overline{\otimes} M$. Since $P \prec_{M \overline{\otimes} M} \Delta(M)$, we then also get that $M \overline{\otimes} M \prec_{M \overline{\otimes} M} \Delta(M)$. A fortiori, $M \otimes 1 \prec_{M \overline{\otimes} M} \Delta(M)$.

Since the group $G$ is infinite, we can take a sequence $g_n \in G$ such that $g_n \to \infty$. I claim that $\|E_{\Delta(M)}(a (u_{g_n} \otimes 1) b)\|_2 \to 0$ for all $a,b \in M \overline{\otimes} M$. By density and uniform continuity, it suffices to prove this claim for elements $a$ and $b$ of the form $u_k \otimes u_h$ with $k,h \in G$, in which case the result is immediate.

From the claim, it however follows that $M \otimes 1 \not\prec_{M \overline{\otimes} M} \Delta(M)$ and we have reached a contradiction.