$$\dim(im(f)) \cdot \|f\|^2 \cdot | {\rm supp}(f)| \geq \tau(f^*f) \cdot |{\rm supp}(f)| \geq |G| \cdot \|f\|^2_1.$$
Here, $\tau \colon \mathbb C[G] \to \mathbb C$ is the non-normalized trace on $\mathbb C[G]$, coming from the inclusion $\mathbb C[G] \subset M_{|G|} \mathbb C$. It is best decribed by $\tau(\sum_{g$\tau(\sum_{g \in G} a_g g) = |G| \cdot a_e$a_e.$$Also, f \mapsto f^* denotes the usual involution, i.e.$$(\sum_{g \in G}a_g g)^{*} = \sum_{g \in G} \bar a_g g^{-1}.$$The first inequality, follows since \tau(f^*f) is just the sum of the eigenvalues of f^*f, which is obviously bounded by the number of non-zero eigenvalues (which equals the dimension of image of f) times the size of the largest eigenvalue (which equals \|f\|^2). Note that by direct computation$$\tau\left((\sum_{g \in G} a_g g)^* (\sum_{g \in G} a_g g)\right)=|G| \cdot \sum_{g \in G} |a_g|^2.$$The second inequality follows from this observation and the Cauchy-Schwarz inequality applied to f\cdot \chi_{{\rm supp} f}, where the product is here the pointwise product of coefficients and \|f\|_1 denotes the usual 1-norm on \mathbb C[G]. Now, since each group element acts as a unitary (and hence with operator norm 1) on M_{|G|} \mathbb C, we get \|f\|_1 \geq \|f\| and hence$$\dim(im(f)) \cdot | {\rm supp}(f)| \geq |G|.$$This even has an extension to all (possibly infinite) groups with essentially the same proof. The appropriate statement is then that for the normalized Murray-von Neumann dimension (with respect to the group von Neumann algebra LG) of the closure of the image of \lambda(f) acting on \ell^2 G via the left-regular representation \lambda, we have$$\dim_G \left (\overline{im(\lambda(f))} \right) \cdot |{\rm supp}(f)| \geq 1.$$2 added 368 characters in body; added 11 characters in body; added 67 characters in body The answer is yes, this always holds. Note that$$\dim(im(f)) \cdot \|f\|^2 \cdot | {\rm supp}(f)| \geq \tau(f^*f) \cdot |{\rm supp}(f)| \geq |G| \cdot \|f\|^2_1.$$Here, \tau \colon \mathbb C[G] \to \mathbb C is the non-normalized trace on \mathbb C[G], coming from the inclusion \mathbb C[G] \subset M_{|G|} \mathbb C. It is best decribed by \tau(\sum_{g \in G} a_g g) = |G| \cdot a_e. The first inequality, follows since \tau(f^*f) is just the sum of the eigenvalues of f^*f, which is obviously bounded by the number of non-zero eigenvalues (which equals the dimension of image of f) times the size of the largest eigenvalue (which equals \|f\|^2). Note that by direct computation$$\tau\left((\sum_{g \in G} a_g g)^* (\sum_{g \in G} a_g g)\right)=|G| \cdot \sum_{g \in G} |a_g|^2.$$The second inequality follows from this observation and the Cauchy-Schwarz inequality applied to f\cdot \chi_{{\rm supp} f}, where the product is here the pointwise product of coefficients and \|f\|_1 denotes the usual 1-norm on \mathbb C[G]. Now, since each group element acts as a unitary (and hence with operator norm 1) on M_{|G|} \mathbb C, we get \|f\|_1 \geq \|f\| and hence$$\dim(im(f)) \cdot | {\rm supp}(f)| \geq |G|.$$This even has an extension to all (possibly infinite) groups with essentially the same proof. The appropriate statement is then that for the normalized Murray-von Neumann dimension (with respect to the group von Neumann algebra LG) of the closure of the image of \lambda(f) acting on \ell^2 G via the left-regular representation \lambda, we have$$\dim_G \left (\overline{im(\lambda(f))} \right) \cdot |{\rm supp}(f)| \geq 1.$$1 The answer is yes, this always holds. Note that$$\dim(im(f)) \cdot \|f\|^2 \cdot | {\rm supp}(f)| \geq \tau(f^*f) \cdot |{\rm supp}(f)| \geq |G| \cdot \|f\|^2_1.$$Here, \tau \colon \mathbb C[G] \to \mathbb C is the non-normalized trace on \mathbb C[G], coming from the inclusion \mathbb C[G] \subset M_{|G|} \mathbb C. It is best decribed by \tau(\sum_{g \in G} a_g g) = |G| \cdot a_e. The first inequality, follows since \tau(f^*f) is just the sum of the eigenvalues of f^*f, which is obviously bounded by the number of non-zero eigenvalues (which equals the dimension of image of f) times the size of the largest eigenvalue (which equals \|f\|^2). Note that by direct computation$$\tau\left((\sum_{g \in G} a_g g)^* (\sum_{g \in G} a_g g)\right)=|G| \cdot \sum_{g \in G} |a_g|^2.$$The second inequality follows from this observation and the Cauchy-Schwarz inequality applied to f\cdot \chi_{{\rm supp} f}, where the product is here the pointwise product of coefficients and \|f\|_1 denotes the usual 1-norm on \mathbb C[G]. Now, since each group element acts as a unitary (and hence with operator norm 1) on M_{|G|} \mathbb C, we get \|f\|_1 \geq \|f\| and hence$$\dim(im(f)) \cdot | {\rm supp}(f)| \geq |G|.$\$