I'm getting lost with a simple computation -- these are exponent manipulations in a holder-type situation:
First define convolution in this setting: let convolution be defined for real- or complex-valued measurable functions on some locally compact topological group $G$ equipped with a left-invariant Haar measure μ. The definition of convolution between two such functions $f$ and $g$ is as follows:
$$f \star g = \int_G f(y)g(y^{-1}\cdot x)\,d\mu(y)$$
Note that the relations between $p,q,r$ is as follows: $$\frac 1 p + \frac 1 q = 1 + \frac 1 r$$
There are the following two steps:
This tells me that $\frac{1}{r+1}\left(\frac 1 p + \frac 1 q \right) = 1/p$. I'm not getting this though. Am I missing something somewhere?
You have $$ \frac1{r+1} \left( \frac1p + \frac1q \right) = \frac1r $$ and this gives the correct exponent for the convolutions.
The exponents for the norms follow as follows. From the exponent relation, we get (multiply by $p/r$): $$ \frac{p}{q r} + \frac{1}{r} = \frac{p}{r} + \frac{p}{r^2}, $$ hence, $$\frac{r+1}{r}\left(1 - \frac pr\right) = 1 + \frac1r - \frac pr - \frac p{r^2} = 1 - \frac p {qr}, $$ hence, $$ r - p = \frac{r^2}{r+1} \left( 1 - \frac{p}{q r}\right).$$ This can be used immediately in your inequality. $$