convolution, $\frac{1}{p}+\frac{1}{q}\neq 1+\frac{1}{r}$

Question

I need help/a hint for the following task. Let $p,q,r\in [1,\infty)$ with $\frac{1}{p}+\frac{1}{q}\neq 1+\frac{1}{r}$ and $C>0$. Show that there are $f\in L^p(\mathbb R^d), g\in L^q(\mathbb R^d)$ with $\|f*g\|_r>C\|f\|_p\|g\|_q$. ($f*g$ is the convolution)
$Hint:$ Look at the functions $f_t:\mathbb R^d\rightarrow \mathbb K, x\mapsto f(tx)$ and $g_t:\mathbb R^d\rightarrow \mathbb K, x\mapsto g(tx)$ for arbitrary non-negative $f\in L^p(\mathbb R^d)$ and $g\in L^q(\mathbb R^d)$ and $t>0$.
I calculated $\|f*g\|_r=\Big (\int_{\mathbb R^d}\big |\int_{\mathbb R^d}f(tx-ty)g(ty)dy\big |^rdx\Big )^{1/r}$.
How can I use this to show the statement? Thanks for help.

Answer 1

Using the hint, let's relate properties of $f_t$ and $g_t$ to those of $f$ and $g$. Define the operator $\Phi_t$ which maps a function $h$ to the function $h_t:x\mapsto h(tx)$, so that $f_t=\Phi_t(f)$ and $g_t=\Phi_t(g)$. You have $$(f_t*g_t)(x)=\int_{\mathbb R^d}f(tx-ty)g(ty)dy=\frac1{t^d}\int_{\mathbb R^d}f(tx-y)g(y)dy=\frac1{t^d}(f*g)(tx),$$ so $f_t*g_t=t^{-d}\Phi_t(f*g)$. Also, for any $s>0$ and any $h\in L^s(\mathbb R^d)$, $$\|\Phi_t h\|_s=\left(\int_{\mathbb R^d}|h(tx)|^sdx\right)^{1/s}=\left(\frac 1{t^d}\int_{\mathbb R^d}|h(x)|^sdx\right)^{1/s}=t^{-d/s}\|h\|_s.$$ Can you use these properties to compute how $$\frac{\|f_t*g_t\|_r}{\|f_t\|_p\|g_t\|_q}$$ varies as $t$ varies?