Norm of an operator and topological groups

Question

$G=(0,+\infty)$ the multiplicative topological group (with respect to the standard topology of the real line) equipped with the positive Haar measure $\mu=\frac{1}{x} dx$

For $1 \leq p< \infty$, we define the operator $T:L^p(G) \to L^p(G)$ to be $T(f)(x)=(f\star K)(x)=\int_0^{\infty}K(t)f(\frac{x}{t})\frac{dt}{t}$, where $K \in L^1(G)$ is nonnegative.

We must compute the norm of $T$.

Clearly from Minkowski's inequality we have that $||T|| \leq ||K||_1$

How can i prove that $||T||=||K||_1$?

Can you give me a hint?

$\text{EDIT}$

Ι can present my attempt when $p=1$

Let $0<\epsilon<1$ and $N>1$

Let $g_{N,\epsilon}=1_{(\epsilon,N)}$ and $K_N=K1_{(0,N)}$

$$(g_{N,\epsilon}\star K_N)(x)=\int_0^{\infty}K_N(t)1_{(\frac{x}{N},\frac{x}{\epsilon})}(t) \frac{dt}{t}$$

$||g_{N,\epsilon}||_1=\log{N}-\log{\epsilon}$

Also $1_{(\frac{x}{N},\frac{x}{\epsilon})}(t)=1$ if and only if $1_{(t\epsilon,tN)}(x)=1$

From this and Tonneli's theorem we have that

$$||g_{N,\epsilon}\star K_N||_1=\int_0^{\infty}\int_0^{\infty}K_N(t)1_{(\frac{x}{N},\frac{x}{\epsilon})}(t) \frac{dt}{t}\frac{dx}{x}$$ $$=\int_0^{\infty}\int_0^{\infty}K_N(t)1_{(t\epsilon,tN)}(x) \frac{dx}{x}\frac{dt}{t}=||K_N||_1(\log{N}-\log{\epsilon})$$

Using the definition of the norm of an operator and letting $N \to +\infty$ we have the desired inequality.

Is this correct ?

If it is, can someone help me adapting a similar idea to solve this for $p>1$?

Thank you in advance.

Answer 1

I believe I came up with a solution for general $p \geq 1$.

In general from the definition of a left Haar measure and using a change of variables, we have for $x>0$ that:

$$(f \star g)(x)=\int_0^{\infty}f(tx)g(\frac{1}{t})\frac{dt}{t}=\int_0^{\infty}f(tx)g(\frac{x}{xt})\frac{dt}{t}=\int_0^{\infty}f(t)g(\frac{1}{tx})\frac{dt}{t}.$$

Now let $K_N=K1_{(\frac{1}{N},N)}$ and $s>2, s \in \Bbb{N}$ and $g_{N,s}=1_{(\frac{1}{N^s},N)}$.

We have that $g_{N,s}=1_{(\frac{1}{N^s},N)}(\frac{1}{tx})=1 \Longleftrightarrow \frac{1}{Nx} <t<\frac{N^s}{x}.$

Thus for $x \in (1,N^{s-1})$, we have that $(K_N \star g_{N,s})(x)=||K_N||_1$.

So \begin{align*} \|T\|^p &\ge \frac{||K*g_{N,s}||^p_p}{||g_{N,s}||_p^p} \geq \frac{||K_N*g_{N,s}||^p_{L^p(1,N^{s-1})}}{||g_{N,s}||_p^p}=||K_N||_1^p \frac{(s-1)\log{N}}{(s+1)\log{N}}\\ &\ge||K_N||_1^p \frac{s-1}{s+1}. \end{align*}

Sending firstly $s \to +\infty$ and then $N \to +\infty$, we have the desired conclusion.