# Norm of an operator and topological groups

$$G=(0,+\infty)$$ 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 $$p>1$$?

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}

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.

• Hey Marios. Your solution looks great to me. I've submitted a proposed edit to correct a typo and add the operator norm to the line of inequalities, since I think it makes that last line a little clearer. Hope you don't mind. – Josh Keneda Jul 14 at 21:01