If $T$ sends $L^r(\mathbb R^n)$ to itself with norm $B$ for some $1< r\le \infty$ then what can we say about the norm of its adjoint $T^*$ which sends $L^{r'}(\mathbb R^n)$ to itself where $r'=\frac{r}{r-1}$
$T$ is given by convolution with $K$, i.e. $T(f)=\int K(x-y)f(y)dy$ and the adjoint is defined by $\langle T(f)|g\rangle=\int T(f)\bar g\ dx=\int f \overline{T^*(g)}\ dx=\langle f|T^*(g)\rangle$
The adjoint of an operator always has the same norm as the operator itself. Note that for $g \in L^{r'}(\mathbf R^n)$, we have \begin{align*} \def\abs#1{\left|#1\right|}\def\<#1>{\left<#1\right>} \def\norm#1{\left\|#1\right\|} \norm{T^*(g)} &= \sup_{\norm{f}_r \le 1} \abs{\<T^*g,f>}\\ &= \sup_{\norm f_r\le 1}\abs{\<g,Tf>}\\ &\le \sup_{\norm f_r\le 1}\norm g\norm{Tf}\\ &= \norm g \norm T. \end{align*} Hence $\norm{T^*}\le \norm{T}$.
As $T = T^{**}$, we have $$ \norm T = \norm{T^{**}} \stackrel!\le \norm{T^*} \le \norm{T}$$ where for the first inequality (marked with !), we applied the above for $T^*$.
Therefore $\norm T = \norm{T^*}$.