Studying the off-diagonal case of this operator $f \mapsto \int_{B(0,1)} f(x-y)dy $ I am trying to show that 

the operator $T$ defined on Schwarz class functions  (on $\Bbb R^n$) by, $$f \mapsto \int_{B(0,1)} f(x-y)dy $$ (where $B(0,1)=\{t \in \Bbb R^n : ||t|| <1\}$ ) , is not strong type $(p,q)$ $\forall p \ne q$ (i.e. not strong type in the off-diagonal case!), $\forall f \in \mathscr S(\Bbb R^n) $ (Schwarz class of functions )

My approach is by dilation.
For $\lambda > 0$ , define $f^{\lambda}(x):=f(\lambda x)$
Then $$||f^\lambda||_p={(\int_{\Bbb R^n} {|f(\lambda x)|}^pdx)}^{1 \over p}=\lambda^{-n \over p} ||f||_p \dots(*)$$
And $$Tf^\lambda (x)=\int_{B(0,1)}f^\lambda(x-y)dy=\int_{B(0,1)}f(\lambda x-\lambda y)dy$$
Now if we substitute, $t=\lambda y$ in the above we get, $$Tf^\lambda(x)=\lambda^{-n} \int_{B(0,\lambda)} f(\lambda x-t)dt \dots (**)$$
My objective is to get the integral in $(**)$ exactly in the form of the integral given in the definition of the opertaor $T$ so that I can write $(**)$ as in some simple form of $f$ or $f^\lambda$ and then I'll compute the $L^q$ norm as in $(*)$ and finally do the dimensional analysis! 
Everything in $(**)$ seems okay except the fact that I am doing the integral on $B(0,\lambda)$ instead of $B(0,1)$ . Anyway, can I manipulate this while computing the $L^q$ norm of $Tf^\lambda$ ?
Please help me complete my argument
 A: First we deal with $(q,1)$ for $q>1$. Assume that $f\ne 0$ is a nonnegative Schwartz function, then it is easy to compute that $\|Tf^{\lambda}\|_{L^{1}}=\lambda^{-n}\|f\|_{L^{1}}$, it it were $\|Tf^{\lambda}\|_{L^{1}}\leq C_{q}\|f^{\lambda}\|_{L^{q}}$, then $\lambda^{-n(1-1/q)}\|f\|_{L^{1}}\leq C_{q}\|f\|_{L^{q}}$ and we can let $\lambda\downarrow 0$ to get a contradiction.
Now assume that $f \in C_{c}^{\infty}$ non-zero, non-negative and supported in $B(0,1)$, and we are to deal with $(q,p)$ for $p\ne q$ and $p>1$. Note that $f^{\lambda}$ is supported in $B(0,1/\lambda)$, whereas $Tf^{\lambda}$ is supported in $B(0,1)+B(0,1/\lambda)\subseteq B(0,1+1/\lambda)$. We have with $1/p+1/p'=1$ that
\begin{align*}
\|T(f^{\lambda})\|_{L^{1}}&\leq\|T(f^{\lambda})\|_{L^{p}}|B(0,1+1/\lambda)|^{1/p'}\\
&\leq C_{p,q}\|f^{\lambda}\|_{L^{q}}(1+1/\lambda)^{n/p'}\\
&=C_{p,q}\|f\|_{L^{q}}\lambda^{-n/q}\lambda^{-n/p'}(1+\lambda)^{n/p'}\\
&=C_{p,q}\|f\|_{L^{q}}\lambda^{n(1/p-1/q-1)}(1+\lambda)^{n/p'},
\end{align*}
and hence
\begin{align*}
\lambda^{-n}\|f\|_{L^{1}}&\leq C_{p,q}\|f\|_{L^{q}}\lambda^{n(1/p-1/q-1)}\\(1+\lambda)^{-n/p'}\lambda^{n(1/q-1/p)}\|f\|_{L^{1}}&\leq C_{p,q}\|f\|_{L^{q}}.
\end{align*}
If $1/q-1/p<0$, then we let $\lambda\downarrow 0$ to get a contradiction.
To deal with $1/q-1/p>0$, we consider the adjoint operator $T^{\ast}$ defined by $\left<Tf,g\right>=\left<f,T^{\ast}g\right>$, one may check that $T^{\ast}$ is formally $T$ itself. And if $T:L^{q}\rightarrow L^{p}$ bounded, then so is $T^{\ast}:L^{p}\rightarrow L^{q}$, but our previous result contradicts the boundedness of $T^{\ast}$.
