Translation-invariant operator Let $T$ be a translation invariant bounded linear operator $L^p(\mathbb{R}^d)\rightarrow L^q(\mathbb{R}^d)$, i.e. $T(\tau_c f)=T f$ where $\tau_cf(x)=f(x+c)$ for $c\in\mathbb{R}^d$. Then I have read in this marvellous post by Tao that necessarily $q\ge p$ ("the larger exponents are always on the left"). He says one can see this by considering $$f(x)=\sum_{n=1}^N g(x+x_n)$$ where $g$ is say, smooth and compactly supported. Then supposedly $$\|Tf\|_q\sim N^{1/q} \|Tg\|_q$$
But when I apply $T$ I get $Tf(x)=N\cdot Tg(x)$ and hence
$$\|Tf\|_q=N\|Tf\|_q$$
Where is my mistake? Please help me, I'm utterly confused.
 A: 
i.e. $T(\tau_cf)=Tf$ where

You're misunderstanding translation-invariance here.
A translation-invariant operator $T \colon L^p(\mathbb{R}^d) \to L^q(\mathbb{R}^d)$ is an operator that commutes with translations, i.e. $T \circ \tau_c = \tau_c \circ T$ for all $c \in \mathbb{R}^d$.
Then, with a bit of ho-humming (that can be made precise, see below), when the $x_k$ are spread far enough apart, the $T(\tau_{x_k}g) = \tau_{x_k}Tg$ have the bulk of their weights separated, so
$$\lVert Tf\rVert_q = \lVert \sum \tau_{x_k}Tg\rVert_q \approx \left(\int \sum \left\lvert (Tg)(x-x_k)\right\rvert^q\,dx \right)^{1/q} \approx \left(\sum \int \lvert(Tg)(x-x_k)\rvert^q\,dx \right)^{1/q} \approx N^{1/q} \lVert Tg\rVert_q$$
For $f$ itself, with the assumption of compact support there is no problem seeing $\lVert f\rVert_p = N^{1/p}\lVert g\rVert_p$.
To conclude that that implies $q \geqslant p$, one needs $T \neq 0$.

Making the ho-humming precise:
Let
$$\chi_r(x) = \begin{cases}
1,\quad \lVert x\rVert \leqslant r\\
0,\quad \lVert x\rVert > r.
\end{cases}$$
Fix a (smooth) $g \in L^p(\mathbb{R}^d)$ with compact support in $B_R(0)$.
Then, for $f = \sum\limits_{k = 1}^N \tau_{x_k}g$ and $r > 0$, if $\lVert x_i - x_j\rVert \geqslant 2r$ for all $i \neq j$, you have
$$\begin{align}
\lVert Tf\rVert_q &= \left\lVert\left(\sum_{k=1}^N \tau_{x_k}\bigl(\chi_r\cdot(Tg)\bigr)\right) + \left(\sum_{k=1}^N \tau_{x_k}\bigl((1-\chi_r)\cdot(Tg)\bigr)\right) \right\rVert_q\\
&\leqslant \left\lVert\left(\sum_{k=1}^N \tau_{x_k}\bigl(\chi_r\cdot(Tg)\bigr)\right)\right\rVert_q + N\cdot \lVert (1-\chi_r)(Tg)\rVert_q\\
&= \left(\int \sum_{k=1}^N \left\lvert\chi_r(x-x_k)(Tg)(x-x_k)\right\rvert^q\,dx\right)^{1/q} + N\cdot \lVert (1-\chi_r)(Tg)\rVert_q\\
&= \left(N\cdot \lVert \chi_r\cdot (Tg)\rVert^q\right)^{1/q} + N\cdot \lVert (1-\chi_r)(Tg)\rVert_q\\
&= N^{1/q} \lVert \chi_r\cdot (Tg)\rVert_q + N\cdot \lVert (1-\chi_r)(Tg)\rVert_q\\
&\leqslant N^{1/q} \lVert Tg\rVert_q + N\cdot \lVert (1-\chi_r)(Tg)\rVert_q.
\end{align}$$
The first inequality follows from the triangle inequality for $\lVert\cdot\rVert_q$. The next equalities follow from the disjointness of the supports of the $\tau_{x_k}\bigl(\chi_r\cdot(Tg)\bigr)$ and the translation-invariance (note: since the result is a number, and not a function, translation-invariance means $\lambda(\tau_c M) = \lambda(M)$ for all $c$ and measurable $M$ here) of the Lebesgue measure. The final inequality from $\lVert\chi_r\cdot h\rVert_q \leqslant \lVert h\rVert_q$ for all $h$.
Using the triangle inequality in the form $\lVert a + b \rVert \geqslant \lVert a\rVert - \lVert b\rVert$ for the split between the $\chi_r(Tg)$ and $(1-\chi_r)(Tg)$, we obtain
$$\lVert Tf\rVert_q \geqslant N^{1/q} \lVert \chi_r\cdot (Tg)\rVert_q - N\cdot \lVert (1-\chi_r)(Tg)\rVert_q$$
Now, given $g$, $N$, and an arbitrary $\varepsilon > 0$, by the dominated convergence theorem, you can choose $r_0 > 0$ so that $\lVert (1-\chi_r)\cdot(Tg)\rVert_q < \varepsilon/N$ for all $r \geqslant r_0$. Choose additionally $r_0 > 2R$, so that the $\tau_{x_k}g$ have disjoint support.
If the $x_k$ are then chosen far enough apart, you find
$$N^{1/q}\lVert Tg\rVert_q - \frac{\varepsilon}{N^{1-1/q}} - N\cdot\frac{\varepsilon}{N} \leqslant \lVert Tf\rVert_q \leqslant N^{1/q}\lVert Tg\rVert_q + \varepsilon,$$
so $\lVert Tf\rVert_q \approx N^{1/q}\lVert g\rVert_q$, and
$$N^{1/q}\lVert Tg\rVert_q - 2\varepsilon \leqslant \lVert Tf\rVert_q \leqslant \lVert T\rVert \cdot \lVert f\rVert_p = N^{1/p}\lVert T\rVert\cdot\lVert g\rVert_p.$$
Since $\varepsilon$ could be arbitrarily chosen,
$$N^{1/q}\lVert Tg\rVert_q \leqslant N^{1/p}\lVert T\rVert\cdot \lVert g\rVert_p$$
for all $N$ and (smooth) $g$ with compact support. If $T \neq 0$, a smooth $g$ with compact support and $Tg \neq 0$ exists ($\mathscr{C}_c(\mathbb{R}^d)$ is dense in $L^p(\mathbb{R}^d)$ for $p < \infty$). Then, taking the limit for $N \to \infty$ would lead to a contradiction if $p > q$.
A: The ho-humming process of Daniel Fischer can be simplified further.
Suppose $T$ is not identically zero, then there exists $0\neq f\in L^p(\mathbb{R}^d)$ such that $0\neq Tf\in L^q(\mathbb{R}^d).$ Then by density we may assume that $f\in C_c^\infty(\mathbb{R}^d).$ Again by density, for every $N\in\mathbb{N}$ we can find $0\neq g\in C_c^\infty(\mathbb{R}^d)$ such that $\Vert Tf-g\Vert_{L^q(\mathbb{R}^d)}<\frac{1}{N^2}\Vert Tf\Vert_{L^q(\mathbb{R}^d)}.$ Choose $R>0$ such that $B_{R}(0)\supset \text{supp}f\cup \text{supp}g.$ Now $x_1,\ldots,x_N$ be $N$ points in $\mathbb{R}^d$ such that $\Vert x_i-x_j\Vert>2R$ for all $i\neq j.$
Let $f_i = \text{Trans}_{x_i}f$ and $g_i = \text{Trans}_{x_i}g$ for each $i=1,2,\ldots,N.$ Then $\Vert Tf_i-g_i\Vert_{L^q(\mathbb{R}^d)}<\frac{1}{N^2}\Vert Tf_i\Vert_{L^q(\mathbb{R}^d)}$ since $T$ commutes with $\text{Trans}_{x_i}$'s.
Then we have $$\Vert \sum_{i=1}^N f_i\Vert_{L^p(\mathbb{R}^d)}=N^{1/p}\Vert f\Vert_{L^p(\mathbb{R}^d)}$$ since the supports of $f_i$'s are mutually disjoint. Also we have \begin{align*}
        \Vert \sum_{i=1}^N Tf_i\Vert_{L^q(\mathbb{R}^d)}&\geq \Vert \sum_{i=1}^N g_i\Vert_{L^q(\mathbb{R}^d)}-\sum_{i=1}^N\Vert  Tf_i-g_i\Vert_{L^q(\mathbb{R}^d)}\\&\geq N^{1/q}\Vert  g\Vert_{L^q(\mathbb{R}^d)}-\frac{1}{N}\Vert Tf\Vert_{L^q(\mathbb{R}^d)}\\&\geq N^{1/q}\frac{N^2-1}{N^2}\Vert Tf\Vert_{L^q(\mathbb{R}^d)}-\frac{1}{N}\Vert Tf\Vert_{L^q(\mathbb{R}^d)}.
    \end{align*}
Hence we get the inequality $$N^{1/q}\frac{N^2-1}{N^2}\Vert Tf\Vert_{L^q(\mathbb{R}^d)}-\frac{1}{N}\Vert Tf\Vert_{L^q(\mathbb{R}^d)}\leq N^{1/p}\Vert f\Vert_{L^p(\mathbb{R}^d)}. $$
$$\therefore \frac{N^2-1}{N^2}\Vert  Tf\Vert_{L^q(\mathbb{R}^d)}-\frac{1}{N^{1+1/q}}\Vert Tf\Vert_{L^q(\mathbb{R}^d)}\leq N^{1/p-1/q}\Vert f\Vert_{L^p(\mathbb{R}^d)}.$$
Letting $N\to\infty,$ we get $\Vert  Tf\Vert_{L^q(\mathbb{R}^d)}=0$ since $p>q,$ which is a contradiction.
