If $T$ is a bounded linear operator from $L^p(\mathbb{R})$ to $L^q(\mathbb{R})$ and $T$ is non-zero, then $p \leq q$

Question

I'm a student in a Fourier Analysis class with a broad background in Functional Analysis. When this question was posed to me, I wasn't sure how to begin. Poking around, I looked at $L^p$ embeddings but $T$ may not be injective into $L^q$. Trying to solve this problem naively, I considered $f$ to have $L^p$ norm 1 (the sphere) and considered $$(\int_{\mathbb{R}} |Tf(x)|^q)^{\frac{1}{q}} \leq C $$ where $C$ is the norm of $T$. Can I "normalize" T and assume $C=1$ to get rid of that $\frac{1}{q}$ exponent. How would $f$ having norm 1 help me here?

Edit: Would it be possible to "mod out" by the kernel of $T$ and consider that as an injective linear operator from $\frac{L^p(\mathbb{R}) }{kerT}$ to $L^q(\mathbb{R})$? I don't think I'd be able to use any embedding results here though.

Answer 1

As stated, this is not true at all. For example, let $\phi$ be any bounded linear functional on $L^p$, $g$ any nonzero member of $L^q$, and define $T f = \phi(f) g$. That defines a nonzero bounded linear operator from $L^p(\mathbb R)$ into $L^q(\mathbb R)$. Perhaps the assumption was supposed to be that $T$ is injective?