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Given some linear operator $T: V \mapsto W$, we can talk about the operator norm between the spaces V and W, i.e. $$ \| T \|_{V \mapsto W} \ = \ \sup_{g} \| Tg \|_W \ , \quad \mbox{ with } \| g \|_V \leq 1 $$

I am wondering whether given such an operator norm, and under certain conditions, we can say more general things about the operator norm between other spaces?

For example, if $V = W = L_2(0,1)$, then can we say anything about the norm $\| \cdot \|_{L_\infty \mapsto L_1}$ etc?

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  • $\begingroup$ You need to fix a number of problems for your question to make sense. Your definition of operator norm is incorrect (the inequality cannot hold for all $g$, one takes a supremum over $g$ of norm $1$); for non-linear operator this seems unusual; if the operator is unbounded you won't have much to study; and your example seems self-contradictory ($L^2$ becomes $L^\infty$ and $L^1$, which is confusing). $\endgroup$
    – Benoît Kloeckner
    Commented Dec 20, 2017 at 8:24
  • $\begingroup$ My apologies Benoit. In terms of the non-linearity it was not necessary but I thought someone might have some interesting viewpoint (I have now restricted to linear operators). In terms of the operator norm, I should have been more precise (this has been fixed). $\endgroup$
    – Atransportconfusion
    Commented Dec 20, 2017 at 8:32
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    $\begingroup$ I don't quite see what you can expect. Constructing operators that violate any inequality you may formulate is pretty standard. For example, the identity from $L^2$ to $L^2$ extends to $L^\infty$ but sends most bounded function to non-$L^1$ function. $\endgroup$
    – Benoît Kloeckner
    Commented Dec 20, 2017 at 13:01

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This is not the right way to think about operator norms. Instead, you can say that if $T:L^2\rightarrow L^2$ and $T:L^\infty\rightarrow L^1$ (as you consider) are bounded, then $T:L^p\rightarrow L^{p'}$ is bounded for every $p\ge2$ and $\frac1p+\frac1{p'}=1$. More generally, the Riesz-Thorin Theorem tells you that if $T:L^{p_0}\rightarrow L^{q_0}$ and $T:L^{p_1}\rightarrow L^{q_1}$ are bounded, then $T:L^{p}\rightarrow L^{q}$ is bounded whenever $$\frac1p=\frac{1-\theta}{p_0}+\frac\theta{p_1},\qquad\frac1q=\frac{1-\theta}{q_0}+\frac\theta{q_1}$$ and $\theta\in(0,1)$. In addition $$\|T\|_{p\rightarrow q}\le\|T\|_{p_0\rightarrow q_0}^{1-\theta}\|T\|_{p_1\rightarrow q_1}^\theta.$$ The canonical example is that of Fourier transform: $\cal F$ is bounded from $L^p({\mathbb R}^d)$ to $L^{p'}({\mathbb R}^d)$ whenever $p\in[1,2]$.

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  • $\begingroup$ Hi Denis. Thank you for the quick response. I think you may have misunderstood my problem. In the above, you're essentially talking about interpolation-type arguments (Holder, CS, Riesz-Thorin etc) when you know $T$ is bounded on two different mappings ($L_2 \mapsto L_2$ and $L_\infty \mapsto L_1$). What I am actually referring to is this - if we only have one known mapping e.g. $T: L_2 \mapsto L_2$, can we say anything else about $T$? Or is the above exactly the point, that we need to know some other mapping property of $T$ to know anything additional (by interpolation)? $\endgroup$
    – Atransportconfusion
    Commented Dec 20, 2017 at 10:08

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