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Recall that ~o show continuity of a linear map, you only need to
show that it is continuous at $0$
If you estimate
$\lVert f\cdot 1_A \rVert_q^{q/p} + \lVert f\cdot 1_{A^c} \rVert_q^{q/r}$
by
$\lVert f \rVert_q^{q/p} + \lVert f\rVert_q^{q/r}$,
then we have
$$
\| f \|_{L^p+L^r} \to 0
\quad\text{for}\; \|f\|\to0.
$$
Thus, the inclusion is continuous at $0$ and therefore continuous.
No, there is no almost multiplicative norm on $\mathbb R[X].$ The argument is an adaptation of the argument for multiplicative norms, using the following statement (copy and pasted from https://math.stackexchange.com/a/2674259/3643)
Theorem (Gelfand–Mazur). Let $A$ be a real normed division algebra. Then $A$ is isomorphic with $\mathbb{R}$, $\mathbb{C}$, or the algebra of quaternions $\mathbb{H}$.
For a proof, see theorem 14.7 in [F.F. Bonsall, J. Duncan, Complete Normed Algebras, Springer–Verlag, Berlin Heidelberg New York 1973].
Given $\|.\|,$ we will define a norm $\|.\|'$ on the field $\mathbb R(X)$ by:
$$\|p/q\|'=\inf_{g}\sup_{h}\|ghp\|/\|ghq\|\tag{•}$$
where the inf and sup range over $\mathbb R[X]\setminus\{0\}.$
We need to check (•) does not depend on the representatives $p,q.$ I will write it as $\inf_g S(g,p,q)$ where $$S(g,p,q):=\sup_h \|ghp\|/\|ghq\|.$$
Note:
- $S(g,p,q)\geq S(gk,p,q)$ for any non-zero polynomial $k\neq 0$
- Hence $\inf_g S(g,p,q)=\inf_g S(gk,p,q)$
Also, $S(gq,p',q')=S(gq',p,q)$ if $pq'=p'q.$ Together with property 2, this implies that $\|.\|'$ is well-defined.
For submultiplicativity of $\|.\|'$ consider non-zero $p,q,p',q'.$ Let $\epsilon>0$ be arbitrary.
Pick $g_1,g_2$ such that $S(g_1p',p,q)\leq \|p/q\|'+\epsilon$ and $S(g_2q,p',q')\leq \|p'/q'\|'+\epsilon$ (making use of property 2). Set $g=g_1g_2.$ Using property 1 we get the same inequalities with $g$ instead of $g_1$ and $g_2.$
The identity $\frac{\|ghpp'\|}{\|ghqq'\|}=\frac{\|ghpp'\|}{\|ghqp'\|}\frac{\|ghqp'\|}{\|ghqq'\|}$ implies $S(g,pp',qq')\leq S(gp',p,q)S(gq,p',q'),$ so $\|pp'/qq'\|'\leq (\|p/q\|'+\epsilon)(\|p'/q'\|'+\epsilon).$ Since $\epsilon$ was arbitrary we get $\|pp'/qq'\|'\leq\|p/q\|'\|p'/q'\|'.$
So $\|.\|'$ makes $\mathbb R(X)$ an infinite-dimensional normed division algebra, which is impossible by the Gelfand-Mazur theorem.
Best Answer
(1) Yes, there are supermultiplicative norms. (2) No, no positive multiple of the norm defined by $\Big\lVert\sum_ka_kx^k\Big\rVert=\max_k(k!|a_k|)$ is supermultiplicative.
Proofs:
(1) We’ll construct a norm of the form $\|\sum a_kx^k\|=\max_k(c_k|a_k|).$ We construct the $c_k$ to ensure supermultiplicativity, with a slightly stronger property to help the induction.
Take $c_0=c_1=\tfrac 1 2.$ For each $n\geq 1,$ suppose for induction that we have constructed $c_0,\dots,c_n>0$ such that, if we define a seminorm by $\|\sum a_kx^k\|_n=\max_{k\leq n}(c_k|a_k|),$ then $$\|fg\|_n\geq (1+2^{-\deg(fg)})\|f\|_n\cdot\|g\|_n\tag{†}$$ holds whenever $0\leq \deg(fg)\leq n.$
Consider the set $S$ of pairs of polynomials $(f,g)$ such that: $\deg(f),\deg(g)\leq n,$ and $\deg(fg)\leq n+1,$ and $\|f\|_n=\|g\|_n=1,$ and $\|fg\|_n\leq 1+2^{-(n+1)}.$ The set $S$ is compact (with the “usual” topology, as a subspace of a finite dimensional vector space). Let $m$ be the minimum absolute value of the coefficient of $x^{n+1}$ in $f(x)g(x),$ over all $(f,g)\in S.$ This is attained by compactness of $S.$ We must have $m\neq 0,$ because $m=0$ would imply (†) which contradicts $(f,g)\in S.$ Taking $c_{n+1}=(1+2^{-(n+1)})/m$ works: (†) holds with $n$ replaced by $n+1.$ (We only needed to check factors of degree $\leq n$ because (†) holds when $f$ or $g$ is constant and $c_0\leq \tfrac 1 2.$)
(2) Consider $f_N(x)=\prod_{n=1}^N(1-x/\pi n)$ and $g_N(x)=xf_N(-x)=x\prod_{n=1}^N(1+x/\pi n).$ The $x$ coefficient of $f_N(x)$ tends to $-\infty$ as $N\to\infty,$ and similarly the $x^2$ coefficient of $g_N(x)$ tends to $\infty.$ So $\|f_N\|\cdot\|g_N\|$ is unbounded.
But $\|f_Ng_N\|\leq 1.$ Let $a_{N,k}$ be the $x^k$ coefficient of $f_N(x)g_N(x)=x \prod_{n=1}^N(1-x^2/\pi^2 n^2).$ If we expand the product, all the terms contributing to $a_{N,k}$ have the same sign: there are no terms for even $k$ and the sign for odd $k$ is the same as $(-1)^{(k-1)/2}.$ So $|a_{N,k}|$ converges from below. The sequence $f_Ng_N$ converges locally uniformly to $\sin x$ - this is a standard exercise with the Hadamard factorization thorem. Locally uniform convergence gives convergence of Taylor series coefficients by Cauchy’s differentiation formula. Hence $|a_{N,k}|\leq 1/k!$ as required.