Equivalent Norms for Intermediate Subspaces

Question

Let $(X,\left\|\cdot\right\|)$ be a Banach space, and let $\left\{T(t) : t\geq 0\right\}$ be an equibounded strongly continuous semi-group on $X$. Define a functional $\left\|\cdot\right\|_{\alpha,r;q}:X\rightarrow[0,\infty)$ by

$$\left\|f\right\|_{\alpha,r;q}:=\begin{cases} \displaystyle{\left\|f\right\|+\left(\int_{0}^{\infty}(t^{-\alpha}\left\|[T(t)-I)]^{r}f\right\|^{q}\dfrac{dt}{t}\right)^{q}} & {(0<\alpha<r;1\leq q<\infty)}\\ \\ \displaystyle{\left\|f\right\|+\sup_{0<t<\infty}(t^{-\alpha}\left\|(T(t)-I)^{r}f\right\|)} & {(0\leq\alpha\leq r; q=\infty, r\in\mathbb{N})}\end{cases} $$

where for $r\in\mathbb{N}$, $[T(t)-I]^{r}f$ is the $r^{th}$ difference.

Definition. We say that an element $f\in X$ belongs to the intermediate space $X_{\alpha,r;q}$ if $\left\|f\right\|_{\alpha,r;q}<\infty$.

One can show that the spaces $(X_{\alpha,r;q},\left\|\cdot\right\|_{\alpha,r;q})$ are Banach space and that they are continuously embedded between $D(A^{r})$ (the domain of the $r^{th}$ power of the infinitesimal generator of the semigroup) and $X$.

For $f\in X$, define the $r^{th}$ modulus of continuity $\omega_{r}(t;f)$ by

$$\omega_{r}(t;f):=\sup_{0\leq h\leq t}\left\|[T(h)-I]^{r}f\right\|$$

Problem. I am trying to show that the norms

$$\left\|f\right\|+\left(\int_{0}^{\infty}(t^{-\alpha}\left\|[T(t)-I]^{r}f\right\|)^{q}\dfrac{dt}{t}\right)^{1/q} \tag{1}$$

and

$$\left\|f\right\|+\left(\int_{0}^{\infty}(t^{-\alpha}\omega_{r}(t;f))^{q}\dfrac{dt}{t}\right)^{1/q} \tag{2}$$

are equivalent for the intermediate subspace $X_{\alpha,r;q}$, where $0<\alpha<r, 1\leq q<\infty$. This problem stems from Theorem 3.4.2 in Butzer and Berens, Semi-Groups of Operators and Approximation, in which the authors assert the equivalence saying "it is not hard to see". I have shown the case $r=1$ (see answer below), but I am not seeing how my argument generalizes to the case $r>1$. I think I may be missing some algebraic identity.

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Edit 2: Generalized the formulation of the original question.

Answer 1

In what follows below, I will denote the norm defined in (1) by $\left\|\cdot\right\|_{\alpha,r;q}^{(1)}$ and the norm defined in (2) by $\left\|\cdot\right\|_{\alpha,r;q}^{(2)}$.

So the algebraic identity that I was missing is the following lemma, which allows us to write the $r^{th}$ difference $[T(t)-I]^{r}$ as a linear combination of $[T(2t)-I]^{r}$ and $[T(t)-I]^{r+1}$. Using this lemma together with the equiboundedness of the semigroup argument allows us to show that for any positive integers $s,r>\alpha$, the norms $\left\|\cdot\right\|_{\alpha,r;q}^{(i)}$ and $\left\|\cdot\right\|_{\alpha,s;q}^{(i)}$ are equivalent. Using the algebraic lemma, we can also show that $\left\|\cdot\right\|_{\alpha,r;q}^{(1)}$ dominates $\left\|\cdot\right\|_{\alpha,2r;q}^{(2)}$. The rest of the proof is just bookkeeping.

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Lemma 1. There exists a polynomial $P$ with real coefficients such that, for any $m\in \mathbb{N}$, $$(z-1)^{m}=2^{-m}(z^{2}-1)^{m}+(z-1)^{m+1}P(z)$$

Proof. Write $Q(z):=(z-1)^{m}-2^{-m}(z^{2}-1)^{m}$, which we can factor as $Q(z)=(z-1)^{m}(1-2^{-m}(z+1))$. The second factor has a root at $z=1$, whence $Q$ is divisble by $(z-1)^{m+1}$ (i.e. $P$ exists). $\Box$

We can apply the lemma replacing $z$ with $T(t)$ and $1$ with $I$ to obtain $$[T(t)-I]^{m}=2^{-m}[T(t)^{2}-I]^{m}+[T(t)-I]^{m+1}P(T(t))=2^{-m}[T(2t)-I]^{m}+P(T(t))[T(t)-I]^{m+1}$$

Lemma 2. If $r\in\mathbb{N}$ is $>\alpha$, then $\left\|\cdot\right\|_{\alpha,r;q}^{(1)}$ and $\left\|\cdot\right\|_{\alpha,r+1;q}^{(1)}$ are equivalent. Analogously, $\left\|\cdot\right\|_{\alpha,r;q}^{(2)}$ and $\left\|\cdot\right\|_{\alpha,r+1;q}^{(2)}$ are equivalent.

Proof. We only prove the second assertion, as the arguments are the same. Since $P(T(t))$ is a finite linear combination of powers of $T(t)$ and $\left\{T(t)\right\}$ is equibounded, it follows that there is a some constant, which only depends on $P$, such that $$\left\|P(T(t))[T(t)-I]^{m+1}\right\|\leq C\left\|[T(t)-I]^{m+1}\right\|, \qquad\forall t\geq 0$$ Whence, if $\delta>0$ and $t\leq\delta$, then $$\left\|[T(t)-I]^{r}\right\|\leq 2^{-r}\left\|[T(2t)-I]^{r}\right\|+C\left\|[T(t)-I]^{r+1}\right\|\leq 2^{-r}\omega_{r}(f;2\delta)+C\omega_{r+1}(f;\delta)$$ Taking the supremum of the LHS over all $t\leq\delta$, we conclude that $$\omega_{r}(f;\delta)\leq2^{-r}\omega_{r}(f;2\delta)+C\omega_{r+1}(f;\delta),\qquad\forall\delta>0$$ Whence, $$\left(\int_{0}^{\infty}h^{-\alpha q}\omega_{r}(f;h)^{q}\dfrac{dh}{h}\right)^{1/q}\leq2^{-r}\left(\int_{0}^{\infty}h^{-\alpha q}\omega_{r}(f;2h)^{q}\dfrac{dh}{h}\right)^{1/q}+C\left(\int_{0}^{\infty}h^{-\alpha q}\omega_{r+1}(f;h)^{q}\dfrac{dh}{h}\right)^{1/q}$$ Making the change of variable $2h\mapsto h$ in the first integral on the RHS, we obtain that $$2^{-r}\left(\int_{0}^{\infty}h^{-\alpha q}\omega_{r}(f;2h)^{q}\dfrac{dh}{h}\right)^{1/q}=2^{-r+\alpha}\left(\int_{0}^{\infty}h^{-\alpha q}\omega_{r}(f;h)^{q}\dfrac{dh}{h}\right)^{1/q}$$ Since $r>\alpha$, $2^{-r+\alpha}<1$, from which we conclude $$\left(\int_{0}^{\infty}h^{-\alpha q}\omega_{r}(f;h)^{q}\dfrac{dh}{h}\right)^{1/q}\leq\dfrac{C}{(1-2^{-r+\alpha})}\left(\int_{0}^{\infty}h^{-\alpha q}\omega_{r+1}(f;h)^{q}\dfrac{dh}{h}\right)^{1/q}$$ The reverse inequality is trivial, as $$\left\|[T(t)-I]^{r+1}f\right\|\leq 2M\left\|[T(t)-I]^{r}f\right\|\leq 2M\omega_{r}(f;\delta)$$ for all $t\leq \delta$. Taking the supremum of the LHS over $t\leq\delta$ completes the proof. $\Box$

Lemma 3. For $r\in\mathbb{N}$ with $r>\alpha$, $\left\|\cdot\right\|_{\alpha,r;q}^{(1)}$ and $\left\|\cdot\right\|_{\alpha,2r;q}^{(2)}$ are equivalent.

Proof. First, observe that by the triangle inequality, \begin{align*} I&=\left(\int_{0}^{\infty}h^{-\alpha q}\omega_{r}(f;h)^{q}\dfrac{dh}{h}\right)^{1/q}\\ &\leq\left(\int_{0}^{\infty}h^{-\alpha q}\left(\sup_{h/2\leq t\leq h}\left\|[T(t)-I]^{r}f\right\|\right)^{q}\dfrac{dh}{h}\right)^{1/q}+\left(\int_{0}^{\infty}h^{-\alpha q}\omega_{r}(f;h/2)^{q}\dfrac{dh}{h}\right)^{1/q}\\ &=\left(\int_{0}^{\infty}h^{-\alpha q}\left(\sup_{h/2\leq t\leq h}\left\|[T(t)-I]^{r}f\right\|\right)^{q}\dfrac{dh}{h}\right)^{1/q}+2^{-\alpha}I, \end{align*} where the final equality comes from making the change of variable $h/2\mapsto h$ in the second integral. Since $2^{-\alpha}<1$, rearranging shows that it suffices to estimate the first term. Let $h/2\leq t\leq h$, and consider the interval $(h/8,h/4)$. Write $t$ as the sum $t=t_{1}+t_{2}$, where $t_{1}\in (h/8,h/2)$, and $$T(t)-I=T(t_{1})[T(t_{2})-I]+[T(t_{1})-I]$$ Using the binomial theorem, we obtain \begin{align*} [T(t)-I]^{2r}&=\sum_{j=0}^{2r}{2r\choose j}(T(t_{1}))^{j}[T(t_{2})-I]^{j}[T(t_{1})-I]^{2r-j} \end{align*} Observe that on the RHS either the exponent of $[T(t_{1})-I]$ is $\geq r$ or the exponent of $[T(t_{2}-I)]$ is $\geq r$. So it follows from equiboundedness, the triangle inequality, and the estimate $(a+b)^{q}\leq 2^{q-1}(a^{q}+b^{q})$ that there exist constant $c>0$ such that $$\left\|[T(t)-I]^{2r}f\right\|^{q}\leq c\left\|[T(t_{1})-I]^{r}f\right\|^{q}+c\left\|[T(t_{2})-I]^{r}f\right\|^{q},\qquad\forall f\in X$$ Integrating both sides with respect to $t_{1}$ over $(h/8/,h/4)$ and noting that $7h/8\geq t_{2}\geq h/4$, we obtain \begin{align*} (h/8)\left\|[T(t)-I]^{2r}f\right\|^{q}&\leq c\int_{h/8}^{h/4}\left\|[T(t_{1})-I]^{r}f\right\|^{q}dt_{1}+c\int_{h/8}^{h/4}\left\|[T(t-t_{1})-I]^{r}f\right\|^{q}dt_{1}\\ &\leq c\int_{h/8}^{h}\left\|[T(s)-I]^{r}f\right\|^{q}ds \end{align*} Dividing both sides by $h/8$ and making the change of variable $s\mapsto s/h$, we obtain $$\left\|[T(t)-I]^{2r}f\right\|^{q}\leq c'\int_{1/8}^{1}\left\|[T(s h)-I]^{r}f\right\|^{q}ds$$ Since the RHS is independent of $t\in [h/2,h]$, we take the supremum of the LHS over such $t$ to get an upper bound for $(\sup_{h/2\leq t\leq h}\left\|[T(t)-I]^{2r}f\right\|)^{q}$. \begin{align*} \int_{0}^{\infty}h^{-\alpha q}(\sup_{h/2\leq t\leq h}\left\|[T(t)-I]^{2r}f\right\|)^{q}\dfrac{dh}{h}&\leq c'\int_{0}^{\infty}h^{-\alpha q}\left(\int_{1/8}^{1}\left\|[T(sh)-I]^{r}f\right\|^{q}ds\right)\dfrac{dh}{h}\\ &=c'\int_{1/8}^{1}\left(\int_{0}^{\infty}\left\|[T(sh)-I]^{r}f\right\|^{q}h^{-\alpha q}\dfrac{dh}{h}\right)ds\\ &=c'\int_{1/8}^{1}s^{\alpha q}\left(\int_{0}^{\infty}\left\|[T(h)-I]^{r}f\right\|^{q}h^{-\alpha q}\dfrac{dh}{h}\right)ds\\ &\leq c''\int_{0}^{\infty}\left\|[T(h)-I]^{r}f\right\|^{q}h^{-\alpha q}\dfrac{dh}{h}, \end{align*} where we use Fubini's theorem together with the fact that $dh/h$ is invariant under dilation (it's the Haar measure for the multiplicative group $\mathbb{R}^{+}$). $\Box$

Applying Lemmas 2 and 3, we conclude that there are constants $c_{1},c_{2}>0$, which may depend on $\alpha,q,r$, such that \begin{align*} \left\|f\right\|_{\alpha,r;q}^{(2)}\leq c_{1}\left\|f\right\|_{\alpha,2r,q}^{(2)}\leq c_{2}\left\|f\right\|_{\alpha,r;q}^{(1)}\leq\left\|f\right\|_{\alpha,r;q}^{(2)},\qquad\forall f\in X \end{align*}

Answer 2

In what follows, $M>0$ is such that $\left\|T(t)\right\|\leq M$ for all $0\leq t<\infty$ (such a constant exists by hypothesis that $\left\{T(t):t\geq 0\right\}$ is equibounded).

We state without proof the following lemma. A proof can be found in Proposition 3.1.4, pg. 162 of Butzer and Huber.

Lemma 1. For any $r\in\mathbb{N}$, $D(A^{r})$ is dense in $X_{\alpha,r;q}$.

To avoid a technical issue, we will also need the following lemma.

Lemma 2. For each $f\in D(A^{r})$, $\left\|f\right\|_{\alpha,r;q}^{'}<\infty$.

Proof. For $f\in D(A^{r})$, one can show that $$\left\|[T(t)-I]^{r}f\right\|\leq M\left\|A^{r}f\right\|t^{r},\quad\forall 0\leq t<\infty$$ Furthermore, for any $f\in X$, one has $$\left\|[T(t)-I]^{r}f\right\|\leq(M+1)^{r}\left\|f\right\|,\quad\forall 0\leq t<\infty$$ Combining these estimates, we obtain that $$\left\|[T(t)-I]^{r}f\right\|\leq(M+1)^{r}\min\left\{1,t^{r}\right\}\left(\left\|f\right\|+\left\|A^{r}f\right\|\right), \quad\forall 0\leq t<\infty $$ for every $f\in D(A^{r})$. Whence, $$\omega_{r}(t;f)=\sup_{0\leq h\leq t}\left\|[T(h)-I]^{r}f\right\|\leq (M+1)^{r}\min\left\{1,t^{r}\right\}\left(\left\|f\right\|+\left\|A^{r}f\right\|\right), \quad\forall 0\leq t<\infty$$ and therefore, $$\left(\int_{0}^{\infty}(t^{-\alpha}\omega_{r}(t;f))^{q}\dfrac{dt}{t}\right)^{1/q}\leq(M+1)^{r}\left(\int_{0}^{1}(t^{-\alpha}t^{r})^{q}\dfrac{dt}{t}+\int_{1}^{\infty}t^{-\alpha q}\dfrac{dt}{t}\right)^{1/q}<\infty$$

$\Box$

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Suppose we have shown that there exists a constant $C>0$ such that $$\left\|f\right\|_{\alpha,r;q}^{'}\leq C\left\|f\right\|_{\alpha,r;q},\quad\forall f\in D(A^{r}) \tag{*}$$ I claim the inequality then holds for all $f\in X_{\alpha,r;q}$. Indeed, let $\left\{f_{n}\right\}$ be a sequence in $D(A^{r})$ such that $\left\|f-f_{n}\right\|_{\alpha,r;q}\rightarrow 0$ as $n\rightarrow\infty$. Then $$\limsup_{n\rightarrow\infty}\left\|f_{n}\right\|_{\alpha,r;q}^{'}\leq C\limsup_{n\rightarrow\infty}\left\|f_{n}\right\|_{\alpha,r;q}=C\left\|f\right\|_{\alpha,r;q}$$

Since $f_{n}\rightarrow f$ in $X$ implies $\omega_{r}(t;f_{n})\rightarrow\omega_{r}(t;f)$, for any $t>0$ fixed, we have by Fatou's lemma that \begin{align*} \int_{0}^{\infty}(t^{-\alpha}\omega_{r}(t;f))^{q}\dfrac{dt}{t}\leq\liminf_{n\rightarrow\infty}\int_{0}^{\infty}(t^{-\alpha}\omega(t;f_{n}))^{q}\dfrac{dt}{t}&\leq\liminf_{n\rightarrow\infty} C\int_{0}^{\infty}(t^{-\alpha}\left\|[T(t)-I]^{r}f_{n}\right\|)^{q}\dfrac{dt}{t}\\ &=C\int_{0}^{\infty}(t^{-\alpha}\left\|[T(t)-I]^{r}f\right\|)^{q}\dfrac{dt}{t} \end{align*}

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We now prove (*) for $r=1$. Let $h>0$, and let $h\leq t\leq 2h$. Since $$[T(t)-I]=[T(t-h)T(h)-T(t-h)]+[T(t-h)-I]=T(t-h)[T(h)-I]+[T(t-h)-I],$$ we see that for any $f\in X$, $$\left\|[T(t)-I]f\right\|\leq M\left\|[T(h)-I]f\right\|+\left\|[T(t-h)-I]f\right\|\leq M\left\|[T(h)-I]f\right\|+\omega(h;f),$$ since $0\leq t-h\leq h$. It is evident that the above inequality also holds for $t\leq h$. Taking the supremum of the LHS over $t\leq 2h$, we obtain $$\omega(2h;f)\leq M\left\|[T(h)-I]f\right\|+\omega(h;f)$$

Now for $f\in D(A^{r})$, $$\left(\int_{0}^{\infty}(h^{-\alpha}\omega(2h;f))^{q}\dfrac{dh}{h}\right)^{1/q}\leq M\left(\int_{0}^{\infty}(h^{-\alpha}\left\|[T(h)-I]f\right\|)^{q}\dfrac{dh}{h}\right)^{1/q}+\left(\int_{0}^{\infty}(h^{-\alpha}\omega(h;f))^{q}\dfrac{dh}{h}\right)^{1/q}$$ Making the change of the variable $2h\mapsto h$, we obtain the inequality $$2^{\alpha}\left(\int_{0}^{\infty}(h^{-\alpha}\omega(h;f))^{q}\dfrac{dh}{h}\right)^{1/q}\leq M\left(\int_{0}^{\infty}(t^{-\alpha}\left\|[T(h)-I]f\right\|)^{q}\dfrac{dh}{h}\right)^{1/q}+\left(\int_{0}^{\infty}(h^{-\alpha}\omega(h;f))^{q}\dfrac{dh}{h}\right)^{1/q}$$ Whence, $$\left(\int_{0}^{\infty}(h^{-\alpha}\omega(h;f))^{q}\dfrac{dh}{h}\right)^{1/q}\leq\dfrac{M}{2^{\alpha}-1}\left(\int_{0}^{\infty}(t^{-\alpha}\left\|[T(h)-I]f\right\|)^{q}\dfrac{dh}{h}\right)^{1/q},\quad\forall f\in D(A)$$

This completes the proof in the case $r=1$. I'm failing to see an algebraic identity at the moment that would lend this argument to the case $r>1$.