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This is Exercise 8, Chapter 2 in Stein and Shakarchi's Functional Analysis. It looks easy but I have not found a way to prove it.

Suppose $T$ is a bounded linear transformation mapping the space of real-valued $L^p$ functions into itself with $$\|T(f)\|_{L^p} \le M \|f\|_{L^p}.$$

(a) Let $T′$ be the extension of T to complex-valued functions: $T′ (f_1+ if_2) = T(f_1) + iT(f_2)$. Then $T′$ has the same bound: $\|T'(f)\|_{L^p} \le M \|f\|_{L^p}$.

(b) More generally, fix any $N$, then $$\|(\sum_{j=1}^N |T(f_j)|^2)^{1/2}\|_{L^p} \le M \|(\sum_{j=1}^N |f_j|^2)^{1/2}\|_{L^p}$$

[Hint: For part (b), let $\xi$ denote a unit vector in $\mathbb{R}^N$, and let $F_\xi = \sum_{j=1}^N \xi_j f_j , \xi = (\xi_1 , \dots , \xi_N )$. Then $\int | (TF_\xi)(x)|^p \le M^p \int | F_\xi (x) |^p$. Integrate this inequality for $\xi$ on the unit sphere.]

EDIT: Following Oliver Díaz's kind advice, here's some context. I am self studying functional analysis using the aforementioned textbook. I try to solve every exercise and problem. Most exercises are straightforward enough. I turn to stackexchange if I fail to get a solution within 24 hours.

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3 Answers 3

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The tricky part in the hint is how to estimate a particular integral over the unit sphere $\mathbb{S}^{d-1}=\{\boldsymbol{\xi}\in\mathbb{R}^d:|\boldsymbol{\xi}|_2=1\}$, namely $$\int_{\mathbf{S}^{d-1}}|\boldsymbol{a}\cdot\boldsymbol{u}|^p\,\sigma_{d-1}(d\boldsymbol{u})$$ where $\boldsymbol{a}\in\mathbb{R}^d$, $d\sigma_{n-1}$ is the spherical Lebesgue measure, and $\cdot$ is the standard inner product on $\mathbb{R}^d$. It suffices to assume that $\boldsymbol{a}\neq\boldsymbol{0}$. Let $U$ be any unitary trasformation of $\mathbb{R}^d$ such that $U\mathbf{a}=|\boldsymbol{a}|_2\mathbf{e}_d$, where $\mathbf{e}_j$, $1\leq d$ are the canonical orthogonal basis in $\mathbb{R}^d$. As $|\operatorname{det}(U)|=1$, and $U$ maps the unit ball $B(0;1)$ onto itself, $$\int_{B(0;1)}|\boldsymbol{a}\cdot \boldsymbol{x}|^p\,d\boldsymbol{x}=\int_{B(0;1)}|\boldsymbol{a}\cdot U^\intercal \boldsymbol{x}|^p\,d\boldsymbol{x}=|\boldsymbol{a}|^p_2\int_{B(0;1)}|x_d|^p\,d\boldsymbol{x}$$ Finally, by using spherical coordinates we obtain \begin{align} \int_{\mathbb{S}^{n-1}}|\boldsymbol{a}\cdot \mathbf{u}|^p\,\sigma_{d-1}(d\boldsymbol{u})=|\boldsymbol{a}|^p_2 k_{p,d}\tag{1}\label{one} \end{align} where $k_{p,d}:=\int_{\mathbb{S}^{n-1}}|u_d|^p\,\sigma_{d-1}(d\boldsymbol{u})$ is a dimensional constant.


Suppose now $T$ is a real-bounded operator in $\mathcal{L}(L_p(\mu), L_p(\nu))$ where $(X,\mathscr{F},\mu)$ and $(Y,\mathscr{E},\nu)$ are $\sigma$-finite measure spaces. Given real valued function $f_1,\ldots, f_d$ in $L_p(\mu)$ and define the map $F:X\times\mathbb{S}^{d-1}\rightarrow\mathbb{R}$ by $$F(x,\boldsymbol{u})=[f_1(x),\ldots, f_d(x)]^\intercal\cdot \boldsymbol{u}$$. To ease notation, denote $F(x,\boldsymbol{u})=F_\boldsymbol{u}(x)$. Then $$\int_X\Big|\sum^d_{k=1}u_kTf_k\Big|^{p}\,d\nu=\int_X|TF_\boldsymbol{u}|^p\,d\nu\leq\|T\|^p_{p,p}\int_X|F_\boldsymbol{u}|^p\,d\mu=\|T\|^p_{p,p}\int_X\Big|\sum^d_{k=1}u_kf_k\Big|^p\,d\mu$$ Integrating over the unit sphere $\mathbb{S}^{d-1}$, applying Fubini's theorem, and using \eqref{one} yields $$ k_{p,d} \int_X\Big(\big(\sum^d_{k=1}(Tf_k)^2\big)^{1/2}\Big)^p\,d\nu\leq k_{p,d}\|T\|^p_{p,p}\int_X\Big(\big(\sum^d_{k=1}f^2_k\big)^{1/2}\Big)^p\,d\mu $$

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Oliver Díaz's answer is wonderful. I have figured out another approach.

For (a), we can first restrict to the case where $f$ is a finite sum $\sum a_j \chi_{E_j}$, where $a_j$ are complex valued and the sets $E_j$ are disjoint and of finite measure, and use the following lemma, which is valid for complex-valued functions.

Suppose $1 \le p, q \le \infty$ are conjugate exponents. $f$ is integrable on all sets of finite measure, and $$\sup_{\|g\|_{L^q}\le 1\\ g \textrm{simple}} \left|\int fg\right| = M < \infty.$$ Then $f\in L^p$, and $\|f\|_{L^p} = M$.

For any simple function $g = \sum b_k \chi_{F_k}$, where $b_k$ are also complex valued and the sets $F_k$ are disjoint and of finite measure, we have $$ \begin{align} & \left|\int T'(f)g\right| = \left|\sum_{j,k} a_j b_k \int T(\chi_{E_j}) \chi_{F_k}\right| \\ \le& \sum_{j,k} |a_j| |b_k| \int T(\chi_{E_j}) \chi_{F_k} = \int T(\sum_j |a_j|\chi_{E_j}) \sum_k |b_k| \chi_{F_k} \\ \le& M \left\|\sum_j|a_j|\chi_{E_j} \right\|_{L^p} \left\|\sum_k|b_k|\chi_{F_k} \right\|_{L^q} = M \|f\|_{L^p} \|g \|_{L^q}. \end{align} $$

The last step is valid because $\chi_{E_j}$ are disjoint and so are $\chi_{F_k}$. This gives us $\|T'(f)\|_{L^p} \le M \|f\|_{L^p}$.

We then use the fact that simple functions are dense to get the general result.

For (b), we should be able to extend every related theorem for $L^p$ to incorporate vector-valued functions and get a similar result. I have not checked this rigorously but intuitively it seems to make sense.

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  • $\begingroup$ You proof looks fine to me. Just a couple of comments: (1) in the last inequality (third line in the row of inequalities, it should be $\|\sum_j|a_j|\mathbb{1}_{E_j}\|_{L_p}$ similar for the the neighboring factor; (2) Since your posting is about to get closed -some users may think this is a question without "context", on which I disagree completely- maybe you want to copy you answer below your question as an edit: Edit: blah,blah ... so that those who review questions (I often do that myself) decide not close yours, get some retractions and even if closed, get it reopen quickly. $\endgroup$
    – Mittens
    Nov 9, 2022 at 21:50
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    $\begingroup$ Thanks! How can I accept an answer? I see the following options: share, cite, edit, follow, flag. There is not one called accept. $\endgroup$ Nov 9, 2022 at 22:15
  • $\begingroup$ Funny! I never expected it to be clickable. Done! $\endgroup$ Nov 9, 2022 at 22:21
  • $\begingroup$ It makes answerers happier when it happens! Thanks! Notice some of your answers had been accepted as well $\endgroup$
    – Mittens
    Nov 9, 2022 at 22:22
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Although not needed for this problem, the constant $k_{p,d}$ in my previous posting concerning the hint in the textbook referred by the OP can be evaluated by using an explicit parametrization for the unit sphere $\mathbb{S}^{d-1}$. One such parametrization is discussed in this posting for example. For $d\geq2$, and $\boldsymbol{x}\in\mathbb{R}^d\setminus\boldsymbol{0}$

\begin{align} x_d&=\rho\cos\varphi_{d-1}, \quad x_k=\rho\Big(\prod^{d-1}_{j=k}\sin\varphi_j\Big) \, \cos\varphi_{k-1},\quad 2<k\leq d-1,\\ x_2&=\rho\prod^{d-1}_{j=1}\sin\varphi_j, \quad x_1=\rho\Big(\prod^{d-1}_{j=2}\sin\varphi_j\Big)\,\cos\varphi_1 \end{align} where $\rho=|\boldsymbol{x}|_2\geq0$ and $(\varphi_1,\ldots,\varphi_{n-1})\in[0,2\pi]\times[0,\pi]^{n-2}$. It is easy to check that the parameterization $\Phi:(0,\infty)\times (0,2\pi)\times(0,\pi)^{d-2}\rightarrow\mathbb{R}^d\setminus(\{0\}\times\mathbb{R}^{d-2}_+\times\mathbb{R})$ defined above is a diffeomorphism, and that \begin{align} |\det(\Phi')|=\rho^{d-1}\,\prod^{d-1}_{j=2} \sin^{j-1}\varphi_j \end{align}

Then \begin{align} k_{p, n}&=\int_{\mathbb{S}^{d-1}}|u_d|^p\,\sigma_{d-1}(\boldsymbol{u})\\ &=\int^{\pi}_0|\cos\varphi_{d-1}|^p\sin^{d-2} \phi_{d-1}\,d\varphi_{d-1}\int_{(0,\pi)^{d-3}}\int^{2\pi}_0\prod^{d-2}_{j=2}\sin^{j-1}\varphi_j d\varphi_1\ldots \,d\varphi_{d-2}\\ &=\sigma_{d-2}(\mathbb{S}^{d-2})\int^{\pi}_0|\cos t|^p\sin^{d-2} t\,dt \end{align}

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