# Complexification of a real bounded linear transformation from $L^p$ to $L^p$

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.

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 $$k_{p,d} \int_X\Big(\big(\sum^d_{k=1}(Tf_k)^2\big)^{1/2}\Big)^p\,d\nu\leq k_{p,d}\|T\|^p_{p,p}\int_X\Big(\big(\sum^d_{k=1}f^2_k\big)^{1/2}\Big)^p\,d\mu$$

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.

• 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. Nov 9, 2022 at 21:50
• Thanks! How can I accept an answer? I see the following options: share, cite, edit, follow, flag. There is not one called accept. Nov 9, 2022 at 22:15
• Funny! I never expected it to be clickable. Done! Nov 9, 2022 at 22:21
• It makes answerers happier when it happens! Thanks! Notice some of your answers had been accepted as well Nov 9, 2022 at 22:22

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 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}