# $L^{p_0} \cap L^{p_1}$ and $L^{q_0} + L^{q_1}$ are duals of each other

This is from Exercise 26, Chapter 1, in Stein and Shakarchi's Functional Analysis.

Suppose $$1 < p_0, p_1 < \infty$$ and $$1/p_0+ 1/q_0 = 1$$ and $$1/p_1 + 1/q_1 = 1$$. Show that the Banach spaces $$L^{p_0} \cap L^{p_1}$$ and $$L^{q_0} + L^{q_1}$$ are duals of each other up to an equivalence of norms.

Below are the definitions of $$L^{p_0} \cap L^{p_1}$$ and $$L^{p_0} + L^{p_1}$$.

Define the norm of $$f \in L^{p_0} \cap L^{p_1}$$ as $$\|f\|_{L^{p_0} \cap L^{p_1}} = \|f\|_{L^{p_0}} + \|f\|_{L^{p_1}}.$$

$$L^{p_0}+L^{p_1}$$ is defined as the vector space of measurable functions $$f$$ on a measure space $$X$$, that can be written as a sum $$f=f_0+f_1$$ with $$f_0\in L^{p_0}$$ and $$f_1\in L^{p_1}$$. Define $$\|f\|_{L^{p_0}+L^{p_1}}=\inf\big\{\|f_0\|_{L^{p_0}}+\|f_1\|_{L^{p_1}}\big\},$$ where the infimum is taken over all decomposition $$f=f_0 + f_1$$ with $$f_0\in L^{p_0}$$ and $$f_1\in L^{p_1}$$.

What is meant by "dual space" is as follows.

For every bounded linear functional $$l$$ on $$L^{p_0}+L^{p_1}$$ there is a unique $$g \in L^{q_0} \cap L^{q_1}$$ so that $$l(f) = \int_X f(x)g(x) d\mu(x), \quad \text{for all f \in L^{p_0}+L^{p_1}}$$

Moreover, $$\|l\|_{(L^{p_0}+L^{p_1})*} = \| g \|_{L^{p_0} \cap L^{p_1}}$$.

By modifying Lemma 4.2 in the textbook, it is not too hard to prove the case above, i.e., $$L^{p_0} \cap L^{p_1}$$ is the dual space of $$L^{p_0} + L^{p_1}$$. I have trouble to prove the opposite, that $$L^{p_0} + L^{p_1}$$ is the dual space of $$L^{p_0} \cap L^{p_1}$$.

• Can you show that the spaces are reflexive? Oct 6, 2022 at 23:18
• Hint: Consider $X \cap Y$ as the subspace $\{(f,f) \colon f \in X \cap Y\}$ of $X \times Y$. Here, $X = L^{p_0}$ and $Y = L^{p_1}$. Can you take it from here? Oct 7, 2022 at 8:08
• @PhoemueX Thanks for the hint. Could you please take a look at my answer based on the hint? Also, would you mind recommending a book where the direct product of two Banach spaces are properly defined and discussed. Oct 7, 2022 at 22:02
• @PetraAxolotl: I posted a solution. Some facts are used: (a) $(L_p(\mu),\|\;\|_p)'=(L_p',\|\;\|_{p'})$ for all $1<p<\infty$, $\frac1p+\frac1q=1$; (b) The carrier $\{x: f(x)\neq0\}$ of an $L_p$ function ($0<p<\infty)$ is $\sigma$-finite; (c) If $f$, and $f$ are in some $L_p$ and $\int_Bf=\int_Bg$ for all measurable $B$ contained in a $\sigma$--finte set $A$, then $f=g$ in $A$. For this last assertion I did not provide a proof, however I can see that a monotone class type of argument justifies that. Let me know what you think. Kind regards. Oct 8, 2022 at 15:05

Let me state some general facts:

• Consider $$X=L_{p_0}(\mu)$$ equipped with norm $$\|\;\|_{p_0}$$, $$Y=L_{p_1}(\mu)$$ equipped with norm $$\|\;\|_{p_1}$$. The product linear space $$X\times Y$$ equipped with the norm $$(f, g)\mapsto\|(f, g)\|:=\|f\|_{p_0}+\|g\|_{p_1}$$ is a Banach space which contains isometric copies of $$X$$ and $$Y$$.

• The space $$X\cap Y$$ equipped with the norm $$f\mapsto\|f\|_I=\|f\|_{p_1}+\|f\|_{p_0}$$ is a Banach space, and the map $$f\stackrel{\iota}{\mapsto} (f, f)$$ is an isometry from $$X\cap Y$$ into $$X\times Y$$. Let $$\Delta=\{(f,f):f\in X\cap Y)\}$$. The map $$\iota^{-1}:(f, f)\mapsto f$$ is a linear bounded operator.

Let $$p'_0$$ and $$p'_1$$ be the dual conjugates of $$p_0$$ and $$p_1$$ respectively, i.e., $$p'_0$$ and $$p'_1$$ are number such that $$\tfrac{1}{p_0}+\tfrac{1}{p'_0}=1=\frac{1}{p_1}+\frac{1}{p'_1}$$.

• Since $$X'=L_{p'_0}(\mu)$$ and $$Y'=L_{p'_1}(\mu)$$ are the dual spaces of $$X$$ and $$Y$$ respectively, $$X'\times Y'$$ equipped with the norm $$\|(f, g)\|_{d,\infty}:=\max\big(\|\;\|_{p'_0},\|\;\|_{p'_1}\big)$$ is the dual space of $$X\times Y$$. This follows by restricting $$\Phi\in (X\times Y)'$$ to $$X\times\{0\}$$ and $$\{0\}\times Y$$ yielding a unique $$(\phi,\psi)\in X'\times Y'$$ such that $$\Phi(f, g)=\int f\phi +g\psi$$. Then $$\|\Phi\|=\inf_{\|f\|_{p_0}+\|g\|_{p_1}=1}\Big|\int f\phi +g\psi\,d\mu\Big|=\max(\|\phi\|_{p'_0},\|\psi\|_{p'_1})$$

Observe that the norms $$\|(f, g)\|_{d, s}:=\big(\|f\|^s_{p_0}+\|g\|^s_{p_1}\big)^{1/s}$$, $$1\leq s$$ and $$\|f, g)\|_{d,\infty}$$ on $$X'\times Y'$$ are all equivalent.

Solution to the OP:

1. Suppose first that $$\Lambda$$ is a bounded linear functional on $$X\cap Y$$. Then $$=\Lambda\circ \iota^{-1}$$ is a bounded functional on $$\Delta$$ and can be extended to a functional $$\Lambda'$$ on $$X\times Y$$ so that $$\|\Lambda\circ \iota^{-1}\|=\|\Lambda'\|$$ by the Hahn-Banach theorem. Then, there exists $$(\phi, \psi)\in L_{p'_0}(\mu)\times L_{p'_1}(\mu)$$ such that $$\Lambda'(f, g)=\int f\phi\,d\mu+\int g\psi\,d\mu$$. It follows that $$\Lambda(f)=\int (\phi+\psi)f\,d\mu, \qquad f\in X\cap Y.$$ If $$(\phi',\psi')$$ is another representation of $$\Lambda$$, i.e. $$\int f(\phi+\psi)\,d\mu=\int f(\phi'+\psi')\,d\mu$$ for all $$f\in X\cap Y$$, then $$\int_B(\phi+\psi)\,d\mu=\int_B(\phi'+\psi')\,d\mu$$ for all $$B$$ of finite measure and so, $$\phi+\psi=\phi'+\psi'$$ $$\mu$$-a.s. on sets of finite measure. Notice that \begin{align} |\Lambda(f)|&\leq \int|\phi f|\,d\mu+|\int \psi f|\,d\mu\leq \|\phi\|_{p'_0}\|f\|_{p_0}+\|\psi\|_{p'_1}\|f\|_{p_1}\\ &\leq \max(\|\phi\|_{p'_0},\|\psi\|_{p'_1})\|f\|_I\leq (\|\phi\|_{p'_0}+\|\psi\|_{p'_1})\|f\|_I \end{align} Hence $$|\Lambda(f)|\leq\|\psi+\phi\|_{L_{p'_0}+L_{p'_1}}\|f\|,\qquad f\in X\cap Y$$

2. Suppose now that $$L$$ is a bounded linear on the space $$Z:=L_{p_0}(\mu)+L_{p_1}(\mu)$$ equipped with the norm $$\|f\|_Z=\inf\{\|\phi\|_{p'_0}+\|\psi\|_{p'_1}: f= \phi+\psi\}$$ (a complete norm on $$Z$$). The map $$\kappa: X'\times Y'\rightarrow Z$$ given by $$(\phi, \psi)\mapsto \phi+\psi$$ is bounded since $$\|\phi+\psi\|_H\leq\|\phi\|_{p'_0} + \|\psi\|_{p'_1}$$. Then $$\tilde{L}=L\circ \kappa$$ îs a linear funcțional on $$X'\times Y'$$. Hence, there exists $$(f, g)\in X\times Y$$ such that $$\tilde{L}(\phi,\psi)=L(\phi+\psi)=\int f\phi\,d\mu + \int g\psi\,d\mu$$.
For $$\phi=\psi\in X'\cap Y'$$ we have that $$\tilde{L}(0,\phi)=L(\phi)=\tilde{L}(\phi,0)$$ and so, $$\int f\phi\,d\mu=\int g\phi\,d\mu, \qquad \phi\in X'\cap Y'$$ Thus $$\int_B f\,d\mu=\int_B g\,d\mu$$ for all $$B$$ of finite measure; hence $$f=g$$ $$\mu$$-a.s on any integrable set. Consequently, $$f\in X\cap Y$$ and $$L(h)=\int fh\,d\mu, \qquad h\in Z$$

Below is my attempt following the hint provided by @PhoemueX.

Edited following PhoemueX's and Oliver Díaz's comments.

We first define the subspace $$S \equiv \{(f,f):f \in X \cap Y\}$$ of $$X \times Y$$, where $$X=L^{p_0}$$ and $$X=L^{p_1}$$

For any linear functional $$l$$ on $$X\cap Y$$, we can define a linear functional on $$S$$ $$\lambda_0\left((f_0,f_1)\right) = l\left(\frac{f_0+f_1}{2}\right),$$ and a sub-linear function $$p$$ on $$X \times Y$$ $$p\left((f_0,f_1)\right) = \|l\|_{(X\cap Y)^*} \frac{\|f_0\|_X + \|f_1\|_Y}{2}.$$

On $$S$$, we have $$\lambda_0\left((f_0,f_1)\right) \leq p\left((f_0,f_1)\right)$$. So we can use the Hahn-Banach theorem to extend $$\lambda_0$$ to $$X \times Y$$.

It is easy to show that the dual of $$X \times Y$$, with the norm definition $$\left\|(f_0,f_1)\right\|_{X \times Y} = \|f_0\|_X + \|f_1\|_Y$$ is simply $$X^* \times Y^*$$, with the norm definition $$\left\|(g_0,g_1)\right\|_{X^* \times Y^*} = \max\big(\|g_0\|_{X^*}, \|g_1\|_{Y^*}\big)$$. (There are many other norm definitions possible. This one is simply chosen for the original problem.)

So that we can find $$(g_0,g_1) \in X^* \times Y^*$$, and have $$\lambda\left((f_0,f_1)\right) = \int_X \bigg(f_0(x)g_0(x) + f_1(x)g_1(x) \bigg) d\mu(x).$$

Setting $$f_0 = f_1 = f \in X \cap Y$$ and $$g = g_1 + g_2 \in L^{q_0} + L^{q_1}$$, we get the desired the result.

EDIT: I have found another proof that requires heavier machinery (i.e. Orlicz spaces) but gives us a more general result. I will post it one another day.

• On $X \times Y$ there are many equivalent norms, e.g. $\|(x,y)\|=\|x\|_X + \|y\|_Y$, but one can also take the max instead of the sum. The fact about the dual can be seen by writing $l((x,y))=l((x,0))+l((0,y))$. Oct 8, 2022 at 4:55
• The choice of $q_1,g_2$ depends on principle of which extension we choose in the Hahn Banach theorem. It remains to show that $g-g_1+g_2$ does not depends on the choice $(g_1,g_2)$ Oct 8, 2022 at 17:24
• @Petra: Do you have a reference for the Orlicz spaces machinery? Dec 23, 2023 at 2:59