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 $L^{p_0} \cap L^{p_1}$.

  • $\begingroup$ Can you show that the spaces are reflexive? $\endgroup$ Oct 6, 2022 at 23:18
  • $\begingroup$ 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? $\endgroup$
    – PhoemueX
    Oct 7, 2022 at 8:08
  • $\begingroup$ @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. $\endgroup$ Oct 7, 2022 at 22:02
  • $\begingroup$ @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. $\endgroup$
    – Mittens
    Oct 8, 2022 at 15:05

2 Answers 2


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.

  • 1
    $\begingroup$ 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 $l((x,y))=l((x,0))+l((0,y))$. $\endgroup$
    – PhoemueX
    Oct 8, 2022 at 4:55
  • $\begingroup$ 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 $(g_1,g_2)$ $\endgroup$
    – Mittens
    Oct 8, 2022 at 17:24
  • $\begingroup$ @Petra: Do you have a reference for the Orlicz spaces machinery? $\endgroup$
    – Mambo
    Dec 23, 2023 at 2:59

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