# $\lim_{p\to r} \|f\|_p=\|f\|_r$

Let $(X,\tau,\mu)$ a measure space and $f$ positive measurable function, such that $f\not=0$ a.e. Let $I_f=\left\{p\in [1,\infty]; \|f\|_p<\infty\right\}$

a) Show that $I_f$ is a interval. (I've already tried it.)

b) Suppose $I_f\neq \emptyset.$ Let $r\leq s<\infty$ in $I_f$ and let $p=\alpha r+(1-\alpha)s$ with $\alpha\in (0,1)$. Show that ${\|f\|_p}^p\leq {\|f\|_r}^{\alpha r}{ \|f\|_s}^{(1-\alpha)s}$. (I've already tried it.)

Deduce that the map $p\to \ln({\|f\|_p}^p)$ is convex on $I_f\setminus\left\{\infty\right\}$ and $p\to \|f\|_p$ is continuous on $int(I_f)$. Hint: All convex function on interval J is continuous in $int(J)$. (I've already tried it.)

x) Show that if $r$ is a extreme point of $I_f$, then $\|f\|_p \to \|f\|_r$ when $p\in I_f$, $p\to r$.

c) Deduce, for all $r<p<s$ in $I_f$,then $||f||_p\leq \max(\|f\|_r,\|f\|_s).$ Conclude $L^r(X,\tau,\mu)\cap L^s(X,\tau,\mu)\subseteq \bigcap_{p\in (r,s)} L^p(X,\tau,\mu).$

How can the alternative x be demonstrated? I have ${\|f\|_p}^p\leq {\|f\|_r}^{\alpha r}{\|f\|_s}^{(1-\alpha)s}$ then $\lim_{p\to r} {\|f\|_p}^p\leq {\|f\|_r}^{\alpha r}{\|f\|_r}^{(1-\alpha)}$ because $\displaystyle {\lim_{p\to r} p}=r=\displaystyle \lim_{p\to r} \alpha r+(1-\alpha)s$, i.e., $s=r$ Therefore $\lim_{p\to r} {\|f\|_p}^p\leq {\|f\|_r}^{\alpha r}{\|f\|_r}^{(1-\alpha)}={\|f\|_r}^r$ Therefore $\lim_{p\to r} {\|f\|_p}^p\leq {\|f\|_r}^r$ It is correct?

• Did you just try or did you solve the other items? Jul 9, 2018 at 22:11
• yes. For a), let $E=\left\{x:|f|>1\right\} then \int |f|^p=\int_{E} |f|^q+\int_{E^c} |f|^q \leq \int_E |f|^r+\int_{E^c}|f|^s<\infty+\infty=\infty$ For b), by Holder Jul 9, 2018 at 22:52
• @eraldcoil: Your argument is incorrect. If $s=r$ at the very beginning, then $p=s=r$ and you will get nothing. Jul 24, 2018 at 22:12

We assume that $\mu(X)>0$. Otherwise, the whole question does not make sense.

(a) (The solution has been pointed out by the eraldcoil. However, it seems that the case $s=\infty$ is missing.) Let $r,s\in I_{f}$ with $r<s$. We go to show that $(r,s)\subseteq I_{f}$. Let $p\in(r,s)$ and let $E=\{x\mid f(x)>1\}$. We consider two cases.

Case 1: $s\neq\infty$. In this case, as pointed out by eraldcoil, we have \begin{eqnarray*} \int f^{p} & = & \int_{E}f^{p}+\int_{E^{c}}f^{p}\\ & \leq & \int_{E}f^{s}+\int_{E^{c}}f^{r}\\ & < & \infty \end{eqnarray*} because $\int f^{s}<\infty$ and $\int f^{r}<\infty$.

Case 2: $s=\infty$. Note that $\mu(E)\leq\int_{E}f^{r}\leq\int f^{r}<\infty$. $f\in L^{\infty}$ implies that there exists $M>0$ such that $\mu(\{x\mid f(x)>M\})=0$. Therefore \begin{eqnarray*} \int f^{p} & = & \int_{E}f^{p}+\int_{E^{c}}f^{p}\\ & \leq & M^{p}\mu(E)+\int_{E^{c}}f^{r}\\ & < & \infty. \end{eqnarray*}

This show that $p\in I_{f}$.

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(b) Let $r,s\in I_{f}$ with $r\leq s<\infty$. Let $\alpha\in(0,1)$ and define $p=\alpha r+(1-\alpha)s$. Define $p'=\frac{1}{\alpha}$, $q'=\frac{1}{1-\alpha}$, then $p',q'\in(1,\infty)$ and $\frac{1}{p'}+\frac{1}{q'}=1$. Now \begin{eqnarray*} ||f||_{p}^{p} & = & \int f^{\alpha r}\cdot f^{(1-\alpha)s}\\ & \leq & ||f^{\alpha r}||_{p'}\cdot||f^{(1-\alpha)s}||_{q'}\\ & = & \left(\int f^{r}\right)^{\alpha}\cdot\left(\int f^{s}\right)^{(1-\alpha)}\\ & = & ||f||_{r}^{\alpha r}\cdot||f||_{s}^{(1-\alpha)s}. \end{eqnarray*} Note that (b) also gives a partial proof of (a) except for the case that $s=\infty$.

Define $\phi:I_{f}\setminus\{\infty\}\rightarrow\mathbb{R}$ by $\phi(t)=\ln\left(||f||_{t}^{t}\right)$. If $I_{f}\setminus\{\infty\}$ is a degenerated interval (i.e. empty set or a singleton), there is nothing to prove. Suppose that $I_{f}\setminus\{\infty\}$ is a non-degenerated interval. Let $r,s\in I_{f}\setminus\{\infty\}$ with $r<s$. Let $\alpha\in(0,1)$. Define $p=\alpha r+(1-\alpha)s$. We go to show that $\phi(p)\leq\alpha\phi(r)+(1-\alpha)\phi(s)$. Firstly, note that since $f>0$ a.e., we have $0<\int f^{t}\leq\infty$ for any $t\in[1,\infty)$. In particular, $||f||_{p}$, $||f||_{r}$, $||f||_{s}$ are positive real numbers. Taking logarithm on both sides of $||f||_{p}^{p}\leq||f||_{r}^{\alpha r}\cdot||f||_{s}^{(1-\alpha)s}$ yields $\phi(p)\leq\alpha\phi(r)+(1-\alpha)\phi(s).$ It follows that $\phi$ is a convex function. Let $\psi:I_{f}^{o}\rightarrow\mathbb{R}$ be defined by $\psi(t)=||f||_{t}$. ($I_{f}^{o}$ denotes the interior of the set $I_{f}$.) Since $\phi$ is convex, $\phi\bigg\vert_{I_{f}^{o}}$, the restriction of $\phi$ on $I_{f}^{o}$ is continuous. Note that for $t\in I_{f}^{o}$, $\psi(t)=\exp(\phi(t)/t)$ and that $t\geq1$, so $\psi$ is a continuous function.

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(x) Write $I_{f}=\langle a,b\rangle$ for some $a,b\in[1,\infty].$ Here $\langle$ may be $($ or $]$ and $\rangle$ may be $)$ or $]$. Moreover, we assume that $a<b$ (otherwise, $I_f$ is a degenerated interval and there is nothing to prove).

Case 1: $I_{f}=\langle a,b]$. We go to show that $||f||_{p}\rightarrow||f||_{b}$ as $p\rightarrow b-$. Here, we need to distinguish the cases $b=\infty$, $b<\infty$. Suppose that $b<\infty$. Let $(p_{n})_{n}$ be an arbitrary sequence in $I_{f}$ such that $p_{n}\rightarrow b$ as $n\rightarrow\infty$. Choose $c\in I_{f}$ such that $c\leq p_{n}$ for all $n$. Define $E=\{x\mid f(x)>1\}$. Define $g=1_{E}f^{b}+1_{E^{c}}f^{c}$. Clearly $f^{p_{n}}\leq g$ and $g$ is integrable. By Lebesgue dominated convergence theorem, we have $\lim_{n}\int f^{p_{n}}=\int f^{b}$. Since $(p_{n})_{n}$ is arbitary, by Heine theorem, it follows that $\lim_{p\rightarrow b-}\int f^{p}=\int f^{b}$. Recall that the function $(0,\infty)\times(0,\infty)\rightarrow\mathbb{R}$, $(x,y)\mapsto x^{y}$ is continuous (with respect to the product topology). Now $\int f^{p}\rightarrow\int f^{b}$ and $\frac{1}{p}\rightarrow\frac{1}{b}$ with $(\int f^{b},\frac{1}{b})\in(0,\infty)\times(0,\infty)$. Therefore $(\int f^{p})^{\frac{1}{p}}\rightarrow(\int f^{b})^{\frac{1}{b}}$. That is, $||f||_{p}\rightarrow||f||_{b}$.

Then, we conside the case that $b=\infty$. Note that for any $p\in I_{f}$, both the maps $\alpha\mapsto||\alpha f||_{p}$ and $\alpha\mapsto||\alpha f||_{\infty}$ are homogeneous in the sense that $||\alpha f||_{p}=\alpha||f||_{p}$ and $||\alpha f||_{\infty}=\alpha||f||_{\infty}$ for any $\alpha\geq0$. Since $f>0$ a.e., we have that $M=||f||_{\infty}>0$. By replacing $f$ with $2f/||f||_{\infty}$, without loss of generality, we may assume that $M=2$. Fix $q\in I_{f}$ with $q<\infty$. Define $E=\{x\mid f(x)\geq1\}$. We have that $\mu(E)\leq\int_{E}f^{q}<\infty$. Let $(p_{n})$ be an arbitrary strictly increasing sequence in $I_{f}$ such that $p_{n}\rightarrow\infty$. Without loss of generality, we may assume that $q<p_{n}$ for all $n$. Define $A=\{x\mid f(x)>1.5\}$. Since $M=2$, we have that $\mu(A)>0$. It follows that $\int_{E}f^{p_{n}}\geq\int_{A}f^{p_{n}}\geq(1.5)^{p_{n}}\mu(A)\rightarrow\infty$ as $n\rightarrow\infty$. On the other hand, the sequence of functions $(1_{E^{c}}f^{p_{n}})$ is dominated by the integrable function $1_{E^{c}}f^{q}$ and $1_{E^{c}}f^{p_{n}}\rightarrow0$ pointwisely. By Lebesgue dominated convergence theorem, we have $\int_{E^{c}}f^{p_{n}}\rightarrow0$ as $n\rightarrow\infty$. Choose $N$ such that $\int_{E^{c}}f^{p_{n}}\leq\int_{E}f^{p_{n}}$ whenever $n\geq N$. For any $n\geq N$, we have \begin{eqnarray*} \int f^{p_{n}} & = & \int_{E}f^{p_{n}}+\int_{E^{c}}f^{p_{n}}\\ & \leq & 2\int_{E}f^{p_{n}}\\ & \leq & 2M^{p_{n}}\mu(E). \end{eqnarray*} Hence $||f||_{p_{n}}\leq M\cdot(2\mu(E))^{\frac{1}{p_{n}}}$. It follows that $\limsup_{n}||f||_{p_{n}}\leq\lim_{n}M\cdot(2\mu(E))^{\frac{1}{p_{n}}}=M$.

To show the reverse direction $\liminf_{n}||f||_{p_{n}}\geq M$, let $B_{n}=\{x\mid f^{q}(x)\geq\frac{1}{n}\}$. Clearly $B_{1}\subseteq B_{2}\subseteq\ldots$ and $\cup_{n}B_{n}=\{x\mid f^{q}(x)>0\}$ whose complement has measure zero. Since $f^{q}$ is integrable, we have $\mu(B_{n})<\infty$. (In short, the support of $f$ is $\sigma$-finite) Let $\beta\in(0,1)$ be arbitrary. Define $C_{\beta}=\{x\mid f(x)>\beta M\}$. Since $\beta M$ is not an essential upper bound of $f$, $\mu(C_{\beta})>0$. By continuity of measure, $0<\mu(C_{\beta})=\lim_{n}\mu(C_{\beta}\cap B_{n})$. Hence, there exists $n_{0}$ such that $0<\mu(C_{\beta}\cap B_{n_{0}})<\infty$. Denote $D=C_{\beta}\cap B_{n_{0}}$. We have \begin{eqnarray*} ||f||_{p_{n}} & \geq & \left(\int_{D}f^{p_{n}}\right)^{\frac{1}{p_{n}}}\\ & \geq & \beta M\cdot\mu(D)^{\frac{1}{p_{n}}}. \end{eqnarray*} Letting $n\rightarrow\infty$, we have $\liminf||f||_{p_{n}}\geq\beta M$. Finally, since $\beta$ is arbitrary, we have $\liminf||f||_{p_{n}}\geq M$.

This shows that $\lim_{n}||f||_{p_{n}}=M$. Since the sequence $(p_{n})$ is arbitrary, by Heine theorem, it follows that $\lim_{p\rightarrow\infty}||f||_{p}=||f||_{\infty}$.

Case 2: $I_{f}=[a,b\rangle$. This can be proved similarly to case 1, using Lebesgue dominated convergence theorem and the joint continuity of the function $(x,y)\mapsto x^{y}$.

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(c) Let $r,s\in I_{f}$ with $r<s$. Let $p\in(r,s)$ be arbitrary. Firstly, we consider the case that $s<\infty$. In this case, there exists a unique $\alpha\in(0,1)$ such that $p=\alpha r+(1-\alpha)s$. Since the two sides of the inequality are homogeneous, by replacing $f$ with $f/||f||_{p}$, we may assume that $||f||_{p}=1$. By (b), we have \begin{eqnarray*} 1 & \leq & ||f||_{r}^{\alpha r}\cdot||f||_{s}^{(1-\alpha)s}\\ & = & \left(\int f^{r}\right)^{\alpha}\cdot\left(\int f^{s}\right)^{1-\alpha}. \end{eqnarray*} If $\int f^{r}=\int f^{s}$, then we have $\int f^{r}=\int f^{s}=1$ and hence $||f||_{r}=||f||_{s}=1$. If $\int f^{r}<\int f^{s}$, then $\int f^{s}>1$. It follows that $\max(||f||_{r},||f||_{s})\geq||f||_{s}>1=||f||_{p}$. If $\int f^{r}>\int f^{s}$, then $\int f^{r}>1$. We also have $\max(||f||_{r},||f||_{s})\geq||f||_{p}$.

Next, consider the case that $s=\infty$. Let $t>p$ be arbitrary. Note that $r<p<t$ and $r,p,t\in I_{f}$. By the previous discussion, we have $||f||_{p}\leq\max(||f||_{r},||f||_{t})$. By (x), $||f||_{t}\rightarrow||f||_{\infty}$ as $t\rightarrow\infty$. Since the function $x\mapsto\max(||f||_{r},x)$ is continuous, by letting $t\rightarrow\infty$, we have $||f||_{p}\leq\lim_{t\rightarrow\infty}\max(||f||_{r},||f||_{t})=\max(||f||_{r},||f||_{\infty})$. The inclusion $L^{r}(X,\tau,\mu)\cap L^{p}(X,\tau,\mu)\subseteq\cap_{p\in(r,s)}L^{p}(X,\tau,\mu)$ follows trivially from the above inequality.