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Suppose $1\leq p<\infty$ and $f\in L^p(\mathbb{R})$. Define $g(x)=\int_x^{x+1} f(t)dt$.

Show that $g$ is continuous. What if $f\in L^{\infty} (\mathbb{R})$?


Two attempts to prove $g$ is continuous:

  1. I tried to apply Holder's inequality to g and 1, but I don't see how to use Holder's inequality to prove $d(g(x),g(x_0)< \epsilon$. \begin{align*} \int_R g\cdot1 d\mu &\leq \bigg( \int_R g^p d\mu \bigg)^{1/p} \bigg(\int_R1d\mu\bigg)^{1/q}\\ &\leq \bigg[ \int_R \bigg(\int_x^{x+1} f(t)dt\bigg)^p d\mu \bigg]^{1/p} [\mu(R)]^{1/q}\\ &\leq \cdots\\ \end{align*}

  2. (This has been considered as an incorrect proof.)

Let $x_0 \in \mathbb{R}$ and $\epsilon > 0$. $f\in L^p(\mathbb{R}) \implies \big (\int_{\mathbb{R}}|f(t)|^pdt\big)^{1/p} <\infty \implies f(t)\leq M$(are these implications correct?).

Let $\delta = \frac{\epsilon}{2M}$. Suppose $x\in \mathbb{R}$ and $d(x,x_0)=|x-x_0| < \delta$.

Then, \begin{align*} |g(x)-g(x_0)|&= \bigg|\int_x^{x+1}f(t)dt-\int_{x_0}^{x_0+1}f(t)dt \bigg| \tag{1}\\ &= \bigg|\int_x^{x_0}f(t)dt +\int_{x_0}^{x+1}f(t)dt-\int_{x_0}^{x+1}f(t)dt -\int_{x+1}^{x_0+1}f(t)dt \bigg| \tag{2}\\ &= \bigg|\int_x^{x_0}f(t)dt - \int_{x+1}^{x_0+1}f(t)dt \bigg| \tag{3}\\ &\stackrel{true? why?}{=} \bigg| \delta\ f(x_1) - \delta\ f(x_1+1) \bigg| \text{ for some $x_1$ in $[x,x_0]$} \tag{4} \\ &= \bigg| \delta\ \big[\ f(x_1) - f(x_1+1)\big] \bigg| \tag{5} \\ &\leq | \delta|\times 2M = \epsilon. \tag{6} \end{align*} Therefore $h$ is continuous.

Can someone verify this proof? Or give me some suggestions to improve it. Any hints for proving the case when $f\in L^\infty(\mathbb{R})$? Thanks.

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  • $\begingroup$ The proof is incorrect: An $L^p$ function does not need to be bounded! Try using Holder's inequality. $\endgroup$ – user296602 Mar 21 '18 at 0:17
  • $\begingroup$ @user296602 Thank you for pointing out. I tried to apply Holder's inequality to g and 1, but I don't see how to use Holder's inequality to prove $d(g(x),g(x_0)< \epsilon$. \begin{align*} \int_R g\cdot1 d\mu &\leq \bigg( \int_R g^p d\mu \bigg)^{1/p} \bigg(\int_R1d\mu\bigg)^{1/q}\\ &\leq \bigg[ \int_R \bigg(\int_x^{x+1} f(t)dt\bigg)^p d\mu \bigg]^{1/p} [\mu(R)]^{1/q}\\ &\leq \cdots\\ \end{align*} $\endgroup$ – user398843 Mar 21 '18 at 1:16
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Let $\epsilon > 0$, then take $\delta = \left(\frac{\epsilon}{2\|f\|_p}\right)^q$ where $\tfrac1p+\tfrac1q =1$. For fixed $x_0$, take any $x$ such that $|x-x_0|<\delta$, then

\begin{align} |g(x) - g(x_0)| &= \left|\int_x^{x+1} f(t)\;dt - \int_{x_0}^{x_0+1}f(t)\;dt\right|\\ &= \left|\int_x^{x_0} f(t)\;dt + \int_{x_0}^{x+1} f(t)\;dt- \int_{x_0}^{x+1}f(t)\;dt - \int_{x+1}^{x_0+1}f(t)\;dt\right| \\ &\leq \left|\int_x^{x_0} f(t)\;dt\right| + \left|\int_{x+1}^{x_0+1} f(t)\;dt\right|\\ &= \left|\int_{\mathbb{R}} f(t)\mathbb{I}_{(x,x_0)}\;dt\right| + \left|\int_{\mathbb{R}} f(t)\mathbb{I}_{(x+1,x_0+1)}\;dt\right|\\ &\overset{\text{Holders}}{\leq} \left|\int_{\mathbb{R}} f(t)^p\;dt\right|^{1/p} \left|\int_{\mathbb{R}}\mathbb{I}_{(x,x_0)}^q\;dt\right|^{1/q} + \left|\int_{\mathbb{R}} f(t)^p\;dt\right|^{1/p} \left|\int_{\mathbb{R}}\mathbb{I}_{(x+1,x_0+1)}^q\;dt\right|^{1/q}\\ &= \|f\|_p \cdot |x - x_0|^{1/q} + \|f\|_p \cdot |(x+1) - (x_0+1)|^{1/q} \\ &< \|f\|_p \cdot 2\delta^{1/q} = \epsilon \end{align}

using that abuse of notation that $(a, b)$ actually means $(\min(a,b), \max(a,b))$.

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  • $\begingroup$ Thank you very much. What's the meaning of your last sentence? "using that abuse of notation that $(a,b)$ actually means $(\min(a,b), \max(a,b))$" $\endgroup$ – user398843 Mar 21 '18 at 1:48
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    $\begingroup$ For the indicator functions I use $(x,x_0)$, but if $x_0 < x$ then the interval isn't valid by the usual definition so you should use $(x_0, x)$. Essentially the abuse of notation comment just prevents the need to deal with $x<x_0$ and $x>x_0$ separately. $\endgroup$ – adfriedman Mar 21 '18 at 1:53
  • $\begingroup$ I see you left out $p=1.$ $\endgroup$ – zhw. Mar 21 '18 at 18:57
  • $\begingroup$ @zhw. Would setting $\delta = \left(\frac{\epsilon}{2\|f\|_p}\right)^q$ cancel out the $1/q$ power? $\endgroup$ – user398843 Mar 25 '18 at 2:54
  • $\begingroup$ @user398843 If $p=1,q=\infty.$ $\endgroup$ – zhw. Mar 25 '18 at 3:02
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There's a more elementary way to prove this, without using Holder. I'll work on $[0,\infty)$ for convenience.

Basic result: if $f:[0,\infty) \to \mathbb R$ is integrable on each $[0,x], x\ge 0,$ then $F(x) = \int_0^x f$ is continuous on $[0,\infty).$ This follows from the dominated convergence theorem: Suppose $x_n$ is a sequence in $[0,\infty)$ converging to $x_0.$ We have $f\cdot \chi_{[0,x_n]}\to f\cdot \chi_{[0,x]}$ a.e. A dominating function for this setup is $|f|\chi_I,$ where $I$ is a bounded interval containing all $x_n$ and $x_0.$ The result follows.

A corollary is that $x\to \int_x^{x+1}f$ is continuous. That's simply because the integral equals $F(x+1)-F(x).$

Now if $f\in L^1$ or $f\in L^\infty,$ then the integrability hypothesis above holds, and the desired conclusion follows. If $f\in L^p, 1<p<\infty,$ the integrability condition holds as well: For any $x>0,$ let $A_x = [0,x]\cap \{|f(t)|\le 1\},$ $B_x = [0,x]\cap \{|f(t)|>1\}.$ Then

$$\int_0^x|f| = \int_{A_x}|f| + \int_{B_x}|f| \le \int_{A_x}|f| + \int_{B_x}|f|^p.$$

Both of the integrals on the right are finite, and we're done.

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  • $\begingroup$ Thank you for this different approach. It's pretty elegant. But I have some questions. How does dominated convergence theorem imply continuity? Why do we need to use $A_x$ and $B_x$? Why do we need to take the $p$-th power in the very last inequality? Can we just say $[0,x]$ has finite positive measure and $f\in L^p\implies f \in L^1$, and hence the integrals $\int_o^x |f|$ is finite? Sorry for having so many questions $\endgroup$ – user398843 Mar 25 '18 at 15:17
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Ok, I'll tackle the easier case. Suppose $f\in L^{\infty}(\mathbb{R})$, then we estimate $$ \left|\int_y^{y+1}f(t)\mathrm dt-\int_x^{x+1}f(t)\mathrm dt \right| $$ for $|y-x|<\delta<1$ and wlog $y>x$. Then, $$ \left|\int_y^{y+1}f(t)\mathrm dt-\int_x^{x+1}f(t)\mathrm dt \right| = \left|\int_{x+1}^{y+1}f(t)\mathrm dt-\int_x^{y}f(t)\mathrm dt \right|\\ \leq \int_{x+1}^{y+1}|f(t)|\mathrm dt+\int_x^y|f(t)|\mathrm dt\leq||f||_\infty|y-x| $$

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  • $\begingroup$ How can we get the last inequality? Do you use Holder's inequality? If you use Holder's inequality, should we have $||f||_{\infty}\cdot2|y-x|^{1/q}$ where $q \to 0$? Then we can say $||f||_{\infty}$ is bounded and $|y-x|^{1/q} \to 0$ to conclude the proof? $\endgroup$ – user398843 Mar 21 '18 at 2:22
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    $\begingroup$ the final inequality comes from the trivial bound: $\int_a^b|f|\leq |f|_\infty (b-a)$. I.e. monotonicity of the integral. $\endgroup$ – qbert Mar 21 '18 at 2:26

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