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.

  • $\begingroup$ 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

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 $(\min(a,b), \max(a,b))$.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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, $$ \left|\int_y^{y+1}f(t)\mathrm dt-\int_x^{x+1}f(t)\mathrm dt \right| = \left|\int_{x+1}^{y+1}f(t)\mathrm dt-\int_x^{y}f(t)\mathrm dt \right|\\ \leq \int_{x+1}^{y+1}|f(t)|\mathrm dt+\int_x^y|f(t)|\mathrm dt\leq||f||_\infty|y-x| $$

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.