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Let $g \in L^p([a,b])$ for some $p \in (1, \infty]$. Define $$f(x) = f(a) + \int_a^xg(s)\ ds$$ for any $x \in [a,b]$. The claim is that $f$ is $\frac{1}{p^{\prime}}$-Hölder continuous on $[a,b]$, where $\frac{1}{p} + \frac{1}{p^{\prime}} = 1$.

So I need to show that there exists $M > 0$ and $\delta > 0$, such that for all $t$ with $|t|< \delta$, $|f(x+t) - f(x)| \leq M|t|^{1/p^{\prime}}.$ We have $$|f(x+t) - f(x)| = \left|\int_a^{x+t}g(s)\ ds - \int_a^xg(s)\ ds\right| \leq \int_a^{x+t}|g(s)|\ ds + \int_a^x|g(s)|\ ds. $$ What should I do next? I seem to cannot find a way to use the fact that $g \in L^P([a,b])$. I must say that the condition $$\frac{1}{p} + \frac{1}{p^{\prime}} = 1$$ is very similar to Hölder's inequality. Should I use that too somewhere in the proof?

EDIT: So after reading the comments, I decided to add some new discoveries. We have $$|f(x+t) - f(x)| = \left|\int_x^{x+t}g(s)ds\right|\leq \int_x^{x+t}|g(s)|ds.$$ Rewriting $\int_x^{x+t}|g(s)|ds$ into $$\left(\int_x^{x+t}|g(s)|^{p\cdot\frac{1}{p}} \right)^{p\cdot\frac{1}{p}} = || g^{1/p}||_p^p = ||1 \cdot g^{1/p}||_p^p \leq ||g^{1/p}||_p^p||1||_{p^{\prime}}^p.$$ Still, I don't see that this proves $\frac{1}{p^{\prime}}$-Hölder continuous$. I don't think the rewriting part is correct.

Vicky
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  • Is there something better you can do with $\left| \int_a^{x+t} g(s)ds - \int_a^x g(s)ds\right| = \left|\int_x^{x+t} g(s)ds\right|$, than just the crude triangle inequality bound? – Sarvesh Ravichandran Iyer Oct 06 '20 at 15:27
  • And of course, you will have to use Hölder's inequality. A quick reminder: $\lvert g(s) \rvert = 1 \cdot \lvert g(s) \rvert$. – Hyperbolic PDE friend Oct 06 '20 at 15:29
  • @Shaqinho I have added some new things using your comments. But I 'm still not sure how to show it. I have also used Teresa's comment – Vicky Oct 06 '20 at 16:47

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Without loss of generality, let $t>0$: As Teresa said: $$ \lvert f(x+t) - f(x) \rvert \leq \int^{x+t}_x \lvert g(s) \rvert~\mathrm{d}s = \lVert 1 \cdot g \rVert_{L^1[x, x+t]} $$ Now use Hölder's inequality: (Let $M := \lVert g \rVert_{L^{p}[a, b]}$) $$ \lVert 1 \cdot g \rVert_{L^1[x, x+t]} \leq \lVert 1 \rVert_{L^{p'}[x, x+t]} \lVert g \rVert_{L^{p}[x, x+t]} = \lVert g \rVert_{L^{p}[x, x+t]} \underbrace{\lvert t \rvert}_{=t}~^{\frac{1}{p'}} \leq M \lvert t \rvert^{\frac{1}{p'}} $$

Hyperbolic PDE friend
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  • Shouldn't we have $\lVert 1 \rVert_{L^{p'}[x, x+t]} \lVert g \rVert_{L^{p}[x, x+t]}$? But then ofc this will work only if $1/p' + 1/p = 1$. – Vicky Oct 07 '20 at 10:34
  • Yep, you are right, thank you. I will edit that. In theory we do not even know if $\lVert g \rVert_{L^{p'}[a, b]}$ is finite... – Hyperbolic PDE friend Oct 07 '20 at 13:56