I have a final coming up in my Measure Theory class, and I found a question that I couldn't get a clean answer:
show that for all $ f \in L^p[\mathbb{R}]$ there exists $C \in \mathbb{R}$ such that $\forall$ x and y $\in \mathbb{R}$ the following expression holds true $||\int_x^y f|| \le C|x-y|^q$ where $\frac 1p + \frac1q = 1$
I know I should use holders inequality but I cannot get the final expression.
Thanks in advance
$ \def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|} $As, written, the inequality is wrong if $q \ne \frac 1q$ (that is $p < \infty$). Define $F \colon \mathbf R \to \mathbf R$ by $F(x) = \int_0^x f(\xi)\, d\xi$. Then we have $$ \abs{F(y) - F(x)} \le C\abs{x-y}^q \iff \frac{\abs{F(x)- F(y)}}{\abs{x-y}} \le C\abs{x-y}^{q-1}$$ As for $p < \infty$ we have $q > 1$, letting $y \to x$, we have that $F$ is differentiable at $x$ and $ \abs{F'(x)} = 0 $, as $x$ was arbitrary, this implies that $F$ is constant, hence $F(x) = F(0) = 0$ for all $x$. Therefore $f = 0 \in L^p(\mathbf R)$, contradicting the fact that $f$ was an arbitrary $L^p$-function.
But, if you replace $q$ with $\frac 1q$, your equality becomes true: We will use Hölder's inequality, using the fact mentioned in the comments that $\int_x^y f(\xi)\, d\xi = \int_{\mathbf R}\chi_{[x,y]}(\xi)f(\xi)\, d\xi$. We have \begin{align*} \abs{\int_x^y f(\xi)\, d\xi} &= \abs{\int_{\mathbf R} \chi_{[x,y]}(\xi)f(\xi)\, d\xi}\\ &\le \left(\int_{\mathbf R} \abs{f(\xi)}^p\, d\xi\right)^{1/p} \left(\int_{\mathbf R} \chi_{[x,y]}^q(\xi)\, d\xi\right)^{1/q}\\ &= \norm f_p \cdot \abs{x-y}^{1/q} \end{align*}