[proofplan] For $1 < p \leq \infty$ and $u \in W^{1,p}(a,b)$, the inclusion $W^{1,p}(a,b) \subset W^{1,1}(a,b)$ on the bounded interval (which holds because $L^p(a,b) \hookrightarrow L^1(a,b)$ for $1 \leq p \leq \infty$ on a finite measure space) gives access to the absolutely continuous representative $\tilde{u}$ from the [$W^{1,1}$ Equals Absolutely Continuous Functions in 1D](/theorems/3128) theorem. Hence $\tilde{u}(x) - \tilde{u}(y) = \int_y^x Du(t)\, d\mathcal{L}^1(t)$. Applying Hölder's inequality to this integral with exponents $p$ and $p' = p/(p-1)$ (with $p = \infty$ handled by the essential supremum) yields the Hölder bound with constant $C = 1$ and exponent $1 - 1/p$. [/proofplan]

[step:Reduce to the integral representation via the absolutely continuous representative]Let $u \in W^{1,p}(a,b)$ for some $1 < p \leq \infty$. Since the interval $(a,b)$ has finite Lebesgue measure $\mathcal{L}^1(a,b) = b - a < \infty$, the [Inclusion of $L^p$ Spaces on Finite Measure Spaces](/theorems/???) gives the continuous embedding $L^p(a,b) \hookrightarrow L^1(a,b)$ for every $1 \leq p \leq \infty$. In particular $u \in L^1(a,b)$ and $Du \in L^1(a,b)$, so $u \in W^{1,1}(a,b)$. By the [$W^{1,1}$ Equals Absolutely Continuous Functions in 1D](/theorems/3128) theorem, $u$ admits an absolutely continuous representative \begin{align*} \tilde{u}: [a,b] &\to \mathbb{R} \end{align*} with $\tilde{u} = u$ $\mathcal{L}^1$-a.e. on $(a,b)$, and for every $x, y \in [a,b]$, \begin{align*} \tilde{u}(x) - \tilde{u}(y) = \int_y^x Du(t)\, d\mathcal{L}^1(t). \end{align*}[/step]

[guided]The theorem statement is about all $1 < p \leq \infty$, and the natural way to access an absolutely continuous representative is the [$W^{1,1}$ Equals Absolutely Continuous Functions in 1D](/theorems/3128) theorem — but that result requires $u \in W^{1,1}$, not $W^{1,p}$. So the first task is to show that $W^{1,p}(a,b) \subseteq W^{1,1}(a,b)$ for $1 < p \leq \infty$. This holds because the interval has finite measure. The [Inclusion of $L^p$ Spaces on Finite Measure Spaces](/theorems/???) says that for a measure space $(E, \mathcal{E}, \mu)$ with $\mu(E) < \infty$ and $1 \leq q \leq p \leq \infty$, we have $L^p(E) \hookrightarrow L^q(E)$ with $\|f\|_{L^q} \leq \mu(E)^{1/q - 1/p} \|f\|_{L^p}$. Applying this with $q = 1$, $\mu = \mathcal{L}^1$, and $E = (a,b)$, we have $\mathcal{L}^1(a,b) = b - a < \infty$, so $L^p(a,b) \hookrightarrow L^1(a,b)$ for every $1 \leq p \leq \infty$. Apply this to both $u$ and $Du$: since $u \in W^{1,p}(a,b)$, both lie in $L^p(a,b) \subseteq L^1(a,b)$. Therefore $u \in W^{1,1}(a,b)$, and the [$W^{1,1}$ Equals Absolutely Continuous Functions in 1D](/theorems/3128) theorem applies, giving the absolutely continuous representative $\tilde{u}$ together with the integral representation \begin{align*} \tilde{u}(x) - \tilde{u}(y) = \int_y^x Du(t)\, d\mathcal{L}^1(t) \end{align*} for all $x, y \in [a,b]$. (Note: $Du$ is the *weak* derivative; the theorem identifies it with the classical a.e. derivative of $\tilde{u}$.)[/guided]

[step:Apply Hölder with exponents $p$ and $p' = p/(p-1)$]Fix $x, y \in [a,b]$ with $y \leq x$ (the other ordering is symmetric). Taking absolute values in the integral representation, \begin{align*} |\tilde{u}(x) - \tilde{u}(y)| = \left|\int_y^x Du(t)\, d\mathcal{L}^1(t)\right| \leq \int_y^x |Du(t)|\, d\mathcal{L}^1(t) = \int_y^x |Du(t)| \cdot 1 \, d\mathcal{L}^1(t). \end{align*} **Case $1 < p < \infty$.** Let $p' = p/(p-1)$ be the [Hölder Conjugate](/theorems/???) of $p$, so $1/p + 1/p' = 1$. By [Hölder's Inequality](/theorems/???) applied to the pair $|Du| \cdot \mathbb{1}_{[y,x]}$ (in $L^p$) and $1 \cdot \mathbb{1}_{[y,x]}$ (in $L^{p'}$) on $(a,b)$, \begin{align*} \int_y^x |Du(t)|\, d\mathcal{L}^1(t) &= \int_a^b |Du(t)| \cdot \mathbb{1}_{[y,x]}(t)\, d\mathcal{L}^1(t) \\ &\leq \left(\int_a^b |Du(t)|^p \cdot \mathbb{1}_{[y,x]}(t)\, d\mathcal{L}^1(t)\right)^{1/p} \left(\int_a^b \mathbb{1}_{[y,x]}(t)\, d\mathcal{L}^1(t)\right)^{1/p'} \\ &\leq \|Du\|_{L^p(a,b)} \cdot |x - y|^{1/p'}, \end{align*} where in the last line we enlarged the domain of integration of $|Du|^p$ from $[y,x]$ to $(a,b)$ (a strict enlargement in general, valid because $|Du|^p \geq 0$) and computed $\int_a^b \mathbb{1}_{[y,x]}\, d\mathcal{L}^1 = |x - y|$. Using $1/p' = 1 - 1/p$, \begin{align*} |\tilde{u}(x) - \tilde{u}(y)| \leq \|Du\|_{L^p(a,b)} \cdot |x - y|^{1 - 1/p}. \end{align*} **Case $p = \infty$.** Here $1 - 1/p = 1$ and the inequality reduces to a Lipschitz estimate. Bounding $|Du(t)| \leq \|Du\|_{L^\infty(a,b)}$ for $\mathcal{L}^1$-a.e. $t$, \begin{align*} |\tilde{u}(x) - \tilde{u}(y)| \leq \int_y^x \|Du\|_{L^\infty(a,b)}\, d\mathcal{L}^1(t) = \|Du\|_{L^\infty(a,b)} \cdot |x - y|. \end{align*} This is the desired bound with exponent $1 - 1/\infty = 1$. In both cases the constant $C = 1$ works, completing the proof.[/step]

[guided]We now have the integral representation $\tilde{u}(x) - \tilde{u}(y) = \int_y^x Du(t)\, d\mathcal{L}^1(t)$ and want to extract a Hölder bound. The right-hand side is the integral of a function in $L^p$ — one could try the crude bound $\int |Du| \leq |x-y| \cdot \|Du\|_{L^\infty}$, but that requires $p = \infty$. For finite $p$, the natural way to relate an $L^1$-norm of a function on a small set to its $L^p$-norm and the size of the set is Hölder's inequality: the smaller the set, the smaller the $L^1$-norm contribution, with the gain measured by an exponent depending on $p$. Take $y \leq x$ and start with \begin{align*} |\tilde{u}(x) - \tilde{u}(y)| \leq \int_y^x |Du(t)|\, d\mathcal{L}^1(t). \end{align*} Rewrite this integral as a pairing of $|Du|$ with the indicator $\mathbb{1}_{[y,x]}$: \begin{align*} \int_y^x |Du|\, d\mathcal{L}^1 = \int_a^b |Du(t)| \cdot \mathbb{1}_{[y,x]}(t)\, d\mathcal{L}^1(t). \end{align*} Now apply [Hölder's Inequality](/theorems/???) to the pair $f := |Du|$ and $g := \mathbb{1}_{[y,x]}$ on $(a,b)$ with exponents $p$ and $p'$. The hypotheses of Hölder are: $1/p + 1/p' = 1$ (we have $p' = p/(p-1)$) and the functions lie in $L^p$ and $L^{p'}$ respectively. Both are met: $|Du| \in L^p(a,b)$ because $u \in W^{1,p}$, and $\mathbb{1}_{[y,x]} \in L^{p'}(a,b)$ since the indicator of a bounded set lies in every $L^q$. Hölder gives \begin{align*} \int_a^b |Du| \cdot \mathbb{1}_{[y,x]}\, d\mathcal{L}^1 \leq \||Du|\|_{L^p} \cdot \|\mathbb{1}_{[y,x]}\|_{L^{p'}}. \end{align*} The first factor is $\|Du\|_{L^p(a,b)}$ by definition. For the second, \begin{align*} \|\mathbb{1}_{[y,x]}\|_{L^{p'}}^{p'} = \int_a^b \mathbb{1}_{[y,x]}^{p'}\, d\mathcal{L}^1 = \int_a^b \mathbb{1}_{[y,x]}\, d\mathcal{L}^1 = |x - y|, \end{align*} so $\|\mathbb{1}_{[y,x]}\|_{L^{p'}} = |x - y|^{1/p'}$. Combining and using $1/p' = 1 - 1/p$, \begin{align*} |\tilde{u}(x) - \tilde{u}(y)| \leq \|Du\|_{L^p(a,b)} \cdot |x - y|^{1 - 1/p}. \end{align*} For $p = \infty$, the conjugate is $p' = 1$, and the formula $1 - 1/p = 1$ predicts a Lipschitz estimate. Hölder still works ($\|\mathbb{1}_{[y,x]}\|_{L^1} = |x-y|$, $\|Du\|_{L^\infty}$ as the other factor), but it is more direct to use the essential bound $|Du(t)| \leq \|Du\|_{L^\infty}$ a.e. and integrate. Why does the proof need $p > 1$? For $p = 1$, the conjugate $p' = \infty$ and the prediction $1 - 1/p = 0$ gives no decay in $|x - y|$ — i.e. no Hölder regularity. This matches the well-known fact that $W^{1,1}$ functions in 1D are absolutely continuous but not in general (Hölder-)continuous with a uniform modulus.[/guided]