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Show that \begin{equation} (Tu)(x)=\int_{\alpha(x)}^{\beta(x)} u(t)dt \end{equation} is Compact linear operator on $C([0,1],R)$ where $\alpha, \beta:[0,1]\rightarrow [0,1]$ are continuous.

My Attempt

$T$ is obviously linear. Let $M=\max (\alpha(x)-\beta(x))$. Then $|Tu(x)|\leq ||u||_{\infty}M$ and $||T||=M$.

For compactness, let $B=\{u\in (C[0,1],R):||u||_{\infty}\leq 1\}$. We show $T(B)$ is equicontinuous family so that by Arzela Ascoli it is relatively compact.

\begin{equation} |Tu(x)-Tu(y)|=\bigg|\int_{\alpha(x)}^{\beta(x)} u(t)dt-\int_{\alpha(y)}^{\beta(y)} u(t)dt\bigg| \end{equation}

Please how do I subtract this integrals. I'm thinking to make the assumption $\alpha(y)<\beta(y)<\alpha(x)<\beta(x)$.

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  • $\begingroup$ Is there any assumption on the regularity of $\alpha$ and $\beta$ at all? $\endgroup$ – BigbearZzz Jan 12 at 20:24
  • $\begingroup$ They are continuous. I will edit now. $\endgroup$ – Muhammad Mubarak Jan 12 at 20:25
  • $\begingroup$ My answer below can be rewritten without using the characteristic function $\chi$ by arguing case by case. However, I prefer using $\chi$ as it makes some idea clearer. $\endgroup$ – BigbearZzz Jan 12 at 22:39
  • $\begingroup$ Thanks a lot. Its clear now $\endgroup$ – Muhammad Mubarak Jan 13 at 9:56
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The integral from $a$ to $b$ can be viewed as $\int_a^bf=\int_{0}^1\chi_{[a,b]}f$. Recall the relation $$ |\chi_A-\chi_B| = \chi_{A \Delta B} $$ where $A\Delta B$ is the symmetric difference of $A$ and $B$. It is also easy to verify that $$ [a,b] \Delta [c,d] \subset [a,c] \cup[c,a]\cup[b,d]\cup[d,b] $$ where $[x,y]=\emptyset$ if $y<x$, so half of the terms of the right hand side would disappear, depending on the order of $a,b,c,d$. By symmetry of the right hand side, we may assume $a\le c$ and $b\le d$.

Using the above, we can compute that $$\begin{align} \left|\int_a^bf-\int_c^df \right| &\le \int_0^1 |\chi_{[a,b]} - \chi_{[c,d]}| \,|f|\\ &= \int_0^1 \chi_{[a,b] \Delta [c,d]} |f|\\ &\le\int_0^1 \chi_{[a,c]} |f| + \int_0^1 \chi_{[b,d]}|f| \\ &\le (|c-a|+|d-b|)\sup_{x\in[0,1]} |f(x)|. \end{align}$$

Back to your question, we can deduce that $$\begin{align} |Tu(x)-Tu(y)| &\le \big(|\alpha(x)-\alpha(y)|+ |\beta(x)-\beta(y)| \big)\, ||u||_\infty \\ &\le |\alpha(x)-\alpha(y)|+ |\beta(x)-\beta(y)| \end{align}$$ for any $u\in B$. By continuity of $\alpha$ and $\beta$, for any given $\varepsilon>0$ we may find $\delta$ such that $$ |x-y|<\delta \implies |\alpha(x)-\alpha(y)|+ |\beta(x)-\beta(y)|<\varepsilon, $$ which proves what you want.

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