# Show that $(Tu)(x)=\int_{\alpha(x)}^{\beta(x)} u(t)dt$ is Compact linear operator on $C([0,1])$

Show that $$$$(Tu)(x)=\int_{\alpha(x)}^{\beta(x)} u(t)dt$$$$ is Compact linear operator on $$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.

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

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

• Is there any assumption on the regularity of $\alpha$ and $\beta$ at all? – BigbearZzz Jan 12 at 20:24
• They are continuous. I will edit now. – Muhammad Mubarak Jan 12 at 20:25
• 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. – BigbearZzz Jan 12 at 22:39
• Thanks a lot. Its clear now – Muhammad Mubarak Jan 13 at 9:56

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 $$y, so half of the terms of the right hand side would disappear, depending on the order of $$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.