problem 4. suppose f : a,b → ℝ be an l¹ -integrable function. extend f to be outside the interval a,b. let…

Explanation:

Step1: Use the definition of $\phi(x)$

By the given definition $\phi(x)=\frac{1}{2h}\int_{x - h}^{x+h}f(t)dt$. Then $\left|\phi(x)\right|=\left|\frac{1}{2h}\int_{x - h}^{x+h}f(t)dt\right|$. By the property of the absolute - value of an integral $\left|\int_{x - h}^{x+h}f(t)dt\right|\leq\int_{x - h}^{x+h}\left|f(t)\right|dt$, so $\left|\phi(x)\right|\leq\frac{1}{2h}\int_{x - h}^{x+h}\left|f(t)\right|dt$.

Step2: Calculate $\int_{a}^{b}\left|\phi(x)\right|dx$

We have $\int_{a}^{b}\left|\phi(x)\right|dx\leq\int_{a}^{b}\frac{1}{2h}\int_{x - h}^{x+h}\left|f(t)\right|dtdx$.

Step3: Change the order of integration

Using Fubini - Tonelli's theorem (since $\left|f\right|$ is non - negative and $L^1$ integrable). Let's consider the double - integral $\int_{a}^{b}\frac{1}{2h}\int_{x - h}^{x+h}\left|f(t)\right|dtdx$. The region of integration in the $x - t$ plane is described by $a\leq x\leq b$ and $x - h\leq t\leq x + h$. We can rewrite the double - integral as an iterated integral with the order of integration reversed. After changing the order of integration, we get $\int_{a - h}^{b + h}\left|f(t)\right|\left(\frac{1}{2h}\int_{\max(a,t - h)}^{\min(b,t + h)}1dx\right)dt$.

Step4: Analyze the inner integral

Note that $\frac{1}{2h}\int_{\max(a,t - h)}^{\min(b,t + h)}1dx\leq1$. Since $f$ is extended to be $0$ outside $[a,b]$, we have $\int_{a - h}^{b + h}\left|f(t)\right|\left(\frac{1}{2h}\int_{\max(a,t - h)}^{\min(b,t + h)}1dx\right)dt\leq\int_{a}^{b}\left|f(t)\right|dt$. So $\int_{a}^{b}\left|\phi\right|\leq\int_{a}^{b}\left|f\right|$.

Answer:

The inequality $\int_{a}^{b}\left|\phi\right|\leq\int_{a}^{b}\left|f\right|$ is shown as above.