problem 4. suppose f : a,b → ℝ be an l¹ -integrable function. extend f to be 0 outside the interval a,b. let…

Explanation:

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

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

[ \begin{align*} \int_{a}^{b}\left|\phi(x)\right|dx&\leq\int_{a}^{b}\frac{1}{2h}\int_{x - h}^{x + h}\left|f(t)\right|dt\ dx\ \end{align*} ] Change the order of integration. Consider the double - integral $\int_{a}^{b}\frac{1}{2h}\int_{x - h}^{x + h}\left|f(t)\right|dt\ dx$. 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. The limits for $t$ are $a - h\leq t\leq b + h$. For a fixed $t$, the limits for $x$ are $\max(a,t - h)\leq x\leq\min(b,t + h)$. [ \begin{align*} \int_{a}^{b}\frac{1}{2h}\int_{x - h}^{x + h}\left|f(t)\right|dt\ dx&=\frac{1}{2h}\int_{a - h}^{b + h}\left|f(t)\right|\left(\min(b,t + h)-\max(a,t - h)\right)dt \end{align*} ] Since $f$ is $0$ outside $[a,b]$, we have: [ \begin{align*} \frac{1}{2h}\int_{a - h}^{b + h}\left|f(t)\right|\left(\min(b,t + h)-\max(a,t - h)\right)dt&\leq\int_{a}^{b}\left|f(t)\right|dt \end{align*} ] Because $\frac{1}{2h}\left(\min(b,t + h)-\max(a,t - h)\right)\leq1$ for $t\in[a,b]$ and $\left|f(t)\right| = 0$ for $t\notin[a,b]$. 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.