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: Rewrite the left - hand side integral

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

Step2: Apply the absolute - value inequality for integrals

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

Step3: Change the order of integration

Using Fubini - Tonelli's theorem (since $|f|$ is non - negative and $L^1$), we change the order of integration. Let's consider the region of integration. The double integral $\int_{a}^{b}\frac{1}{2h}\int_{x - h}^{x + h}|f(t)|dtdx$ can be rewritten. The region of integration in the $x - t$ plane is described by $a\leq x\leq b$ and $x - h\leq t\leq x + h$. After changing the order, we get $\frac{1}{2h}\int_{a - h}^{b + h}|f(t)|\int_{t - h}^{\min(t + h,b)}dxdt$.

Step4: Evaluate the inner integral

For a fixed $t$, $\int_{t - h}^{\min(t + h,b)}dx$. Case 1: If $t\in[a,b]$, then $\int_{t - h}^{\min(t + h,b)}dx\leq2h$. Case 2: If $t\notin[a,b]$, then $f(t) = 0$ (by the extension of $f$). So, $\frac{1}{2h}\int_{a - h}^{b + h}|f(t)|\int_{t - h}^{\min(t + h,b)}dxdt\leq\int_{a}^{b}|f(t)|dt$.

Answer:

The inequality $\int_{a}^{b}|\phi|\leq\int_{a}^{b}|f|$ holds.