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

problem 4. suppose f : a, b → ℝ be an l¹ -integrable function. extend f to be 0 outside the interval a, b. let φ(x)=1/2h∫ₓ₋ₕˣ⁺ʰ f. show that ∫ₐᵇ|φ|≤∫ₐᵇ|f|.

problem 4. suppose f : a, b → ℝ be an l¹ -integrable function. extend f to be 0 outside the interval a, b. let φ(x)=1/2h∫ₓ₋ₕˣ⁺ʰ f. show that ∫ₐᵇ|φ|≤∫ₐᵇ|f|.

Answer

Explanation:

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

We know that $\phi(x)=\frac{1}{2h}\int_{x - h}^{x+h}f(t)dt$. 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|dtdx \end{align*} We change the order of integration. 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 given by $a\leq x\leq b$ and $x - h\leq t\leq x + h$. We can rewrite the double - integral as $\int_{a - h}^{b + h}\left|f(t)\right|\left(\frac{1}{2h}\int_{\max(a,t - h)}^{\min(b,t + h)}1dx\right)dt$. Since $f$ is $0$ outside $[a,b]$, we have: \begin{align*} \int_{a}^{b}\left|\phi(x)\right|dx&\leq\int_{a}^{b}\left|f(t)\right|dt \end{align*}

Answer:

We have shown that $\int_{a}^{b}\left|\phi\right|\leq\int_{a}^{b}\left|f\right|$.