Math Problems Vault

Problem 10.5.1

(a)(b)

Let $f, g, h \in L^{1}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ .

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(a)

Convolution is commutative in the sense that $f * g = g * f$ $\bar{\lambda }$ -a.e. on $\mathbb {R}$ .

(b)

Convolution is associative in that $f * \left(g * h\right) = \left(f * g\right) * h$ $\bar{\lambda }$ -a.e. on $\mathbb {R}$ .

Problem 10.5.2

This exercise is a lemma for the next exercise. Let $f \in L^{1}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ and $\epsilon > 0$ . There exists a closed interval $F = \left[-\alpha , \alpha \right] \subseteq \mathbb {R}$ such that $\int_{F^{c}} f \, d\bar{\lambda } < \epsilon$ .

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Remark.

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Consider the sequence $\left\{ f \; \mathbb {1}_{[-n, n]}\right\}_{n=1}^{\infty }$ of functions in $L^{1}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ . For every $n \in \mathbb {N}$ we have $\left|f \; \mathbb {1}_{[-n, n]}\right| \leq \left|f\right|$ and $\int_{\mathbb {R}} \left|f\right| \, d\bar{\lambda } < +\infty$ . Therefore $\lim_{n} \int_{\mathbb {R}} f \; \mathbb {1}_{[-n, n]} \, d\bar{\lambda } = \int_{\mathbb {R}} f \, d\bar{\lambda }$ by DCT. Let $F = [-n, n]$ for sufficiently large $n$ .

Problem 10.5.3

In Claim 8 (the result establishing that $f * g$ is meaningfully defined, bounded by $\left\| f\right\|_{p} \left\| g\right\|_{p^{\prime }}$ , and uniformly continuous on $\mathbb {R}$ whenever $1 \leq p < +\infty$ , $f \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ , and $g \in L^{p^{\prime }}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ ), we also have $\lim_{\left|x\right| \rightarrow +\infty } f * g(x) = 0$ .

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Remark.

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Let $\epsilon > 0$ . Use Exercise 2 to find a closed interval $F = \left[-\alpha , \alpha \right] \subseteq \mathbb {R}$ so that

\int _{F^{c}} \left|f\right|^{p} \, d\bar{\lambda } < \left[\frac{\epsilon }{\left\| f\right\| _{p} + \left\| g\right\| _{p^{\prime }}}\right]^{p} = c^{p} \quad \text{and} \quad \int _{F^{c}} \left|g\right|^{p^{\prime }} \, d\bar{\lambda } < \left[\frac{\epsilon }{\left\| f\right\| _{p} + \left\| g\right\| _{p^{\prime }}}\right]^{p^{\prime }} = c^{p^{\prime }} .

Verify that $x \in \mathbb {R}$ and $\left|x\right| > 2\alpha$ gives $\left[x - \alpha , x + \alpha \right] \subseteq F^{c}$ , so that $\left|x\right| > 2\alpha$ yields

\begin{aligned} \left|f * g(x)\right| & \leq \left|\int _{F} f(x - y) g(y) \, d\bar{\lambda }(y)\right| + \left|\int _{F^{c}} f(x - y) g(y) \, d\bar{\lambda }(y)\right| \\ & \leq \left[\int _{F} \left|f(x - y)\right|^{p} \, d\bar{\lambda }(y)\right]^{1 / p} \left\| g\right\| _{p^{\prime }} + \left\| f\right\| _{p} \left[\int _{F^{c}} \left|g(y)\right|^{p^{\prime }} \, d\bar{\lambda }(y)\right]^{1 / p^{\prime }} \\ & \leq \left[\int _{[x - \alpha , x + \alpha ]} \left|f(y)\right|^{p} \, d\bar{\lambda }(y)\right]^{1 / p} \left\| g\right\| _{p^{\prime }} + c \left\| f\right\| _{p} \\ & \leq \left[\int _{F^{c}} \left|f(y)\right|^{p} \, d\bar{\lambda }(y)\right]^{1 / p} \left\| g\right\| _{p^{\prime }} + c \left\| f\right\| _{p} \\ & \leq c \left\| g\right\| _{p^{\prime }} + c \left\| f\right\| _{p} \\ & = \epsilon . \end{aligned}

Problem 10.5.4

$L^{1}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ has no element $u$ such that $u * f = f$ $\bar{\lambda }$ -a.e. on $\mathbb {R}$ for every $f$ in $L^{1}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ .

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Remark.

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Suppose such a $u$ exists. Find $\delta > 0$ with $\int_{[-2\delta , 2\delta ]} \left|u\right| \, d\bar{\lambda } < 1$ . Let $f = \; \mathbb {1}_{[-\delta , \delta ]}$ . For $\bar{\lambda }$ -a.e. $x \in \mathbb {R}$ we have

\begin{aligned} f(x) = u * f(x) & = \int _{\mathbb {R}} u(x - y) f(y) \, d\bar{\lambda }(y) \\ & = \int _{[-\delta , \delta ]} u(x - y) \, d\bar{\lambda }(y) \\ & = \int _{[x - \delta , x + \delta ]} u(t) \, d\bar{\lambda }(t) . \end{aligned}

Since $\bar{\lambda }\left([-\delta , \delta ]\right) > 0$ , there is $x_{0} \in [-\delta , \delta ]$ with $1 = f\left(x_{0}\right) = \int_{[x_{0} - \delta , x_{0} + \delta ]} u(t) \, d\bar{\lambda }(t)$ . Now $[x_{0} - \delta , x_{0} + \delta ] \subseteq [-2\delta , 2\delta ]$ , so

1 = \left|\int _{x_{0} - \delta }^{x_{0} + \delta } u(t) \, d\bar{\lambda }(t)\right| \leq \int _{x_{0} - \delta }^{x_{0} + \delta } \left|u(t)\right| \, d\bar{\lambda }(t) \leq \int _{-2\delta }^{2\delta } \left|u(t)\right| \, d\bar{\lambda }(t) < 1 .

This gives a contradiction.

Problem 10.5.5

A sequence $\left\{ u_{n}\right\}_{n=1}^{\infty }$ of functions in $L^{1}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ is called an approximate unit if

$u_{n} \geq 0$ for each $n \in \mathbb {N}$ ,
$\left\| u_{n}\right\|_{1} = 1$ for each $n \in \mathbb {N}$ ,
for every neighborhood $V$ of $0$ , $\lim_{n} \int_{V^{c}} u_{n} \, d\bar{\lambda } = 0$ .

For example, if $u_{n} = \frac{n}{2} \; \mathbb {1}_{[-\frac{1}{n}, \frac{1}{n}]}$ for each $n \in \mathbb {N}$ , then $\left\{ u_{n}\right\}_{n=1}^{\infty }$ is an approximate unit.

Prove the following: Let $\left\{ u_{n}\right\}_{n=1}^{\infty }$ denote an approximate unit, and let $1 < p < +\infty$ . For every $f \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ , we have $\lim_{n} \left\| f * u_{n} - f\right\|_{p} = 0$ .

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Remark.

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Let $f \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ and $\epsilon > 0$ . The claim is trivial if $\left\| f\right\|_{p} = 0$ , so assume that $\left\| f\right\|_{p} > 0$ . Find a neighborhood $V$ of $0$ such that $y \in V$ forces $\left\| \tau_{-y} f - f\right\|_{p} < \frac{\epsilon }{2}$ . Use (iii) to find $n_{0} \in \mathbb {N}$ such that $n \geq n_{0}$ implies $\int_{V^{c}} u_{n} \, d\bar{\lambda } < \frac{\epsilon }{4 \left\| f\right\|_{p}}$ . Let $n \geq n_{0}$ . We have $f * u_{n} \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ by the results in this section, and for each $h \in L^{p^{\prime }}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ we have

\begin{aligned} & \left|\int _{\mathbb {R}} \left[f * u_{n}(x) - f(x)\right] h(x) \, d\bar{\lambda }(x)\right| \\ & = \left|\int _{\mathbb {R}} \left[\int _{\mathbb {R}} f(x - y) u_{n}(y) \, d\bar{\lambda }(y) - f(x)\right] h(x) \, d\bar{\lambda }(x)\right| \\ & = \left|\int _{\mathbb {R}} \int _{\mathbb {R}} \left[f(x - y) - f(x)\right] u_{n}(y) \, d\bar{\lambda }(y) \, h(x) \, d\bar{\lambda }(x)\right| \\ & \leq \int _{\mathbb {R}} \int _{\mathbb {R}} \left|f(x - y) - f(x)\right| \left|u_{n}(y)\right| \, d\bar{\lambda }(y) \, \left|h(x)\right| \, d\bar{\lambda }(x) \\ & = \int _{\mathbb {R}} \left|u_{n}(y)\right| \int _{\mathbb {R}} \left|f(x - y) - f(x)\right| \left|h(x)\right| \, d\bar{\lambda }(x) \, d\bar{\lambda }(y) \\ & \leq \int _{\mathbb {R}} \left|u_{n}(y)\right| \left\| \tau _{-y} f - f\right\| _{p} \left\| h\right\| _{p^{\prime }} \, d\bar{\lambda }(y) \\ & \leq \int _{V} u_{n}(y) \frac{\epsilon }{2} \left\| h\right\| _{p^{\prime }} \, d\bar{\lambda }(y) + \int _{V^{c}} 2 u_{n}(y) \left\| f\right\| _{p} \left\| h\right\| _{p^{\prime }} \, d\bar{\lambda }(y) \\ & \leq \left(\frac{\epsilon }{2} \left\| h\right\| _{p^{\prime }} \times 1\right) + \left(2 \left\| f\right\| _{p} \left\| h\right\| _{p^{\prime }} \times \frac{\epsilon }{4 \left\| f\right\| _{p}}\right) \\ & = \epsilon \left\| h\right\| _{p^{\prime }} . \end{aligned}

Define $L$ on $L^{p^{\prime }}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ by writing $L(h) = \int_{\mathbb {R}} \left[f * u_{n}(x) - f(x)\right] h(x) \, d\bar{\lambda }(x)$ . Then $L$ is a bounded linear functional and $\epsilon \geq \left\| L\right\| = \left\| f * u_{n} - f\right\|_{p}$ by Claim 1 of Section 8.9.

Problem 10.5.6

Redo Exercise 5 if $p = 1$ .

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Remark.

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Take $h = 1$ above.

Problem 10.5.7

If $f$ and $g$ are elements of $L^{1}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ that depend only on $\left|x\right|$ , then $f * g$ only depends on $\left|x\right|$ .

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Problem 10.5.8

(a)(b)(c)(d)

The following parts give some concrete examples of convolutions courtesy of Jones, pages 278-279.

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(a)

Let $0 < b \leq a$ . Let $f = \; \mathbb {1}_{(-a, a)}$ and $g = \; \mathbb {1}_{(-b, b)}$ . Then for each $x \in \mathbb {R}$

f * g(x) = \begin{cases} 2b & \left|x\right| \leq a - b \\ a + b - \left|x\right| & a - b \leq \left|x\right| \leq a + b \\ 0 & a + b \leq \left|x\right| . \end{cases}

(b)

Let $a$ and $b$ denote distinct real numbers. Let $f(x) = e^{-ax} \; \mathbb {1}_{[0, +\infty )}(x)$ and $g(x) = e^{-bx} \; \mathbb {1}_{[0, +\infty )}(x)$ for each $x \in \mathbb {R}$ . Then

f * g(x) = \begin{cases} \dfrac {e^{-ax} - e^{-bx}}{b - a} & x \geq 0 \\ 0 & x < 0 . \end{cases}

(c)

If in (b) we have $a = b$ , then $f * g(x) = x e^{-ax}$ for each $x \in \mathbb {R}$ .

(d)

If $a + b > 0$ , $f(x) = e^{-ax^{2}}$ , and $g(x) = e^{-bx^{2}}$ [for all $x \in \mathbb {R}$ ], then

f * g(x) = \sqrt{\frac{\pi }{a + b}} \, e^{-\frac{ab}{a + b} x^{2}} \quad \text{for all } x \in \mathbb {R} .

Problem 10.5.9

(a)(b)(c)(d)

Claim 6 (Young's Theorem): Let $1 < p < +\infty$ , let $f \in L^{1}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ with $g \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ . Then $f * g \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ , $g * f = f * g$ $\bar{\lambda }$ -a.e. on $\mathbb {R}$ , and $\left\| f * g\right\|_{p} = \left\| g * f\right\|_{p} \leq \left\| f\right\|_{1} \left\| g\right\|_{p}$ .

This exercise generalizes Young's Theorem and requires the result of Exercise 3 in Section 8.1. Show the following: Let $1 \leq p, q, r \leq +\infty$ be such that $\frac{1}{r} = \frac{1}{p} + \frac{1}{q} - 1$ . Let $f \in L^{p}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ and $g \in L^{q}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ . Then $f * g$ exists $\bar{\lambda }$ -a.e. on $\mathbb {R}$ and is in $L^{r}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ . Also, $\left\| f * g\right\|_{r} \leq \left\| f\right\|_{p} \left\| g\right\|_{q}$ .

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(a)

If $p = +\infty$ , then $q = 1$ and $r = +\infty$ , and the claim follows easily in this case. The same comment applies to the case where $q = +\infty$ [and hence $p = 1$ and $r = +\infty$ ].

(b)

If $r = +\infty$ , then $p$ and $q$ are conjugate exponents, and the claim follows by Hölder's Inequality.

(c)

Assume that $1 \leq r < +\infty$ now. This forces $1 \leq p, q < +\infty$ . Show that it is sufficient to prove the claim when both $f, g \geq 0$ and $\left\| f\right\|_{p} = \left\| g\right\|_{q} = 1$ .

(d)

Prove the claim in (c).

Remark.

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Claim 3: Let $f, g \in L^{1}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ . Then $f * g \in L^{1}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ and $\left\| f * g\right\|_{1} \leq \left\| f\right\|_{1} \left\| g\right\|_{1}$ .

Since $f, g$ are nonnegative and $\mathcal{L} / \mathcal{B}^{*}$ -measurable, we may meaningfully discuss $\int_{\mathbb {R}} f(x - y) g(y) \, d\bar{\lambda }(y) = f * g(x)$ for each $x \in \mathbb {R}$ . Show that $\frac{1}{r} + \frac{1}{q^{\prime }} + \frac{1}{p^{\prime }} = 1$ , and use Exercise 3 in Section 8.1 to obtain

\begin{aligned} f * g(x) & = \int _{\mathbb {R}} f(x - y) g(y) \, d\bar{\lambda }(y) \\ & = \int _{\mathbb {R}} f(x - y)^{\frac{p}{r}} g(y)^{\frac{q}{r}} f(x - y)^{1 - \frac{p}{r}} g(y)^{1 - \frac{q}{r}} \, d\bar{\lambda }(y) \\ & \leq \left[\int _{\mathbb {R}} f(x - y)^{p} g(y)^{q} \, d\bar{\lambda }(y)\right]^{1 / r} \left[\int _{\mathbb {R}} f(x - y)^{q^{\prime } \left(1 - \frac{p}{r}\right)} \, d\bar{\lambda }(y)\right]^{1 / q^{\prime }} \\ & \quad \times \left[\int _{\mathbb {R}} g(y)^{p^{\prime } \left(1 - \frac{q}{r}\right)} \, d\bar{\lambda }(y)\right]^{1 / p^{\prime }} \\ & = \left[\int _{\mathbb {R}} f(x - y)^{p} g(y)^{q} \, d\bar{\lambda }(y)\right]^{1 / r} \times 1 \times 1 \end{aligned}

for each $x \in \mathbb {R}$ . This shows that $\left(f * g\right)^{r} \leq f^{p} * g^{q} \in L^{1}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ by Claim 3. It follows that $f * g \in L^{r}\left(\mathbb {R}, \mathcal{L}, \bar{\lambda }\right)$ , and Claim 3 applied to $f^{p}$ and $g^{q}$ gives

\left\| f * g\right\| _{r}^{r} = \int _{\mathbb {R}} f * g(x)^{r} \, d\bar{\lambda }(x) \leq \int _{\mathbb {R}} f^{p} * g^{q}(x) \, d\bar{\lambda }(x) \leq \left\| f^{p}\right\| _{1} \left\| g^{q}\right\| _{1} = \left\| f\right\| _{p}^{p} \left\| g\right\| _{q}^{q} = 1 .