ExampleConstruction: AI-adaptedVerification: AI-adaptedverified 2026-09-23 (gpt-6-sol)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Young's inequality on an $L^{1} * L^{2}$ pair

Example

Let $f = g = 1_{[0, 1]}$ on $R$ . Then $f \in L^{1} (R)$ , $g \in L^{2} (R)$ , and the convolution satisfies

$∥ f * g ∥_{2} \leq ∥ f ∥_{1} ∥ g ∥_{2} = 1.$

In fact $∥ f * g ∥_{2} = {(\int_{0}^{1} x^{2} d x + \int_{1}^{2} (2 - x)^{2} d x)}^{1 / 2} = \sqrt{\frac{2}{3}} .$

Facts & Assumptions

Given: The indicator $f = g = 1_{[0, 1]}$ .

Verification

technique · direct

1.1givenalgebra

For each $x$ , the convolution integral is the length of $[0, 1] \cap [x - 1, x]$ ; direct interval overlap gives [given, algebra] $(f * g) (x) = {\begin{cases} x, & 0 \leq x \leq 1, \\ 2 - x, & 1 \leq x \leq 2, \\ 0, & otherwise . \end{cases}$

2.1step 1.1algebra

Therefore [step 1.1, algebra] $∥ f * g ∥_{2}^{2} = \int_{0}^{1} x^{2} d x + \int_{1}^{2} (2 - x)^{2} d x = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} .$

3.1step 2.1∎

Since $∥ f ∥_{1} = 1$ and $∥ g ∥_{2} = 1$ , this gives [step 2.1] $∥ f * g ∥_{2} = \sqrt{2 / 3} \leq 1 = ∥ f ∥_{1} ∥ g ∥_{2}$ .

Used by

Nothing in the library uses this result yet.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Richard L. Wheeden and Antoni Zygmund, Measure and Integral: An Introduction to Real Analysis (standard reference, not scraped)

Young's inequality on an L1∗L2L1∗L2 pair