ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-09-01

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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 Young's inequality gives

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

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 $f = g = 1_{[0, 1]}$ .

[L1]

Young's inequality holds (Young's convolution inequality).

Verification

technique · direct

1.1

The previous tent-function computation gives [L1, given, algebra] $(f * g) (x) = {\begin{cases} x, & 0 \leq x \leq 1, \\ 2 - x, & 1 \leq x \leq 2, \\ 0, & otherwise . \end{cases}$

L1givenalgebra

2.1

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

step 1.1algebra

3.1

Since $∥ f ∥_{1} = 1$ and $∥ g ∥_{2} = 1$ , this gives [L1, step 2.1] $∥ f * g ∥_{2} = \sqrt{2 / 3} \leq 1 = ∥ f ∥_{1} ∥ g ∥_{2}$ , exactly as [L1] predicts.

L1step 2.1∎

Depends on

Young's convolution inequality

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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