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Young's inequality (Theorem)

Let $ \phi : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous , strictly increasing function such that $ \phi(0)=0$ . Then the following inequality holds:

$\displaystyle ab \leq \int_{0}^a \phi(x) dx + \int_{0}^b \phi^{-1}(y) dy $
This inequality can be demonstrated by drawing the graph of $ \phi(x)$ and by observing that the sum of the two areas represented by the integrals above is greater than the area of a rectangle of sides $ a$ and $ b$ .



"Young's inequality" is owned by rspuzio. [ full author list (2) | owner history (1) ]
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See Also: Young inequality

Keywords:  Young's Inequality

Attachments:
proof of Young's inequality by picture (Proof) by archibal
another proof of Young inequality (Proof) by a4karo
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Cross-references: sides, rectangle, integrals, areas, sum, graph, inequality, function, strictly increasing, continuous
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This is version 3 of Young's inequality, born on 2002-12-26, modified 2007-01-03.
Object id is 3834, canonical name is YoungsInequality.
Accessed 7132 times total.

Classification:
AMS MSC: 26D15 (Real functions :: Inequalities :: Inequalities for sums, series and integrals)
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