# Questions tagged [young-inequality]

This tag is for questions relating to Young's inequality, a special case of the weighted AM-GM inequality. It is very useful in real analysis, including as a tool to prove Hölder's inequality. It is also a special case of a more general inequality known as Young's inequality for increasing functions.

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### Inequalities for the norms of the fractional Fourier transform

Introduction and definitions When considering the norms of a function $f(x)$ and its Fourier transform, $\tilde{f}$, \begin{equation} \tilde{f}(y)=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}e^{-iyx}f(x) \,, ...
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### Proof of $\big|\int_E (f\cdot g)(x) \text{d}x\big| \leq [\int_E |f(x)|^p\text{d}x]^{(1/p)} \cdot [\int_E |g(x)| ^q\text{d}x]^{(1/q)}$

The answer to this question is given in here, but I cannot fill the gaps between the accepted answer, so here is what I have done with the guide of @Raito. My work: By Young's inequality, we do know ...
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### Young inequality for negative exponents

Let $p\in(1,2)$, and $f, g:[0, \tau]\rightarrow\mathbb{R}$ are a real valued functions. I like to upper bound $\int_{t}^{u} e^{ ps}|f(s)|^{p-2}|g(s)|^{2}ds$ for $t, u \in[0,\tau]$ in the following way:...
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### Origins of Young's Inequality

I see Young's inequality (for products) pop up a lot in analysis, functional analysis, ODE, PDEs, etc... I've seen a few proofs of it too. I'm just wondering, in what context did Young originally ...
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### Equality in Young's inequality

Let's take a look at Young's inequality: If $u,v\geqslant 0$ and $p,q$ - positive real numbers such that $\frac{1}{p}+\frac{1}{q}=1$ then $$\dfrac{u^p}{p}+\dfrac{v^q}{q}\geqslant uv.$$ It's easy to ...
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### Valid proof of Young's inequality?

Part of an exercise to prove Holder's inequality in Rudin involves proving Young's Inequality... That is, given $\frac{1}{p}+\frac{1}{q} = 1$, prove $$ab \leqslant \frac{a^p}{p} + \frac{b^q}{q}.$$ ...
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### Prove that $xy \leq\frac{x^p}{p} + \frac{y^q}{q}$

OK guys I have this problem: For $x,y,p,q>0$ and $\frac {1} {p} + \frac {1}{q}=1$ prove that $xy \leq\frac{x^p}{p} + \frac{y^q}{q}$ It says I should use Jensen's inequality, but I can't figure ...
I am trying to understand how sharp Young's inequality for convolution is. The inequality says $||f \ast g||_r \leq ||f||_p ||g||_q$ where as $1/p+1/q = 1+1/r$. Actually, there are a couple of papers ...