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.

Statement of the Inequality: Let $~p,~q~$ be positive real numbers satisfying $~\frac1{p} + \frac1{q} = 1~$. Then if $~a,~b~$ are non-negative real numbers, $$ab \le \frac{a^p}{p} + \frac{b^q}{q}~,$$ and equality holds if and only if $~a^p=b^q~$.

Young's inequality for products, bounding the product of two quantities, can be used to prove Hölder's inequality. It is also used widely to estimate the norm of nonlinear terms in PDE theory, since it allows one to estimate a product of two terms by a sum of the same terms raised to a power and scaled.

Young's convolution inequality, bounding the convolution product of two functions. An example application is that Young's inequality can be used to show that the heat semigroup is a contracting semigroup using the $~L^2~$ norm (i.e. the Weierstrass transform does not enlarge the $~L^2~$ norm).

Young's inequality for integral operators, is a bound on the $~{\displaystyle L^{p}\to L^{q}}~$ operator norm of an integral operator in terms of $~{\displaystyle L^{r}}~$ norms of the kernel itself.

For more details see

http://mathworld.wolfram.com/YoungsInequality.html

https://brilliant.org/wiki/youngs-inequality

https://www.math.upenn.edu/~brweber/Courses/2011/Math361/Notes/YMandH.pdf

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Purely "algebraic" proof of Young's Inequality

Young's inequality states that if $a, b \geq 0$, $p, q > 0$, and $\frac{1}{p} + \frac{1}{q} = 1$, then $$ab\leq \frac{a^p}{p} + \frac{b^q}{q}$$ (with equality only when $a^p = b^q$). Back when I was in my first course in real analysis, I was…
asmeurer
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Geometric interpretation of Young's inequality

Is there a geometric interpretation of Young's inequality, $$ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}$$ with $\dfrac{1}{p}+\dfrac{1}{q} = 1$? My attempt is to say that $ab$ could be the surface of a rectangle, and that we could also say…
Frank
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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 $$ab \leqslant \frac{a^p}{p} + \frac{b^q}{q}.$$ Here's my attempt at a proof: Let $$f(x) =…
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Equality in Young's inequality for convolution

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 (for example: Sharpness in Young's inequality for…
user3296322
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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 out how to apply it in this case. Any ideas about the…
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Young's inequality without using convexity

I was doing some problems from Rudin's Principles of Mathematical Analysis and came across a problem in which he asks you to prove Hölder's inequality via Young's inequality: If $u$ and $v$ are nonnegative real numbers, and $p$ and $q$ are positive…
Gyu Eun Lee
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Young's inequality for three variables

Let $x, y, z \geqslant 0$ and let $p, q, r > 1$ be such that $$ \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1. $$ How can one show that under these hypotheses we have $$ xyz \leqslant \frac{x^p}{p} + \frac{y^q}{q} + \frac{z^r}{r} $$ with equality if…
Amateur
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Does Young's inequality hold only for conjugate exponents?

Suppose that $ab \leq \frac{1}{p}a^p+\frac{1}{q}b^q$ holds for every real numbers $a,b\ge 0$. (where $p,q>0$ are some fixed numbers). Is it true that $ \frac{1}{p}+\frac{1}{q}=1$? I guess so, and I would like to find an easy proof of that fact.…
Asaf Shachar
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Showing that $xy \leq \frac{x^p}{p} + \frac{y^q}{q}$

Question: Let $x \geq 0$ , $y \geq 0$ and $p > 0$, $q>0$ with $\frac{1}{p} + \frac{1}{q} = 1$. Show that $$xy \leq \frac{x^p}{p} + \frac{y^q}{q} $$ [Suggestion: Without loss of generality suppose $xy = 1$]. Attempt: Let $f, \varphi : U \to…
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Applications of Young's convolution inequality

Recall that the convolution of two functions is given by $$f*g(y)=\int f(x)g(y-x)dx.$$ The well known inequality known as Young's inequality, say that $$\|f*g\|_r\leq\|f\|_p\cdot\|g\|_q $$ provided $\frac 1p + \frac 1q = 1 + \frac 1r$ and $1\le…
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Minkowski's and Holder's inequality confusion

Something doesn't completely makes sense to me in Minkowski's inequality and I'm trying to understand it. All throughout let $a_i, b_i \geq 0$, and $p > 1$, $\frac{1}{p} + \frac{1}{q} = 1$. Holder's inequality states that: $\sum_i a_ib_i \leq…
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Proving that $\frac{u^p}{p}+\frac{v^q}{q}\ge uv$ under the condition $\frac{1}{p}+\frac{1}{q}=1$

The following is a problem (6.10) from Rudin's principles of Mathematical analysis. Let $p$ and $q$ be positive real numbers such that $$\frac{1}{p}+\frac{1}{q}=1.$$ Prove that if $u\ge 0$ and $v\ge 0$, then $$uv\le…
Frog
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Young inequality: Generalization on $\mathbb{T}$ space.

I'm interested in resolving this question that I find but on the $\mathbb{T}$ space. (Show that for any $f\in L^1$ and $g \in L^p(\mathbb R)$, $\lVert f ∗ g\rVert_p \leqslant \lVert f\rVert_1\lVert g\rVert_p$.) Now my question is, is it possible? I…
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What is the content of Young's inequality?

On the one hand, Young's inequality, in the form of $$ ab \leq \frac{a^p}{p}+\frac{b^q}{q} $$ where $p$ and $q$ are Hölder conjugates, can be seen to be easily rearranged to be a restatement of either Bernoulli's inequality, which is a statement…
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Deriving inequalities from other inequalities

My questions come from the proof of Theorem 5.14 in section 5.7 of this book. My first question can be stated as follows: Suppose for positive numbers $V,y,\Delta,x,\delta >0$ we have $V \leq y^{-1}\left[2\Delta y^{-1} + \sqrt{2x} +…
Fei Cao
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