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Questions tagged [young-inequality]

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1answer
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Help with an application of Young's inequality

I am reading through a set of notes about concentration of Gaussian measure, and on page 56, they make the following claim that I am failing to see the proof of: Now we estimate the second summand ...
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1answer
57 views

Is Young's inequality useful here?

I want to prove that for a given $0<\alpha<N$ and for all $0<\varepsilon< N-\alpha$ there exists $C>0$ s.t. $$ \Vert\vert x\vert^{-\alpha}\ast\vert u \vert^2\Vert_{L^\infty(\mathbb{R}^N)...
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2answers
27 views

Young's convolution inequality: Equivalent representations

According to Wikipedia Young's inequality for convolutions states that For functions $f \in L^p$ and $g \in L^q$ one has $|| f*g ||_r \leq ||f||_p ||g||_q$ $\hspace{6.75cm}$ (Eq. 1) with $...
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3answers
566 views

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 ...
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72 views

Having trouble showing this inequality

Given the initial boundary value problem \begin{align*} &u_t = Du_{xx} + f(u), \quad 0<x<1, t>0 \\ &u(0,t) = u(1,t) = 0, \quad t>0 \\ &u(x,0) = u_{0}(x), \quad 0<x<1 \...
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1answer
79 views

Prove convergence in operator norm and compact operator

Let bounded domain $X \subset \mathbb{R^2}$ , $$T:L^2(X) \to L^2(X) \ \ Tf(x)=\int _{X} \frac{f(t)}{|x-t|}dt $$ $X_r (x)=X \cap \{t \in \mathbb{R}; |t-x| \geq r\}$ ,and $$T_r :L^2(X) \to L^2(X) \ \ ...
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1answer
48 views

Why the relation between the exponents in Young's inequality is inevitable?

Suppose someone told you that there exist a pair of positive real numbers $p,q$ such that for every positive real numbers $a,b$, the following inequality holds $$ ab \leqslant \frac{a^p}{p} + \frac{b^...
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0answers
108 views

Connecting Young's Inequality for Increasing Functions w/Young's Inequality for Conjugate Holder Exponents

From the Wikipedia page on Young's Inequality: This above statement of Young's Inequality is the most frequent one that I've encountered in textbooks. Yet consider: This second version ...
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1answer
627 views

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 $p > 1$, $\frac{1}{p} + \frac{1}{q} = 1$. ...
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1answer
174 views

Is this proof valid - Holder's inequality

Given that $a_i, b_i > 0$ and that $ p, q > 1$ and $\frac{1}{p} + \frac{1}{q} = 1$ I want to show Holder's inequality, that $\sum_{i}a_ib_i \leq (\sum_{i} a_i ^p)^{\frac{1}{p}} (\sum_{i} b_i ^q)^...
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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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1answer
69 views

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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0answers
80 views

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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1answer
344 views

Convolution inequality with weak Lp

I had the pleasure to stumble upon a "well-known convolution inequality": $$\iint u(x)u(y)f(x-y)dxdy \leq C_1||u||_r^2||f||_{p,\infty}$$ The integrals are over $\mathbb{R}^n \times \mathbb{R}^n$, $r = ...
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3answers
606 views

Generalization of algebraic Young's inequality to n variables.

The classical formulation of Young's inequality is \begin{equation*} xy \leq \frac{x^p}{p} + \frac{y^q}{q}, \quad \text{where} \quad \frac{1}{p} + \frac{1}{q} = 1. \end{equation*} It's fairly trivial ...
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2answers
173 views

Equality in convolution inequality

As we know the following inequality holds: $\lVert f \ast g \rVert_{L^p}\leq \lVert f\rVert_{L^1} \lVert g \rVert_{L^p}$ for $1\leq p \leq \infty$. My question is when the inequality becomes an ...
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How can we prove it? ${a}{b}≤\frac{a^p}{p}+\frac{b^q}{q}$ [duplicate]

$(i)$ $a>0,b>0,p>0,q>0$ $(ii)$ $\frac{1}{p}+\frac{1}{q}=1$ How can we prove it? ${a}{b}≤\frac{a^p}{p}+\frac{b^q}{q}$
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2answers
665 views

Prove Young's inequality

Suppose that f is continuous increasing function with $f(0)=0$ then prove that for $a,b>0$ we have Young's inequality $$ab\leq \int_{0}^{a} f(x)dx+\int_{0}^{b}f^{-1}(x)dx$$ My attempt is ...
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1answer
128 views

Why this is a generalization of Young's inequality?

The is the standard version of Young's inequality $$ab \leq \frac{a^p}{p}+\frac{b^q}{q}$$ for $a,b,p,q >0$ and $$\frac{1}{p}+\frac {1}{q}=1$$ But there is another formula called generalization of ...
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1answer
58 views

The inequality $xy \leq \frac{x^p}{p}+\frac{y^q}{q}$

I struggled to show that this is true when $x,y >0$, $p>1$ and $q=\frac{p}{p-1}$. But, I managed to show that if we assume that $x \geq 1$ the assertion is possibly false only for a finite ...
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747 views

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 ...
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1answer
302 views

Young's inequality.

I am referring to the inequality: Young's inequality The standard version for increasing functions. I read the article of Young and also a generalization of this claim in Hardy, Littlewood and Polya'...
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1answer
213 views

absolute minimum of function

Let $p$ and $q$ be positive numbers satisfying $\dfrac1p+\dfrac1q= 1$; and let $f ∶ [0,+∞) \to \mathbb{R}$ be the function $f(x) =\dfrac1p x^p− x +\dfrac1q$ Show that $f$ has an absolute minimum at $...
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1answer
615 views

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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1answer
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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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0answers
101 views

Leading up to Young's Inequality

I am trying to prove Young's Inequality by considering the function $$h(u) = \frac{u^p}{p} + \frac{C^q}{qu^q}$$ for $C,u>0$ and $p,q >1$. We also require $$\frac{1}{p}+\frac{1}{q}=1$$ so that $...
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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\...
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5answers
934 views

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 ...
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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 ...
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2answers
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A proof of Young's inequality

I need to prove that if $1 < p < \infty$ and $a, b \geqslant 0$ then $$ ab \leqslant \frac{a^{p}}{p} + \frac{b^{q}}{q}$$ where $\frac 1p+\frac 1q=1$. I fix $b$ and maximize the function $f(a) =...
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2answers
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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} +...
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4answers
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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 ...
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1answer
716 views

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 ...
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2answers
1k views

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 ...
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1answer
600 views

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 $...
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1answer
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Does Young's inequality reverse when $f$ is decreasing?

According to wikipedia, "let $f$ denote a real-valued, continuous and strictly increasing function on [0, c] with c > 0 and f(0) = 0. Let $f^{−1}$ denote the inverse function of $f$. Then, for all a ∈...