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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1answer
33 views

variant of young's inequality

Let $\alpha_{i}\geq 0$ y $p_{i}\geq 0$ for $i=1,2,...,n$ such that: $$\frac{1}{p_1}+\frac{1}{p_2}+...+\frac{1}{p_n}=1$$ proof that: $$\alpha_1\alpha_2...\alpha_n\leq \frac{\alpha_1^{p_1}}{p_1}+\frac{\...
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1answer
109 views

Showing boundedness of a set (preparation for the proof of the Young inequality)

Define the two sets: $\Omega = \{(a,b) \in \mathbb{R}^2 \; | \; a>0, b>0\}, \; \; \; \; \overline{\Omega} = \{(a,b) \in \mathbb{R}^2 \; | \; a\ge0, b\ge0\}$ as well as the functions $g: \...
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38 views

How to proof that $\lambda^{t}\leq \epsilon \lambda^{s}+C_\epsilon\lambda^{t-1}$

i am reading the book Pseudo Differential Operators, singularities and applications (Y. Egorov, B. Schulze) on page 6, the inequality is $\lambda^{t}\leq \epsilon \lambda^{s}+C_\epsilon\lambda^{t-1}$, ...
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25 views

An application of the Hölder's inequality to Young's inequality

Hölder's inequality Let $f_1, f_2,\dots, f_k\colon X\to \overline{\mathbb{R}}$ be a measurable functions and $p,p_1,\dots, p_k\in [1,\infty)$ such that $$\frac{1}{p}=\frac{1}{p_1}+\frac{1}{p_2}+\cdots ...
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2answers
109 views

Application of Hölder's inequality: $(a+b)^t \le 2^t(a^t+b^t)$ for $t\ge 1.$

While searching for a proof of the algebra property of Sobolev spaces ($H^s(\mathbb R^n)$ is an algebra when $s > n/2$), I found these notes. On page two the author states that if $t \ge 1$ then ...
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31 views

Closing an estimate involving the Lp-norm

I'm reading a paper and got stuck on the following We have that $$\int \vert \nabla u \vert^p dx \leq pC+pC(\int \vert \nabla u \vert^p dx)^{1/p}$$ implies $$(\int \vert \nabla u \vert^p)^{1/p} \leq ...
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2answers
54 views

Proving Young's inequality for n numbers whitout using AM-GM inequality

Let $\alpha_1, ..., \alpha_n \geq 0$, and $p_1,...,p_n \geq 0$ such that $\sum_{i=1}^{n} p_i^{-1} =1$. Show that $\alpha_1...\alpha_n \leq p_{1}^{-1}\alpha_1^{p_1}+...+p_{n}^{-1}\alpha_n^{p_n}$. It ...
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1answer
21 views

Inequality for sum to the exponent of $q>1$

In this question: How to use $\lesssim$ when using exponents I already asked about an inequality for a sum to the exponent of $2$. Now I wonder if $\forall q>1$, there exists a constant $C>0$ ...
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1answer
33 views

Young-like inequality

I'm having trouble with the following statement: For positive variables $\{x_1,\dots,x_n\}\subset \mathbb{R_+}$ and $k\geq 1$ an integer, there exists $C=C(k)$ a numerical constant such that \begin{...
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1answer
47 views

How can we show $\left\|g\ast f\right\|_{L^p}\le\left\|g\right\|_{L^1}\left\|f\right\|_{L^p}$?

Let $p\ge1$, $f\in L^p(\mathbb R^d)$ and $g\in L^1(\mathbb R^d)$. I've read that the inequality $$\left\|g\ast f\right\|_{L^p}\le\left\|g\right\|_{L^1}\left\|f\right\|_{L^p}\tag1$$ would follow from ...
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31 views

Sharpness of Young's Inequality for Convolutions

Young's inequality for convolutions states that for $g\in L^r(\mathbb{R}^d) $ and $f\in L^q(\mathbb{R}^d)$, we have $$\|g*f\|_p\leq \|g\|_r\|f\|_q$$ where $1+\frac{1}{p} = \frac{1}{r}+\frac{1}{q}.$ ...
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3answers
50 views

ab < $\epsilon a^2$ + $\frac{1}{\epsilon}b^2$ $\forall \epsilon$ [closed]

Why is this inequality true $\forall \epsilon > 0$? $$ab < \epsilon a^2 + \frac{1}{\epsilon}b^2 $$
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20 views

Computing using holder's and young's inequality

I need help in obtaining the following estimate which am stuck in calculating it in details: $$ \big| ((v^2+2v\langle u\rangle-v, (-\Delta)^{-1}v))\big| $$ $$ \leqslant \dfrac{3c_0}{8}\int_{\Omega}(v^...
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1answer
80 views

Young’s inequality implies $a^ {n/(n+1)}\leq 2a+(1/n^2)$ for any $a>0$ and $n\geq 1$

I’ve seen this statement in a Yano’s article and I can not prove it. I take the Young’s inequality $a^{1/p}b^{1/q} \leq a/p + b/q$ where $1/p +1/q =1$. I’ve prove it in the case $a\geq 1$. In the ...
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33 views

Connection between Bernoulli's inequality and Young's inequality .

Working on a question Proof that if $x,y>0$ and $x+y=1$, then $(2x)^{\frac 1 x}+(2y)^{\frac 1 y}\leq 2$ . I find a connection between Bernoulli's inequality and Young's inequality : Let $0< r\...
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1answer
49 views

How to prove the following Inequality with exponents

Let $X,Y>0$, $p>2$ and $q>p-1$. Then does the following inequality hold: $$ (|X|^{p-2}+|Y|^{p-2})(X^{1-q}+Y^{1-q})\leq C X^{p-1-q}+ D Y^{p-1-q}, $$ for some positive constants $C$ and $D$ may ...
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1answer
78 views

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\...
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1answer
57 views

Hausdorff-Young inequality on T space

I know that this inequality work well on space $L^p(\mathbb{R})$. But is it possible to generalize this inequality to the $\mathbb{T}$ space? I think that on this space I can write this inequality ...
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1answer
71 views

Prove that $a b \leq \frac{a^{3}}{3}+\frac{b^{3 / 2}}{3 / 2}$ [duplicate]

for any $a,b\in R^+$, how to prove that $$a b \leq \frac{a^{3}}{3}+\frac{b^{3 / 2}}{3 / 2}$$ my try: if I can prove that $\frac{a^2+b^2}{2}\leq \frac{a^{3}}{3}+\frac{b^{3 / 2}}{3 / 2}$then this ...
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0answers
46 views

Young's Inequality for Complementary Functions

I was trying to understand this theorem please see in the image. However I have no idea about the equality which I framed with red. Second, will not Young inequality turn into equality when both $y\...
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70 views

Is it possible a refinement for Fenchel–Young inequality?

The idea of this post arises when I've known in past weeks a refinement of Young’s inequality, I say Lemma 2.1 from [1]. On the other hand the Wikipedia Young's inequality for products shows ...
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71 views

Young convolution inequality for other Lp spaces

I´m wondering if there exist types of the Young convolution inequality (https://en.wikipedia.org/wiki/Young%27s_convolution_inequality) for other $L_p$-Spaces than $L_p(R^d)$. I´m asking since I want ...
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1answer
24 views

Reference to this Young Inequality for matrices

In this post: here I saw the usage of this Young Inequality: $$X^TY+Y^TX\leq \frac{1}{2}(X+SY)^TS^{-1}(X+SY)$$ With S any symmetric positive definite matrix. As far as I understood this could help ...
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1answer
93 views

Hardy-Littlewood-Sobolev inequality using generalised Young inequality in Lorenz spaces

I want to prove that $$\left| \left| \frac{1}{|x|^a} \ast f \right| \right|_q \lesssim ||f||_p$$ with $1 < p < q < \infty$ and $a= n \left(1 + \frac{1}{q}- \frac{1}{p} \right)$ using a ...
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0answers
367 views

Is the Fourier transform bounded from $L^\infty$ to $L^\infty$?

I have seen how to extend the Fourier transform as an operator from $L^p(\mathbb{R})$ to $L^p(\mathbb{R})$ with $1 \leq p < \infty$ using Schwartz's class functions and tempered distributions. So I ...
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1answer
31 views

Inequality with variable expoent

Let $a,b \geq 0$, and $n \in\mathbb{N}$, then if $1<p,q <\infty$ with $\frac{1}{p}+\frac{1}{q}=1$ is valid $(a+b)^n \leq p^{n}a^{n}+q^{n}b^{n}$. We would like to know in place of exponent $n$ ...
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128 views

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 ...
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100 views

Young inequality for distributions

Young Inequality: Let $f \in L^p(\mathbb{R}^n)$, $g \in L^q(\mathbb{R}^n)$. Then, $h := f *g \in L^r(\mathbb{R}^n)$. Furthermore, \begin{equation} \| h \|_{L^r(\mathbb{R}^n)} \leq \| f \|_{L^p(\...
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1answer
80 views

Using Young's inequality to prove $\frac{|x|^{a_1}|y|^{a_2}}{|x|^{b_1}+|y|^{b_2}} \geq 1$ where $\frac{a_1}{b_1}+\frac{a_2}{b_2}=1$

Use Young's Inequality prove that if $a_1$, $a_2$, $b_1$,$b_2$ are all positive and $\frac{a_1}{b_1} + \frac{a_2}{b_2} = 1$ then $$\frac{|x|^{a_1}|y|^{a_2}}{|x|^{b_1}+|y|^{b_2}} {\leq} 1$$ for ...
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1answer
94 views

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
84 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
404 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 $1/p + 1/q = ...
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3answers
733 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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0answers
85 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
170 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
78 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
293 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
901 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
377 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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125 views

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
75 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
118 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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0answers
48 views

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 ...
2
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1answer
980 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 = ...
3
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3answers
1k 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 ...
2
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2answers
736 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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0answers
189 views

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
1k 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
180 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
75 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 ...