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I'm currently working through a proof for Young's inequality in the simplest case:

If $a,b > 0$ , then $a \cdot b \leq \frac{a^p}{p} + \frac{b^q}{q}$ for $p,q$ conjugates.

The proof is relatively simple, but relies on the fact that

$$ \log\bigl(\gamma\cdot x+(1-\gamma)\cdot y\bigr) \geq \gamma\cdot \log(x) + (1-\gamma)\cdot \log(y) $$

for $0 < \gamma \leq 1$. I've tried proving this myself with no success, as well as searching online. Any suggestions would be appreciated.

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  • $\begingroup$ See en.wikipedia.org/wiki/Jensen%27s_inequality. $\endgroup$ – Bob Krueger Jul 14 '17 at 19:42
  • $\begingroup$ The proof of the finite form starts by assuming convexity on $\phi$, I'm not sure how to apply that here. Care to elaborate? $\endgroup$ – Guido A. Jul 14 '17 at 19:45
  • $\begingroup$ Convexity is the opposite of what is the logarithm's concavity (log is concave down, since it's second derivative is negative). Thus the inequality you seek will flow the opposite way Jensen's Inequality does. See the picture in the link for a graphical representation of why the inequalities hold. $\endgroup$ – Bob Krueger Jul 14 '17 at 19:51
  • $\begingroup$ I am already convinced it holds, and I see the link between Jensen's inequality and what I need, but nevertheless this only gives intution: as far as I can tell, the Wikipedia article doesn't give much of a proof for this statement, it assumes it and generalizes it (first to the finite case, then to the probability and measure theoretic cases). $\endgroup$ – Guido A. Jul 14 '17 at 19:54
  • $\begingroup$ I see what you mean. Let me come up with an answer. $\endgroup$ – Bob Krueger Jul 14 '17 at 20:03
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I will prove the following statement. Let $f(x)$ be a twice-differentiable function on $I = [a, b]$ such that for every $x \in I$, $f''(x) \leq 0$. Then for any $t \in [0, 1]$ and for any $c, d \in I$, $f(tc + (1-t)d) \geq tf(c) + (1-t)f(d)$.

Proof: Let $c, d \in I$. Let $g(x) = f(x) - \left( \frac{f(d)-f(c)}{d-c}(x-c) + f(c) \right)$. Then $g(c) = g(d) = 0$ and $g''(x) = f''(x) \leq 0$ for all $x \in I$.

We claim that $g(x) \geq 0$ for all $x \in [c, d]$. For the sake of contradiction, suppose that there is some $x \in [c, d]$ such that $g(x) < 0$. Then

$$ 0 > g(x) = \int_c^x g'(s) ds + g(c) = \int_c^x g'(s) ds .$$

This means that for some $y \in [c, x]$, $g'(y) < 0$. Then for any $z \in [y, d]$, since $g''(s) \leq 0$ for any $s \in I$,

$$ g'(z) = \int_y^z g''(s) ds + g'(y) < 0 .$$

But this implies

$$ 0 = g(d) = \int_c^d g'(s) ds = \int_c^x g'(s) ds + \int_x^d g'(s) ds < 0 ,$$

a contradiction. Thus $g(x) \geq 0$ for all $x \in [c, d]$.

In particular, let $x = tc + (1-t)d$. Then

$$ 0 \leq g(tc + (1-t)d) = f(tc + (1-t)d) - \left( \frac{f(d)-f(c)}{d-c}(tc + (1-t)d - c) + f(c) \right) $$

$$\rightarrow \frac{f(d) - f(c)}{d-c}(1-t)(d-c) + f(c) \leq f(tc + (1-t)d) $$

$$\rightarrow tf(c) + (1-t)f(d) \leq f(tc + (1-t)d) .$$

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  • $\begingroup$ It's certainly possible that this can be streamlined with the use of the Mean Value Theorem, although this is what came to me first. $\endgroup$ – Bob Krueger Jul 15 '17 at 2:05
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If $p>0$, $q>0$ and $\frac{1}{p}+\frac{1}{q}=1$ then our inequality it's just AM-GM: $$\frac{a^p}{p}+\frac{a^q}{q}=\frac{1}{p}\cdot{a^p}+\frac{1}{q}\cdot{b^q}\geq\left(a^p\right)^{\frac{1}{p}}\left(b^q\right)^{\frac{1}{q}}=ab$$

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  • $\begingroup$ how did you use AM-GM ??? $\endgroup$ – contestant IMO 2020 Sep 29 '18 at 20:33
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If the convexity of the exponential function is allowed, then $$ A^{1/p}B^{1/q}=\exp\left(\frac1p\log A+\frac1q\log B\right) \leqslant\frac1p e^{\log A}+\frac1q e^{\log B}=\frac{A}{p}+\frac{B}{q} $$ Now let $a=A^{1/p}$ and $b=B^{1/q}$.


Alternatively, consider the function $f(t)=t^\alpha-\alpha t$ on $(0,\infty)$ with fixed $\alpha\in(0,1)$. Simple calculus gives $$ f(t)\leqslant f(1)=1-\alpha,\quad\forall t>0 $$ which implies that $$ t^\alpha\leqslant \alpha t+(1-\alpha),\quad\forall t>0. $$ Now let $t=\frac{A}{B}$ and $\alpha=1/p$. Some algebra gives $$ A^{1/p}B^{1/q}\leqslant \frac{A}{p}+\frac{B}{q}. $$

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Here is one approach, based on the definition $$\log x = \int_{t=1}^x \frac{1}{t} \, dt.$$ Assume $0 < x < y$, so that in particular we have some $\delta > 0$ satisfying $y = x + \delta$. We find $$\lambda x + (1-\lambda) y = x + (1-\lambda)\delta, \quad 0 \le \lambda \le 1, \quad \delta > 0.$$ Then $$\log(\lambda x + (1-\lambda)y) = \log(x + (1-\lambda)\delta) = \log x + \int_{t=x}^{x + (1-\lambda)\delta} \frac{1}{t} \, dt.$$ Next consider $$\lambda \log x + (1-\lambda) \log y = \lambda \int_{t=1}^x \frac{1}{t} \, dt + (1-\lambda) \left( \int_{t=1}^x \frac{1}{t} \, dt + \int_{t=x}^{x+\delta} \frac{1}{t} \, dt\right) \\ = \log x + (1-\lambda) \int_{t=x}^{x+\delta} \frac{1}{t} \, dt.$$ We are interested in the difference between the first and second expressions; namely that it is nonnegative: $$\Delta(x,\lambda,\delta) = \int_{t=x}^{x+(1-\lambda)\delta} \frac{1}{t} \, dt - (1-\lambda) \int_{t=x}^{x+\delta} \frac{1}{t} \, dt.$$ With the substitution $u = t+x$ we can simplify this to $$\Delta(x,\lambda,\delta) = \int_{u=0}^{(1-\lambda)\delta} \frac{1}{u+x} \, du - (1-\lambda) \int_{u=0}^{\delta} \frac{1}{u+x} \, du.$$ Now we scale the first integrand with the transformation $$v = u/(1-\lambda), \quad du = (1-\lambda) dv,$$ giving $$\Delta(x,\lambda,\delta) = \int_{v=0}^\delta \frac{1-\lambda}{(1-\lambda)v + x} \, dv - \int_{u=0}^\delta \frac{1-\lambda}{u+x} \, du = (1-\lambda) \int_{t=0}^\delta \left(\frac{1}{(1-\lambda)t + x} - \frac{1}{t+x} \right) \, dt.$$ But since $1-\lambda > 0$, $\delta > 0$, and the integrand is clearly nonnegative since $$(1-\lambda) t \le t \\ \iff (1-\lambda)t + x \le t + x \\ \iff \frac{1}{(1-\lambda)t + x} \ge \frac{1}{t+x},$$ the claim is proved.

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This inequality is famous and an easy consequence of Mean Value Theorem. We have $a, b, p, q$ as positive and $1/p+1/q=1$. Let $\alpha=1/p,\beta=1/q$ and replace $a^p, b^{q} $ by $a, b$ to get the inequality in the form $$a^{\alpha} b^{\beta} \leq\alpha a+\beta b$$ where $\alpha +\beta=1$. There is equality if $a=b$ and hence let $a<b$ (for $a>b$ we can interchange the roles of $a, b$).

Consider $f(x) =x^{\beta}$ so that $f'(x) = \beta x^{-\alpha} $ and by mean value theorem we can see that $$b^{\beta} - a^{\beta} =(b-a) \beta c^{-\alpha} $$ for some $c\in(a, b) $. And since $c^{-\alpha} <a^{-\alpha} $, it follows that $$b^{\beta} - a^{\beta} <\beta(b-a) a^{-\alpha} $$ or $$a^{\alpha} b^{\beta} <\beta(b-a) +a=\alpha a+\beta b$$ Also note that if $\alpha, \beta$ are rational then there is no need of calculus and the inequality is an immediate consequence of AM-GM inequality.

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From Wikipedia's article on concave functions: "A differentiable function $f$ is concave on an interval if and only if its derivative function $f′$ is monotonically decreasing on that interval, that is, $f''<0$."

Note that $(\log(x))''=(1/x)'=-1/x^{2}<0$, so $\log$ is concave, and your inequality follows.

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    $\begingroup$ I've seen that already, but then again, no proof of that is exhibited and everywhere I've looked where they prove this, they assume a different definition of concavity (which I'm not sure if it's equivalent to the one I'm using) See for example: proofwiki.org/wiki/… $\endgroup$ – Guido A. Jul 14 '17 at 19:50
  • $\begingroup$ They are equivalent as can be seen here for convex functions: proofwiki.org/wiki/… $\endgroup$ – ervx Jul 14 '17 at 20:05
  • $\begingroup$ It's not necessary for the second derivative to be negative. It could have zeros. Consider a linear function: this is certainly concave, but it's second derivative vanishes everywhere. The statement ervx provides is true if we replace concavity with the term "strict concavity." It's quite meaningful to distinguish between the two. $\endgroup$ – fourierwho Jul 15 '17 at 16:25

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