KindiakMath

Real-Analytic Inequalities

Fix p > 1, q > 1 such that \displaystyle \frac 1p + \frac 1q = 1.

Problem 1. Verify that p+q = pq.

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Solution. By algebruh,

\displaystyle 1 = \frac 1p + \frac 1q = \frac{p+q}{pq}\quad \Rightarrow \quad p+q = pq.

Problem 2. Show that for x \geq 0, \displaystyle x \leq \frac{x^p}{p} + \frac 1q with equality if and only if x = 1.

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Solution. Define the function f(x) := x^p/p + 1/q - x. By observation,

\displaystyle f(1) = \frac 1p + \frac 1q - 1 = 0.

Furthermore,

f'(x) = x^{p-1} - 1 \quad \Rightarrow \quad f''(x) = (p-1)x^{p-2} > 0.

By the second derivative test, solving for f'(x) = 0 yields a global minimum:

\displaystyle x^{p-1} - 1 = 0 \quad \Rightarrow \quad x = 1.

Therefore, f(x) \geq f(1) = 0 for any x \geq 0, as required.

Problem 3. Use Problem 2 to prove Young’s inequality: for any a \geq 0, b \geq 0,

\displaystyle ab \leq \frac{a^p}{p} + \frac{ b^q }{q}

with equality if and only if a^p = b^q.

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Solution. Setting x = a/b^{q-1},

\begin{aligned} \frac a{b^{q-1}} &\leq \frac{\left( \frac a{b^{q-1}} \right)^p}{p} + \frac 1q \\ \frac a{b^{q-1}} \cdot b^q &\leq \frac{1}{p} \cdot \frac{a^p}{b^{p(q-1)}} \cdot b^q + \frac {b^q}q \\ ab &\leq \frac{a^p}{p} \cdot b^{p+q-pq} + \frac {b^q}q \\ ab &\leq \frac{a^p}{p} + \frac {b^q}q . \end{aligned}

Furthermore, since p(q-1) = pq - p = q, equality holds if and only if

\displaystyle x = 1 \quad \iff \quad a = b^{q-1} \quad \iff \quad a^p = b^{p(q-1)} \quad \iff \quad a^p = b^q.

Problem 4. Use Young’s inequality to prove Hölder’s inequality for two-dimensional vectors: given non-negative numbers a_1,a_2,b_1,b_2,

\displaystyle a_1 b_1 + a_2 b_2 \leq \left( a_1^p + a_2^p \right)^{1/p} \left( b_1^q + b_2^q \right)^{1/q}.

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Solution. If \left( a_1^p + a_2^p \right)^{1/p} = 0, then 0 \leq a_1^p + a_2^p = 0 implies a_1 = a_2 = 0, and the inequality holds trivially. Without loss of generality, suppose the right-hand side is non-zero. It suffices to prove that

\displaystyle \frac{ a_1 b_1 + a_2 b_2 }{ \left( a_1^p + a_2^p \right)^{1/p} \left( b_1^q + b_2^q \right)^{1/q} } \leq 1.

By one application of Young’s inequality,

\begin{aligned} \frac{ a_1 }{ \left( a_1^p + a_2^p \right)^{1/p} } \cdot \frac{ b_1 }{\left( b_1^q + b_2^q \right)^{1/q} } &\leq \frac{a_1^p}{ p \cdot \left( a_1^p + a_2^p \right)^{p/p} } + \frac{b_1^q}{ q \cdot \left( b_1^q + b_2^q \right)^{q/q} } \\ &\leq \frac{a_1^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_1^q}{ q \cdot \left( b_1^q + b_2^q \right) }. \end{aligned}

By a second application of Young’s inequality,

\begin{aligned} \frac{ a_2 }{ \left( a_1^p + a_2^p \right)^{1/p} } \cdot \frac{ b_2 }{\left( b_1^q + b_2^q \right)^{1/q} } &\leq \frac{a_2^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_2^q}{ q \cdot \left( b_1^q + b_2^q \right) }. \end{aligned}

Summing the inequalities,

\begin{aligned} \frac{ a_1 b_1 + a_2 b_2 }{ \left( a_1^p + a_2^p \right)^{1/p} \left( b_1^q + b_2^q \right)^{1/q} } &\leq \frac{ a_1 }{ \left( a_1^p + a_2^p \right)^{1/p} } \cdot \frac{ b_1 }{\left( b_1^q + b_2^q \right)^{1/q} } + \frac{ a_2 }{ \left( a_1^p + a_2^p \right)^{1/p} } \cdot \frac{ b_2 }{\left( b_1^q + b_2^q \right)^{1/q} } \\ &= \frac{a_1^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_1^q}{ q \cdot \left( b_1^q + b_2^q \right) } + \frac{a_2^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_2^q}{ q \cdot \left( b_1^q + b_2^q \right) } \\ &= \frac{a_1^p + a_2^p}{ p \cdot \left( a_1^p + a_2^p \right) } + \frac{b_1^q + b_2^q}{ q \cdot \left( b_1^q + b_2^q \right) } = \frac 1p + \frac 1q = 1.\end{aligned}

Problem 5. Use Hölder’s inequality to prove Minkowski’s inequality for two-dimensional vectors: given real numbers a_1,a_2,b_1,b_2,

\displaystyle (|a_1 + b_1|^p + |a_2 + b_2|^p)^{1/p} \leq (|a_1|^p + |a_2|^p )^{1/p} + (|b_1|^p + |b_2|^p )^{1/p}.

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Solution. By the vanilla triangle inequality,

\begin{aligned} &|a_1 + b_1|^p + |a_2 + b_2|^p \\ &= |a_1 + b_1| |a_1 + b_1|^{p-1} + |a_2 + b_2| |a_2 + b_2|^{p-1} \\ &\leq (|a_1| + |b_1|) |a_1 + b_1|^{p-1} +(|a_2| + |b_2|) (a_2 + b_2)^{p-1} \\ &= |a_1| |a_1 + b_1|^{p-1} + |a_2| |a_2 + b_2|^{p-1} + |b_1| |a_1 + b_1|^{p-1} + |b_2| |a_2 + b_2|^{p-1}. \end{aligned}

By Hölder’s inequality,

\begin{aligned} & |a_1| |a_1 + b_1|^{p-1} + |a_2| |a_2 + b_2|^{p-1} \\ &\leq (|a_1|^p + |a_2|^p)^{1/p} \cdot (|a_1 + b_1|^{q(p-1) }+ |a_2 + b_2|^{q(p-1)})^{1/q} \\ &\leq (|a_1|^p + |a_2|^p)^{1/p} \cdot (|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}, \end{aligned}

so that

\displaystyle \frac{|a_1| |a_1 + b_1|^{p-1} + |a_2| |a_2 + b_2|^{p-1} }{(|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}} \leq (|a_1|^p + |a_2|^p)^{1/p}.

Similarly,

\displaystyle \frac{|b_1| |a_1 + b_1|^{p-1} + |b_2| |a_2 + b_2|^{p-1} }{(|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}} \leq (|b_1|^p + |b_2|^p)^{1/p}.

Combining the displays,

\displaystyle \frac{ |a_1 + b_1|^p + |a_2 + b_2|^p} {(|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}} \leq (|a_1|^p + |a_2|^p)^{1/p} + (|b_1|^p + |b_2|^p)^{1/p}.

The result follows by the observation

\displaystyle \frac{ |a_1 + b_1|^p + |a_2 + b_2|^p} {(|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/q}} = (|a_1 + b_1|^{ p }+ |a_2 + b_2|^{ p })^{1/p}.

Remark 1. Denoting

\left\| \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} \right\|_p := (|a_1|^p + |a_2|^p )^{1/p}

and defining \mathbf a := \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} , Hölder’s inequality reduces to

\| \mathbf a \cdot \mathbf b \|_1 \leq \| \mathbf a \|_p \| \mathbf b \|_q,

and Minkowski’s inequality reduces to

\| \mathbf a + \mathbf b \|_p \leq \| \mathbf a \|_p + \| \mathbf b \|_p.

Setting p = q = 2 reduces to the usual Cauchy-Schwarz inequality and triangle inequality for two-dimensional vectors. Furthermore, these results hold for n-dimensional vectors, and even “infinite-dimensional” (sufficiently nice) functions.

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Published by

joelkindiak

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