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While investigating counterexamples for certain conjectures in matrix analysis, I stumbled across the following calculus problem.

I need to show (rigorously) that for $1<p<2$, $$\left\lVert\begin{pmatrix}\sqrt 2 + 1 \\ \sqrt 2 - 1 \end{pmatrix}\right\rVert^2_p < \left\lVert\begin{pmatrix}2 \\ 0 \end{pmatrix}\right\rVert^2_p + \left\lVert\begin{pmatrix}1 \\ 1 \end{pmatrix}\right\rVert^2_p $$ where $\left\lVert \begin{pmatrix} x \\ y \end{pmatrix} \right\rVert_p =(|x|^p + |y|^p)^{\frac{1}{p}}.$

Here is the graph of LHS - RHS.

enter image description here


I could proceed in the following direction with some simple and possibly helpful observations.

  • Take $a = \sqrt 2 + 1$, then $\sqrt 2 - 1 = \dfrac{1}{a}$.
  • Using some norm inequalities the above problem can be "reduced" to proving the following inequality. $$a^p + a^{-p} < (2 \sqrt 2)^p \left[\dfrac{1}{2} + \dfrac{1}{2^p} \right] $$ The following is the graph of LHS - RHS of the above function.

enter image description here

  • In both the equations, as you can see in the graph, at $p = 1$ and $2$, the LHS = RHS. So I have tried to show that the difference as a function of $p$ is convex in the interval $(1,2)$. But that involves a more complicated equation than the one we started with.

  • I am also trying a slight variant of the above argument. That is, to show that the LHS $\neq$ RHS as a function of $p$ in the interval $(1,2)$, and LHS - RHS takes a negative value for some $p$ in that interval.

I have been unsuccesful in all my attempts so far, and any help would be highly appreciated.

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  • $\begingroup$ There was a small mistake in the original question. I have corrected it now. I had missed the squares that appear in each term of the first inequality. $\endgroup$ Jan 4 at 10:52
  • $\begingroup$ GIven the complexity of the most recent answer, I think a computer assisted proof with estimates on e.g. Lipschitz constants would be easier to swallow $\endgroup$ Jan 8 at 6:40
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    $\begingroup$ Any proof without complicated calculations? $\endgroup$
    – River Li
    Sep 23 at 3:17

2 Answers 2

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Sketch of a proof:

For convenience, denote $a = \sqrt2 - 1$.

We need to prove that $$a^{-2}(1 + a^{2p})^{2/p} < 4 + 2^{2/p}. \tag{1}$$

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Fact 1: Let $u > 0$ and $v \in [0, 1]$. Then $u^v \ge \frac{u}{u + v - uv}$.

Fact 2: Let $0 \le r \le 1$ and $0 \le x < 1$. Then $(1 - x)^r \le 1 - rx - \frac12 (r - r^2)x^2$.

Fact 3: Let $0\le s \le 1$. Then $2^s \ge \frac{1}{24}(s^4 - 2s^3 + 11s^2 + 14s + 24)$.

$\phantom{2}$

Using Fact 1, letting $u = 1 + a^{2p}$ and $v = 2 - 2/p$, we have $$(1 + a^{2p})^{2 - 2/p} \ge \frac{1 + a^{2p}}{1 + a^{2p} + 2 - 2/p - (1 + a^{2p})(2 - 2/p)}$$ which results in $$(1 + a^{2p})^{2/p} = \frac{(1 + a^{2p})^2}{(1 + a^{2p})^{2 - 2/p}} \le (1 + a^{2p})^2(2/p - 1) + (1 + a^{2p})(2 - 2/p). \tag{2}$$

Using Fact 2, letting $r = p - 1$ and $x = 1 - a$, we have $$a^{p - 1} \le 1 - (p - 1)(1 - a) - \frac12 (p - 1)(2 - p)(1 - a)^2$$ which results in $$a^{2p} = a^2(a^{p - 1})^2 \le a^2A^2 \tag{3}$$ where $$A = 1 - (p - 1)(1 - a) - \frac12(p - 1)(2 - p)(1 - a)^2.$$

Using (2) and (3), we have $$(1 + a^{2p})^{2/p} \le (1 + a^2A^2)^2(2/p - 1) + (1 + a^2A^2)(2 - 2/p). \tag{4}$$

Using Fact 3, letting $s = 2/p - 1$, we have $$2^{2/p - 1} \ge \frac{(2/p - 1)^4 - 2(2/p - 1)^3 + 11(2/p - 1)^2 + 14(2/p - 1) + 24}{24}$$ which results in $$2^{2/p} = 2\cdot 2^{2/p - 1} \ge \frac{6p^4 - 9p^3 + 23p^2 - 12p + 4}{3p^4}. \tag{5}$$

Using (4) and (5), it suffices to prove that \begin{align*} &a^{-2}\Big[(1 + a^2A^2)^2(2/p - 1) + (1 + a^2A^2)(2 - 2/p)\Big]\\ <\,& 4 + \frac{6p^4 - 9p^3 + 23p^2 - 12p + 4}{3p^4} \tag{6} \end{align*} or (after clearing the denominators) \begin{align*} &(-10089 + 7134\sqrt{2})p^{10} + (-95190\sqrt{2} + 134619)p^9 + (533862\sqrt{2} - 754995)p^8\\ &\qquad + (-1628238\sqrt{2} + 2302677)p^7 + (2911752\sqrt{2} - 4117842)p^6\\ &\qquad + (-3050004\sqrt{2} + 4313364)p^5 + (1733064\sqrt{2} - 2450934)p^4\\ &\qquad + (-412656\sqrt{2} + 583596)p^3 - 6p^2 + 3p - 2 > 0 \end{align*} or (letting $p = 1 + \frac{1}{1 + t}$ for $t > 0$, clearing the denominators) \begin{align*} &(-276\sqrt{2} + 391)t^{10} + (-120\sqrt{2} + 181)t^9 + (-2676\sqrt{2} + 3858)t^8\\ &\quad + (-27546\sqrt{2} + 39219)t^7 + (-24480\sqrt{2} + 35208)t^6 + (154908\sqrt{2} - 218202)t^5\\ &\quad + (372192\sqrt{2} - 525486)t^4 + (313200\sqrt{2} - 442344)t^3\\ &\quad + (125472\sqrt{2} - 177189)t^2 + (24576\sqrt{2} - 34691)t + 1920\sqrt{2} - 2708 > 0. \end{align*} Note that $\mathrm{LHS}$ is a polynomial with non-negative coefficients and positive constant term. So, the inequality is true.

We are done.

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We want to prove that for $1<p<2$, $$ \left((\sqrt{2}+1)^p + (\sqrt{2}-1)^p\right)^{\frac 1 p} < 2^{\frac 1 p} + 2. $$

Note that whenever $a, b \ge 0$ and $\gamma > 1$, we have $a^\gamma + b^\gamma < (a + b)^{\gamma}$. Indeed, for $t > 0$, $$t^{\gamma - 1} < (a + t)^{\gamma -1 }$$ and integrating from $0$ to $b$ gives the desired result.

Now put $a = \sqrt{2}+1, b=\sqrt{2}-1$, and $\gamma = p$. Then we have $$ \left((\sqrt{2}+1)^p + (\sqrt{2}-1)^p\right)^{\frac 1 p} < 2\sqrt{2} < 2\cdot 2^{\frac 1p} < 2^{\frac 1 p} + 2, $$ since $1 < p < 2$.

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    $\begingroup$ I am very sorry. There was a mistake in the questioin. I am updating it immediately. There is a square on each term that I missed. Very sorry for this. $\endgroup$ Jan 4 at 10:48

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