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$$ ||f+g||_p \le ||f||_p + ||g||_p \quad\text{for $1\le p \le \infty$}$$ I have used Minkowski inequality for a long time without really knowing why it should be true. I kind of take it for grant every time I am dealing with $l^p$ or $L^p$ spaces. Years have passed by and now I am taking a better look at it again. It's not like I don't understand its proofs or anything, the standard proof goes like this:

Consider the case $1<p<\infty$, it is not hard to see that $$\left|\frac 12f+\frac 12 g\right|^p\le \left|\frac 12|f|+\frac 12|g|\right|^p$$ for any $f,g\in L^p$. Since $x \mapsto x^p$ is convex, we have $$ \left(\frac 12|f|+\frac 12|g|\right)^p \le \frac 12|f|^p+\frac 12|g|^p. $$ Substituting $f,g$ in places of $\frac 12f,\frac12 g$ yields $$ \left|f+g\right|^p \le 2^{p-1}\left(|f|^p+|g|^p \right) $$ so $f+g\in L^p$.

Observe that for $q$ satisfying $\frac 1q + \frac 1p=1$, $$\begin{align} |f+g|^p &\le |f+g|^{p-1}(|f|+|g|) \\ \int|f+g|^p &\le \int|f||f+g|^{p-1} + \int|g||f+g|^{p-1} \\ &= ||f\cdot (f+g)^p ||_1 + ||g\cdot (f+g)^p ||_1 \\ ||f+g||^p_p&\le ||f||_p ||(f+g)^{p-1}||_q + ||g||_p ||(f+g)^{p-1}||_q \end{align}$$ by Holder inequality.

Now, since $(p-1)q=p$ we have $$\begin{align} ||(f+g)^{p-1}||_q &= \left(\int |f+g|^{(p-1)q}\right)^{\frac 1q} \\ & =\left(\int |f+g|^p\right)^{\frac 1q} \\ &= ||f+g||_p^{p/q}. \end{align}$$ By cancellation on both side of $$||f+g||^p_p\le ||f||_p ||(f+g)^{p-1}||_q + ||g||_p ||(f+g)^{p-1}||_q$$ and noting that $p-p/q=1$, we have $$ ||f+g||_p \le ||f||_p + ||g||_p $$ as required.

Yes, each step is clear and not hard to understand. However, it seems magical to me at several steps and I cannot intuitively follow the logical flow of the proof. The proof of Holder inequality via Jensen inequality seems quite natural to me, unlike this one. I'll try to be more specific:

My questions

  1. Why should I expect that the splitting $|f+g|^p$ into $|f+g|^{p-1}(|f|+|g|)$ should be fruitful?

  2. We utilized the fact that $(p-1)q=p$ and $p-p/q=1$. While it is not hard to verify them algebraically, is there an intuitive reason to anticipate them beforehand?

  3. What is the intuitive/geometric relation between the conjugate pair $p,q$ beside from their algebraic relation $\frac 1q+\frac 1p =1$? I know that $(l^p)^*=l^q$ but I am not sure how it might help.

In general, I just want a better grasp at Minkowski inequality. Any contribution would be really appreciated even if it does not address all of my question.

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  • $\begingroup$ This doesn't really answer your question, but there's a direct proof of Minkowski's inequality by use of convexity on pages 3-4 of this document: andromeda.rutgers.edu/~loftin/ra1/lp.pdf $\endgroup$ Commented Sep 26, 2016 at 10:33
  • $\begingroup$ @OlivierMoschetta It might be interesting nonetheless, thanks! $\endgroup$
    – BigbearZzz
    Commented Sep 26, 2016 at 10:36

1 Answer 1

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The significance of $\frac{1}{p}+\frac{1}{q}=1$ is that this is related to an interpolation. For the exponential, $$ \exp\left( \frac{1}{p}u + \frac{1}{q}v\right) \le \frac{1}{p}e^{u}+\frac{1}{q}e^{v}. $$ Letting $u=\ln(x^p)$, $v=\ln(y^q)$ for $x > 0$ and $y > 0$ gives $$ xy \le \frac{1}{p}x^{p}+\frac{1}{q}y^q. $$ Therefore, if $f\in L^p$ and $g\in L^q$, then $fg\in L^1$ with $$ \int |fg|d\mu \le \frac{1}{p}\|f\|_p^p+\frac{1}{q}\|g\|_q^q \\ \int \frac{|fg|}{\|f\|_p\|g\|_q}d\mu \le \frac{1}{p}+\frac{1}{q}=1 \\ \int |fg|d\mu \le \|f\|_p\|g\|_q. $$ Because of this interpolation, $$ |f+g| = |f+g|^\frac{1}{p}|f+g|^\frac{1}{q}=|f+g|^{\frac{1}{p}}|f+g|^{1-\frac{1}{p}} \\ |f+g|^{p} = |f+g||f+g|^{p-1} \le |f+g|^{p-1}|f|+|f+g|^{p-1}|g| % \\ % (p-1)q = (p-1)\frac{1}{1-1/p}=p $$ That's the reason for the split $|f+g|^{p}=|f+g|^{p-1}|f+g|$, and why you might expect it to yield something useful.

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  • $\begingroup$ Thank you for your answer! I understand the first half of your post about Holder inequality but, still, I fail to understand your last sentence. It doesn't help me understand why we should expect $||(f+g)^{p-1}||_q= ||f+g||_p^{p/q}$ to cancel out $||f+g||_p^p$ to $||f+g||_p$, or perhaps that was not your intention? $\endgroup$
    – BigbearZzz
    Commented Sep 28, 2016 at 21:24
  • $\begingroup$ Oh... I think I now kind of see what you meant there. I'll have to give it some thought before I can truly say that I completely understand it. $\endgroup$
    – BigbearZzz
    Commented Sep 28, 2016 at 21:29

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