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The standard proof of Minkowski's inequality in $L^p$ space using Hölder's inequality seems to be pretty unmotivated (see here: http://en.wikipedia.org/wiki/Minkowski%27s_inequality). Breaking up that factor into two parts, and applying Hölder's to each of them separately makes no geometrical intuitive sense to me. Rather, it would appear that the following proof (which I came up with, though is highly probable isn't in fact mine), appears much more motivated:

We need to show that $(\int_{X} |f+g|^p d\mu)^{\frac{1}{p}} \le (\int_{X} |f|^p d\mu)^{\frac{1}{p}} +(\int_{X} |g|^p d\mu)^{\frac{1}{p}}$.

Let $A=(\int_{X} |f|^p d\mu)^{\frac{1}{p}}$, and $B=(\int_{X} |g|^p)^{\frac{1}{p}} d\mu$.

Then setting $f_1=\frac{f}{A}$, and $g_1=\frac{g}{B}$, we get the equivalent inequalities

$\int_{X} (Af_1+Bg_1)^p d\mu \le (A+B)^p$

$\int_{X} (\frac{Af_1+Bg_1}{A+B})^p d\mu \le 1$

$\int_{X} (\frac{Af_1+Bg_1}{A+B})^p d\mu \le \int_{X} \frac{A|f_1|^p+B|g_1|^p}{A+B} d\mu$

Which follows immediately from the convexity of $x^p$.

Now, there are a few reasons that I find this more appealing. First, each of the steps is motivated. The first step is an attempt at homogenization, then after rewriting it in the form with the R.H.S.=1, it becomes very apparent how to finish it off using the convexity of $x^p$.

Second, a similar line of reasoning is used to prove Hölder's inequality, and is used in the following quite elegant proof of Cauchy-Schwarz for finite sequences:

To prove $(\sum a_i^2)(\sum b_i^2) \ge (\sum a_ib_i)^2$, we first homogenize by setting $\sum a_i^2=\sum b_i^2=1$. Then after square rooting, we see it is equivalent to the inequality $\frac{\sum a_i^2 + \sum b_i^2}{2} \ge \sum a_ib_i$, which is a direct result of AM-GM. Note that the use of homogenization is specifically used so that we can replace the L.H.S. with something much stronger while keeping the R.H.S. the same, almost as if by magic (which was essentially what my proof amounted to).

Third, the equality case almost becomes trivial to see, since only one inequality was applied in the entire process, and that was at the very end of the proof. Further, it's very easy to see geometrically the idea of convexity being used here, so the equality case also seems natural.

My first question is, whether there are merits to the standard proof involving Holder's inequality that I seem to be missing which either make it more general, or make it of more interest than the proof I presented.

My second question is to whether the briefly summarized proof involving dual spaces presented on the same Wikipedia page is in actuality more general than my proof. That is, whether there are spaces to which we simply cannot appeal to a convexity argument, and instead have to resort to the supremum proof. Or whether my proof can somehow be modified to prove the result in a more general class of spaces which encompasses all of those which the supremum argument works for.

Cheers,

Rofler

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    I think the motivation is largely historical. Hardy-Littlewood-Polya gave the proof (due to F. Riesz) via Holder (indeed, their discussion of issues of convexity is a bit messy). It is a bit later that the power of convexity was recognized. Incidentally, the same proof as your was given by Hormander in his book "Notions of Convexity", on page 12, for the summation version of Minkowski's inequality. (Immediately after which he assigns the proof via Holder as an exercise.)2011-06-27
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    I agree with @Willie's assessment. Before Byron Schmuland does so, I'd like to direct your attention to the marvellous book *[The Cauchy-Schwarz Master Class](http://books.google.ch/books?id=7GDyRMrlgDsC)* by J. Michael Steele. Its ninth chapter is entirely devoted to Hölder's inequality and contains a good motivation of the "usual" approach at its beginning. There are also many enlightening historical discussions. Highly recommended.2011-06-27
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    I had a feeling that the motivation was historical, thanks for the comments guys. I still want to find an answer to my second question though, which might be a bit vague... it could be understood as asking which spaces possess such a notion of "convexity" which allows a similar proof to go through.2011-06-27
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    From what I remember, the classical proof of Holder's inequality relies on arguments similar to yours. So I guess most books prove Minkowski's inequality in the usual manner in order to demonstrate how Holder's inequality can be used to produce other inequalities.2011-07-01
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    The "usual" proof has the advantage that you only need the convexity of $e^x$ which is quite easy using only the exponential series (at least in Germany, we introduce $\exp$ much earlier than the first and second derivative). Is it as easy to prove convexity of $x^p$ without calculating the second derivative?2016-01-07

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