Proving Minkowski's inequality with homogenization

Solution 1:

The excellent book The Cauchy-Schwarz Master Class has already been mentioned in the comments by Theo.

Since I cannot stand open questions where a lot of people know the answer to I'll just summarize what is in chapter 9 of the referred book.

You're right that you can prove it the way you do but usually when people take a course in measure and integration theory first Hölder's inequality is proven and then Minkowski's inequality. In that way it can be instructive to use Hölder's inequality to prove Minkowksi's inequality.

There is another advantage to this approach. We can quite easily deduce from the proof when equality arises. To see this let me quickly recall how the proof goes (this can be found in the book by Steele).

First write by using the triangle inequality

$$\sum_{k = 1}^n |x_k + y_k|^p \leq \sum_{k = 1}^n |x_k||x_k + y_k|^{p - 1} + \sum_{k = 1}^n |x_k||x_k + y_k|^{p - 1}.$$

So now we can assume $p > 1$ otherwise we are done. We can now apply Hölder to both of the terms on the right hand side so we find

$$\sum_{k = 1}^n |x_k||x_k + y_k|^{p - 1} \leq \left (\sum_{k = 1} |x_k|^p \right )^{1/p} \left (\sum_{k = 1}^n |x_k + y_k|^{p} \right )^{(p - 1)/p}$$

and

$$\sum_{k = 1}^n |y_k||x_k + y_k|^{p - 1} \leq \left (\sum_{k = 1} |y_k|^p \right )^{1/p} \left (\sum_{k = 1}^n |x_k + y_k|^{p} \right )^{(p - 1)/p}$$

Now we can assume that the left hand side of the first inequality is non-zero so we can divide by $\displaystyle \left (\sum_{k = 1}^n |x_k + y_k|^{p} \right )^{(p - 1)/p}$ to obtain the proof.

Fine. So now if we would have equality in Minkowski's inequality the first inequality written here would also be an equality. This implies $|x_k + y_k| = |x_k| + |y_k|$ for all $1 \leq k \leq n$. Thinking for a bit we can conclude that $x_k$ and $y_k$ must be of the same sign for all $k$. Actually, there is no problem to assume $x_k, y_k \geq 0$ because we can factor the - out and it gets lost in the absolute value.

But equality in Minkowksi's inequality also means that we have equality in the two lines where Hölder's inequality is used. Now you can recall what it means to have equality in Hölder's inequality. We have that there exists $\lambda, \lambda' \geq 1$ such that $$\lambda |x_k|^p = (|x_k + y_k|^{p - 1})^q = |x_k + y_k|^p \text{ and } \lambda' |y_k|^p = (|x_k + y_k|^{p - 1})^q = |x_k + y_k|^p.$$

Dividing both equalities we get that $\frac{\lambda}{\lambda'} |x_k|^p = |y_k|^p$. So this proof can be easily backtraced.

But again, credit must be given where credit is due: This is just what is written in Steele in my own words. I don't think I'm plagiarizing because this method can be considered to be common knowledge.

So check out the book in the library or buy it, it is quite cheap for a math book and it contains fun exercises.