Solution. Statement. If \(p,q>1\) with \(\frac 1p+\frac 1q=1\), then for any complex \(x_1,\dots ,x_n\) and \(y_1,\dots ,y_n\), \[\sum _{i=1}^{n}|x_iy_i|\leq \left [\sum _{i=1}^{n}|x_i|^p\right ]^{1/p} \left [\sum _{i=1}^{n}|y_i|^q\right ]^{1/q}.\]

Proof. If either factor on the right is zero then all the corresponding \(x_i\) or \(y_i\) vanish and both sides are zero, so assume both are positive. Normalise: \[x'_i=\frac {x_i}{\big [\sum _j |x_j|^p\big ]^{1/p}},\qquad y'_i=\frac {y_i}{\big [\sum _j |y_j|^q\big ]^{1/q}},\] so that \(\sum _i|x'_i|^p=1\) and \(\sum _i|y'_i|^q=1\). Young’s inequality applied to each pair gives \[|x'_iy'_i|\leq \frac {|x'_i|^p}{p}+\frac {|y'_i|^q}{q}.\] Summing over \(i\), \[\sum _{i=1}^{n}|x'_iy'_i|\leq \frac {1}{p}+\frac {1}{q}=1 ,\] and multiplying back through by the two normalising factors gives the result.

The finite case is the infinite one with all but \(n\) terms zero, so nothing new is needed — but the normalisation trick is worth remembering on its own.

Solution. Three things must be checked, and one of them is the whole content.

\(l_p\) is closed under addition

If \(x,y\in l_p\) then by Minkowski’s inequality \[\left [\sum _i |x_i+y_i|^p\right ]^{1/p}\leq \left [\sum _i |x_i|^p\right ]^{1/p}+\left [\sum _i |y_i|^p\right ]^{1/p}<\infty ,\] so \(x+y\in l_p\). Closure under scalar multiplication is immediate, since \(\sum |\alpha x_i|^p=|\alpha |^p\sum |x_i|^p\). So \(l_p\) is a linear space.

\(||\cdot ||_p\) is a norm

Non-negativity is clear. If \(||x||_p=0\) then \(\sum |x_i|^p=0\), and a sum of non-negative terms vanishes only if every term does, so \(x=0\). Homogeneity is the computation just made, taking \(p\)-th roots: \(||\alpha x||_p=|\alpha |\,||x||_p\). The triangle inequality is exactly Minkowski’s inequality.

That is the point of the section: the triangle inequality for \(||\cdot ||_p\) is not obvious and is not proved directly — it is Minkowski’s inequality, which needed Hölder’s, which needed Young’s.

Solution. Taking \(p=q=2\) is legitimate because \(\frac 12+\frac 12=1\). Both exponents \(\frac 1p\) and \(\frac 1q\) become \(\frac 12\), so both brackets become square roots and Hölder’s inequality reads \[\sum _{i=1}^{n}|x_iy_i|\leq \sqrt {\sum _{i=1}^{n}|x_i|^2}\ \sqrt {\sum _{i=1}^{n}|y_i|^2}.\]

In \(\mathbb {R}^n\) with the usual inner product this says \[|(x,y)|\leq ||x||\,||y|| .\] Geometrically it is what makes the angle between two vectors well defined: dividing through gives \[-1\leq \frac {(x,y)}{||x||\,||y||}\leq 1 ,\] so the quotient is the cosine of a genuine angle. Without the inequality there would be no reason for that ratio to lie in \([-1,1]\) at all.

Solution. The sequence lies in \(l_p\) exactly when \(\sum _{n=1}^{\infty }n^{-p}\) converges, which by the \(p\)-series test happens precisely for \(p>1\). So \(x\in l_p\) for every \(p>1\) and \(x\notin l_1\).

For the inclusion, suppose \(x\in l_p\) and \(p<r\). Then \(|x_n|^p\rightarrow 0\), so \(|x_n|\leq 1\) for all large \(n\), and for such \(n\) \[|x_n|^r\leq |x_n|^p\] because raising a number in \([0,1]\) to a larger power makes it smaller. Hence \(\sum |x_n|^r\) converges by comparison and \(x\in l_r\); that is, \(l_p\subseteq l_r\).

The inclusion is strict, and the sequence above shows it: taking \(p=1\) and \(r=2\), the sequence \(\left (\frac 1n\right )\) lies in \(l_2\) but not in \(l_1\).

Solution. Fix \(\beta \geq 0\) and consider \[\varphi (\alpha )=\frac {\alpha ^p}{p}+\frac {\beta ^q}{q}-\alpha \beta ,\qquad \alpha \geq 0 .\] Then \(\varphi '(\alpha )=\alpha ^{p-1}-\beta \), which is negative for \(\alpha ^{p-1}<\beta \) and positive after, so \(\varphi \) has a unique minimum at the \(\alpha \) with \(\alpha ^{p-1}=\beta \).

At that point, using \(\frac 1p+\frac 1q=1\) in the form \(q(p-1)=p\), \[\alpha \beta =\alpha \cdot \alpha ^{p-1}=\alpha ^{p},\qquad \beta ^q=\alpha ^{q(p-1)}=\alpha ^{p},\] so \[\varphi =\frac {\alpha ^p}{p}+\frac {\alpha ^p}{q}-\alpha ^p =\alpha ^p\left (\frac 1p+\frac 1q-1\right )=0 .\] So the minimum value of \(\varphi \) is \(0\), attained only there: the inequality holds always, with equality exactly when \(\alpha ^{p-1}=\beta \), that is when \(\alpha ^p=\beta ^q\).

This is worth knowing because it identifies when Hölder’s inequality is sharp — precisely when \(|x_i|^p\) and \(|y_i|^q\) are proportional.

Solution. Take \(p=\frac 12\) and work in two coordinates, with \[x=(1,0),\qquad y=(0,1).\] Then \[||x||_{1/2}=\big [1^{1/2}\big ]^{2}=1,\qquad ||y||_{1/2}=1 ,\] while \(x+y=(1,1)\) gives \[||x+y||_{1/2}=\big [1^{1/2}+1^{1/2}\big ]^{2}=2^{2}=4 .\] So \(||x+y||_{1/2}=4>2=||x||_{1/2}+||y||_{1/2}\), and the triangle inequality fails.

The reason is structural rather than accidental: Minkowski’s inequality is proved from Hölder’s, and Hölder’s requires a conjugate exponent \(q\) with \(\frac 1p+\frac 1q=1\). For \(0<p<1\) that forces \(q<0\), and the argument collapses.