Proof. For \(\alpha \) or \(\beta =0\), \((1)\) is trivially true.

So, let \(\alpha , \beta >0\), consider the curve \(\hspace {0.2cm}\displaystyle {x=t^{p-1}}\hspace {0.2cm}\), we can write this as \(\hspace {0.2cm}\displaystyle {t=x^{q-1}}\hspace {0.2cm}\)

Now \(\hspace {0.3cm}\displaystyle {\int ^{\alpha }_0 t^{p-1}\,dt=\frac {t^p}{p}\Bigg |^{\alpha }_0=\frac {\alpha ^p}{p}}\hspace {0.3cm}\) and \(\hspace {0.3cm}\displaystyle {\int ^{\beta }_0 x^{q-1}\,dx=\frac {\beta ^q}{q}}\)

Now, consider

\[\alpha \beta \leq \frac {\alpha ^p}{p}+\frac {\beta ^q}{q}\] □

Proof.

Set \(\hspace {0.3cm} \displaystyle {x'_i=\frac {x_i}{\Bigg [\sum \limits ^{\infty }_{i=1}|x_i|^p\Bigg ]^{\frac {1}{p}}}}\hspace {0.3cm}\) and \(\hspace {0.3cm} \displaystyle {y'_i=\frac {y_i}{\Bigg [\sum \limits ^{\infty }_{i=1}|y_i|^q\Bigg ]^{\frac {1}{q}}}}\)

Now \(\hspace {0.5cm}\displaystyle {|x'_i|=\frac {|x_i|}{\Bigg [\sum \limits ^{\infty }_{i=1}|x_i|^p\Bigg ]^{\frac {1}{p}}}\hspace {1cm},\hspace {1cm} |y'_i|=\frac {|y_i|}{\Bigg [\sum \limits ^{\infty }_{i=1}|y_i|^q\Bigg ]^{\frac {1}{q}}}}\)

\[\implies |x'_i|^p=\frac {|x_i|^p}{\sum \limits ^{\infty }_{i=1}|x_i|^p}\hspace {1cm},\hspace {1cm} |y'_i|^q =\frac {|y_i|^q}{\sum \limits ^{\infty }_{i=1}|y_i|^q}\]
\[\implies \sum ^{\infty }_{i=1}|x'_i|^p=\frac {\sum \limits ^{\infty }_{i=1}|x_i|^p}{\sum \limits ^{\infty }_{i=1}|x_i|^p}=1\hspace {1cm},\hspace {1cm}\sum ^{\infty }_{i=1}|y'_i|^q=\frac {\sum \limits ^{\infty }_{i=1}|y_i|^q}{\sum \limits ^{\infty }_{i=1}|y_i|^q}=1\] Now, using the inequality \((1)\) in \((2)\), we have for any \(i\) \[|x'_i||y'_i|=|x'_iy'_i|\leq \frac {|x'_i|^p}{p}+\frac {|y'_i|^q}{q}\] Therefore, \begin {align*} \sum _{i=1}^{\infty }|x'_iy'_i| &\leq \frac {\sum \limits ^{\infty }_{i=1}|x'_i|^p}{p}+\frac {\sum \limits ^{\infty }_{i=1}|y'_i|^q}{q} =\frac {1}{p}+\frac {1}{q} =1 \end {align*}

\[\implies \sum ^{\infty }_{i=1}|x'_iy'_i|\leq 1\implies \sum ^{\infty }_{i=1}|x'_i||y'_i|\leq 1\]
\[\implies \sum _{i=1}^{\infty }\Bigg \{\frac {|x_i|}{\Big [\sum \limits ^{\infty }_{i=1}|x_i|^p\Big ]^{\frac {1}{p}}}\Bigg \} \Bigg \{\frac {|y_i|}{\Big [\sum \limits ^{\infty }_{i=1}|y_i|^q\Big ]^{\frac {1}{q}}}\Bigg \} \leq 1\]
Hence we have, \[\sum ^{\infty }_{i=1}|x_i||y_i|=\sum ^{\infty }_{i=1}|x_iy_i|\leq \Bigg [\sum ^{\infty }_{i=1}|x_i|^p\Bigg ]^{\frac {1}{p}}\Bigg [\sum ^{\infty }_{i=1}|y_i|^q\Bigg ]^{\frac {1}{q}}.\] □

Proof. Take \(p=q=2\) in Hölder’s inequality, which is legitimate since \(\frac 12+\frac 12=1\). The exponents \(\frac 1p\) and \(\frac 1q\) both become \(\frac 12\), and the two bracketed sums become square roots, giving the stated form at once. □

Proof. If \(p=1\), this is obvious.

Let \(p>1\), write \(\hspace {0.3cm}\displaystyle {|x_i+y_i|^p=|x_i+y_i||x_i+y_i|^{p-1}}\). Thus, \begin {align*} |x_i+y_i|^p &=|x_i+y_i||x_i+y_i|^{p-1}\\ &\leq \Big [|x_i|+|y_i|\Big ]|x_i+y_i|^{p-1}\\ &=|x_i||x_i+y_i|^{p-1}+|y_i||x_i+y_i|^{p-1} \end {align*}

Now, consider \(\hspace {0.3cm}\displaystyle {\sum ^{\infty }_{i=1}|x_i||x_i+y_i|^{p-1}}\hspace {0.3cm},\) if \(q\) is the conjugate of \(p\) i.e \(\hspace {0.3cm}\dfrac {1}{p}+\dfrac {1}{q}=1\). Then \begin {align*} \sum ^{\infty }_{i=1}|x_i||x_i+y_i|^{p-1} &\leq \Bigg [\sum ^{\infty }_{i=1}|x_i|^p\Bigg ]^{\frac {1}{p}} \Bigg [\sum ^{\infty }_{i=1}\Big (|x_i+y_i|^{p-1}\Big )^q\Bigg ]^{\frac {1}{q}}\\ &=\Bigg [\sum ^{\infty }_{i=1}|x_i|^p\Bigg ]^{\frac {1}{p}}\Bigg [\sum ^{\infty }_{i=1}|x_i+y_i|^p\Bigg ]^{\frac {1}{q}} \end {align*}

Therefore, \[\sum ^{\infty }_{i=1}|x_i||x_i+y_i|^{p-1}\leq \Bigg [\sum ^{\infty }_{i=1}|x_i|^p\Bigg ]^{\frac {1}{p}}\Bigg [\sum ^{\infty }_{i=1}|x_i+y_i|^p\Bigg ]^{\frac {1}{q}}\qquad (*)\] Similarly, \[\sum ^{\infty }_{i=1}|y_i||x_i+y_i|^{p-1}\leq \Bigg [\sum ^{\infty }_{i=1}|y_i|^p\Bigg ]^{\frac {1}{p}}\Bigg [\sum ^{\infty }_{i=1}|x_i+y_i|^p\Bigg ]^{\frac {1}{q}}\dots (**)\] Hence using \((*)\) and \((**)\), we get \[\sum ^{\infty }_{i=1}|x_i+y_i|^p\leq \Bigg [\sum ^{\infty }_{i=1}|x_i|^p\Bigg ]^{\frac {1}{p}}\Bigg [\sum ^{\infty }_{i=1}|x_i+y_i|^p\Bigg ]^{\frac {1}{q}}+\Bigg [\sum ^{\infty }_{i=1}|y_i|^p\Bigg ]^{\frac {1}{p}}\Bigg [\sum ^{\infty }_{i=1}|x_i+y_i|^p\Bigg ]^{\frac {1}{q}}\]

divide both sides by \(\hspace {0.3cm}\displaystyle {\Bigg [\sum ^{\infty }_{i=1}|x_i+y_i|^p\Bigg ]^{\frac {1}{q}}}.\hspace {0.3cm}\) We get \[\Bigg [\sum ^{\infty }_{i=1}|x_i+y_i|^p\Bigg ]^{\frac {1}{p}}\leq \Bigg [\sum ^{\infty }_{i=1}|x_i|^p\Bigg ]^{\frac {1}{p}}+\Bigg [\sum ^{\infty }_{i=1}|y_i|^p\Bigg ]^{\frac {1}{p}}.\]
(It also has the integral form) □

Solution.

i

\(\hspace {0.4cm}\displaystyle {||x||=\Bigg [\sum ^{n}_{i=1}|x_i|^2\Bigg ]^{\frac {1}{2}}\geq 0},\)
\[|x_i|^2=0\implies x_i=0\iff x=(0,0,\dots 0)=0\]

ii

\begin {align*} ||\alpha x|| &=\Bigg [\sum ^n_{i=1}|\alpha x_i|^2\Bigg ]^{\frac {1}{2}} =\Bigg [\sum ^n_{i=1}|\alpha |^2|x_i|^2\Bigg ]^{\frac {1}{2}}\\ &=|\alpha |\Bigg [\sum ^n_{i=1}|x_i|^2\Bigg ]^{\frac {1}{2}}\\ &=|\alpha |||x||\\ \end {align*}

iii

\begin {align*} ||x+y|| &=\Bigg [\sum ^n_{i=1}|x_i+y_i|^2\Bigg ]^{\frac {1}{2}}\\ &\leq \Bigg [\sum ^n_{i=1}|x_i|^2\Bigg ]^{\frac {1}{2}}+\Bigg [\sum ^n_{i=1}|y_i|^2\Bigg ]^{\frac {1}{2}}\\ &=||x||+||y||\\ \end {align*}