Explaining maths

  Young's inequality  (for products) is sometimes called the generalised  AM-GM (arithmetic mean -geometric mean) inequality . Let $a$ and $b$ be non-negative real numbers. Then $(ab)^{1/2} \leq (a+b)/2$, with equality if and only if $a=b$. This can be seen easily by multiplying both sides by $2$, then squaring both sides and noting that $(a+b)^2=(a-b)^2+4ab$. This is the special case of the AM-GM inequality for two non-negative real numbers (and the result generalises to any finite number of non-negative real numbers). However it is also the special case of Young's inequality when $p=q=2$. Young's inequality states that for non-negative real numbers $x$ and $y$ and positive real numbers $p$ and $q$ satisfying $1/p + 1/q = 1$, we have $xy \leq x^p/p + y^q/q$, with equality if and only if $x^p=y^q$. When $p=q=2$ this says that $xy \leq (x^2+y^2)/2$ with equality if and only if $x^2=y^2$, which is the same as the version of AM-GM above above when you set $a=x^2$ and $b=y^2$...

In an earlier post  we looked at the Boundedness Theorem and the Extreme Value Theorem for continuous, real-valued functions on closed and bounded intervals $[a,b]$. However, the proof we gave there generalises to any setting in which the lemma we used is valid. Let $D$ be a non-empty subset of $\R^2$, and consider the following property (B) ( non-standard! ) that $D$ may or may not have: (B)           Every continuous function $f:D\to\R$ is bounded above on $D$. By the reasoning in  the earlier post , we can show that the Extreme Value Theorem holds for any non-empty set $D$ that satisfies condition (B). So the question arises, which subsets of $\R^2$ satisfy condition (B)? By considering the continuous function $f(x,y)=\sqrt{x^2+y^2}$ (or we could use $f(x,y)=x^2+y^2$), it is clear that any subset $D$ of $\R^2$ that satisfies condition (B) must be a bounded subset of $\R^2$ (that is, the set of distances from points of $D$ to $(0,0)$ is bounde...

 In the past I used to confuse students by asking them what the correct value of $((-1)^{2/3})^{3/2}$ is. This is on the assumption that you follow the convention that $x^{2/3}$ is equal to the square of the cube root of $x$ for all real numbers $x$. But perhaps instead I should argue that $1$ must be equal to $\pm \mathrm{i}$, on the grounds that $((-1)^2)^{1/4}$ should be equal to $(-1)^{1/2}$?

 Recall our claim from Part I Cauchy-Schwarz in complex inner product spaces can be deduced from Cauchy-Schwarz in real inner product spaces. This is essentially because every complex inner product space can also be regarded as a real inner product space with the same norm, but where the real inner product is the real part of the complex inner product. The last bit of the deduction uses a fairly standard "rotation trick": for any complex number \(z\), we have  \[|z|=\max\{\Re(\alpha z):\alpha \in \C, |\alpha|=1\,\}\,.\qquad(*)\] Let's check $(*)$ first. Let $z\in\C$, and set \[A=\{\Re(\alpha z):\alpha \in \C, |\alpha|=1\,\} \subseteq \R\,.\]  Clearly $A$ is non-empty.  Let $\alpha \in \C$ with $|\alpha|=1$. Then certainly we have \[\Re(\alpha z) \leq |\alpha z| = |z|\,.\] Thus we have \[\sup A \leq |z|\,.\] We now need to prove that equality holds, and that the supremum is actually a maximum. For this we just need to show that $|z| \in A$. This is trivial if $z=0$, b...

In this part, we take as our starting point the (easy) results discussed so far for real inner product spaces of dimensions less than or equal to $2$. That is, that for a given dimension $n$ (and we only need $n\leq 2$ below), up to "real inner product space isomorphism", all $n$-dimensional inner product spaces over $\R$ are the same as $\R^n$ with the usual dot product. In particular, the Cauchy-Schwarz inequality holds in all real inner product spaces of dimension $\leq 2$, with equality if and only if the two vectors involved are linearly dependent. Now let $V$ be any  inner product space over $\R$, with inner product $\langle \cdot,\cdot \rangle_V$ and associated norm $\|\cdot\|_V$. (Here $V$ can be infinite-dimensional, and need not be separable.) Let $\vx, \vy \in V$, let $W$ be the linear subspace of $V$ spanned by $\vx$ and $\vy$, i.e., $\lin\{\vx,\vy\}$. Clearly $W$ is then itself a real inner product space of dimension at most $2$. Thus the Cauchy-Schwarz inequalit...

 To conclude Part IV, we prove the easy fact that all two-dimensional real inner product spaces are "essentially the same" as $\R^2$ with the usual dot product and the associated (Euclidean) norm $\|\vx\|_2 = \sqrt{\vx\cdot\vx}$. Of course what we need is a linear isomorphism which preserves norms and inner products. The argument is almost identical to that in Part III (and, unsurprisingly, generalises via the usual Gram-Schmidt process to higher dimensions). As an exercise you should check for yourselves the details of any unproven claims below. Now let $V$ be any two-dimensional inner product space over $\R$, with inner product $\langle \cdot,\cdot \rangle_V$ and associated norm $\|\cdot\|_V$. Let $\{\vv_1,\vv_2\}$ be a basis of $V$. Then $\vv_1\neq 0$, so we can normalize by setting \[\Ve_1=\frac{1}{\|\vv_1\|_V} \vv_1\,.\] Then $\{\Ve_1,\vv_2\}$ is easily seen to still be a basis of $V$. Next we write $\vv_2=\va+\vb$ with $\va=\langle \Ve_1,\vv_2\rangle\Ve_1$ being the com...

In this part we will look at two-dimensional real inner product spaces, which are all essentially identical to $\R^2$ with the standard dot product. They are also all essentially identical to $\C$ when $\C$ is regarded as a  real  inner product space, using the  real  inner product (as discussed in Part II). \[\langle z,w \rangle = \Re(\bar z w)= \Re(z) \Re(w) + \Im(z) \Im(w)\,,\] and the Cauchy-Schwarz inequality for all two-dimensional real inner product spaces is immediate from this, because $|\Re(\bar z w)| \leq |\bar z w|$. (Of course, Cauchy-Schwarz in $\R^2$ is easy to prove anyway!) We already discussed the connection between $\R^2$ and $\C$ in Part II. Recall the following from that part (also discussed in intermediate parts). The (linear in the second variable) standard complex inner product on $\C$ is given by $\langle z,w \rangle = \bar z w$, with the associated norm given by modulus. Writing $z=a_1+\i a_2$ and $w=b_1+\i b_2$, with $a_1,a_2,b_1,b_2$ real,...