I was reading the book Mathematics for Machine Learning by Marc Peter Deisenroth, Cheng Soon Ong and A.Aldo Faisal and came across this remark:  Remark(Cauchy-Schwarz Inequality). For an inner product vector space $(V,\langle\cdot,\cdot\rangle)$, the induced norm $||\cdot||$ satisfies the Cauchy-Schwarz inequality $|\langle x,y\rangle|\le ||x||\text{ }||y||$  Now I know how to prove this inequality for the dot product and the induced Euclidean norm. But I want to know if there is some general proof of this inequality that can show the truth of this statement for every possible inner product and induced norm. So, my question is whether such a proof exists, and if yes how it works(some clues, not the proof itself), or whether the inequality has to be proven separately for any considered inner product. Thanks for your time!
5394
How many proofs of the Cauchy-Schwarz inequality are there? Is there some kind of reference that lists all of these proofs?
14182
Let $A$ be a positive definite real $n\times n$ matrix. How can I prove that $$ \int_{\mathbb{R}^n}\mathrm{e}^{-\langle Ax,x\rangle}\text{d}x=\left|\,\det\left(\pi^{-1}{A}\right)\right|^{-1/2}=\pi^{n/2}\lvert\,\det A\rvert^{-1/2}\!, $$ where $\langle\cdot,\cdot\rangle$ denotes the inner product in $\mathbb R^n$, i.e. $\langle x,y\rangle =x^Ty$.
63372
Let $a_1,a_2,\ldots ,a_n$, be positive real numbers and  S = $a_1+a_2+a_3+\cdots+a_n$. Use the Cauchy-Schwarz inequality to prove that   $$\sum_{i=1}^{i=n}\frac{a_i}{S-a_i}\geqslant \frac{n}{n-1}$$ and $$\sum_{i=1}^{i=n}\frac{S-a_i}{a_i}\geqslant n(n-1)$$  Hint: Apply the inequality to:   $\sum \frac{S}{a_i}$ and $\sum \frac{a_i}{S}$ for one case,     and $\sum a_i(S-a_i)$ and $\sum \frac{a_i}{S-a_i}$ for the other case.
73086
Suppose $V$ is an inner product space and $U_1, U_2$ are subsets, then I'd like to show that $U_1 \subset U_2$ if and only if $U^{\perp}_2 \subset U^{\perp}_1$. I can show the first forward direction, by first taking $w \in U^{\perp}_2$. Then, $w$ must be a vector where $&lt;w, u_2&gt; = 0$ for all $u_2 \in U_2$. But, since $U_1 \subset U_2$, then $&lt;w, u_1&gt; = 0$ for all $u_1 \in U_1$. Hence, $w \in U^{\perp}_1$.   I am not sure how to show the opposite direction. My proof is as follows:  Take $u_1 \in U_1$. Now, let $w \in U^{\perp}_2$. Then, $&lt;w,u_2&gt; = 0$ for all $u_2 \in U_2$. But, $U^{\perp}_2 \subset U^{\perp}_1$, and so $w \in U^{\perp}_1$ as well. Hence, $&lt;w,u_1&gt; = 0$ for all $u_1 \in U_1$. After this, I am not sure how to complete the proof. Am I in the right direction? Thanks!
16048
mathematicians! I want to ask to all wise people about a problem I met at the quiz to obtain some ideas. The problem is following. It may not be accurate since the problem is dependent on my memory   Let $f : [0, \infty) \rightarrow \mathbb{R}$ and $f$ is differentiable. $|f'(x) |\; \leq\  |f(x)|$ for all $x \in [0, \infty)$. $f(0) = 0$. Then prove that $f(x) = 0$ for all $x \in [0,\infty)$.  At the quiz, I applied the Mean Value Theorem and the Cauchy-Schwarz inequality.   For any $x$, $|f(x)| = |\{f(x) - f(0)\}(x-0)| = |f'(c_{1})(x)|$ for some $c_{1} \in (0,x)$, and then  $$|f'(c_{1})(x)| \leq |f'(c_{1})||x| \leq |f(c_{1})||x|$$ by Cauchy-Schwarz and given assumption. By doing so consecutively, we can obtain the inequality that  $$ |f(x)| \leq |f(c_{n})||x||c_{1}|\ldots |c_{n-1}| $$ for any $x \in [0,\infty)$. As $\{c_{i}\}, \; i\in \mathbb{N}$ is decreasing sequence and $f$ is continuous, $|f(c_{n})|$ can be arbitrary close to $0$ when $c_{n} \rightarrow 0$. Thus the inequality above can be bounded by $\epsilon$.  I think my answer is right, but a bit uncertain. Could you give me a certainty? Thank you very much.
101405
$U,V$ are unitary $n\times n$ matrices, and the norm is the operator norm (so we can use $\|UV\|\leq\|U\|\|V\|$).  I've noticed that \begin{align} \|UVU^{-1}V^{-1}-I\|&amp;= \|(UV-VU)U^{-1}V^{-1}\|\\ &amp;\leq \|UV-VU\|\|U^{-1}V^{-1}\| \end{align}  I can bound the first term by $\|UV\|+\|VU\|$, but I don't think this is useful.   Hints (rather than complete answers) would be appreciated.  The question comes from  (exercise 1)
64952
Could you please give me a hint to prove that $\mathbb{R}^n$ with the 1-norm $\lvert x\rvert_1=\lvert x_1\rvert+\cdots+\lvert x_n\rvert$ is not isometric to $\mathbb{R}^n$ with the infinity-norm $\lvert x\rvert_\infty = \max_{i=1,\ldots,n}(\lvert x_i\rvert)$ for $n\gt2$.  I do not see how the critical case $n=2$ enters the picture.  Thank you!
25116
The question is to show that for any $x,y \in\Re^n$  $||x+y||\cdot||x-y||\le||x||^2+||y||^2$  with equality if and only if $\langle x,y \rangle= 0.$  I have that $||x+y||^2=||x||^2+2 \langle x,y \rangle+||y||^2$ , which helps me prove equality but I'm lost on how to show this and show that the inequality part.
64861
Suppose that $V$ is an inner product space.  (a) Show that if $\{e_1, . . . , e_n\}$ is an orthonormal basis for $V$ , then $$||v||^2=\sum_{i=1}^{n}|\langle v|e_i\rangle|^2\quad \quad \text{for every $v\in V.$}$$ (b) Suppose that $\{e_1, . . . , e_k\}$ is an orthonormal set in $V$ (not necessarily a basis), and let $v ∈ V$. Show that if $$||v||^2=\sum_{i=1}^{k}|\langle v|e_i\rangle|^2\quad \quad \text{for every $v\in V$}$$ then $v = \sum_{i=1}^{k}⟨v | e_i⟩e_i.$  I tried to use the reconstruction formula but am not getting anywhere. Please help.
94210
Show that a norm on an inner product space satisfies parallelogram law. Hence use the parallelogram law to show that the space of continuous real functions defined on the interval $[a,b]$ is not a Hilbert space. Here I did the first part. Let $X$ be an inner product space and $x,y\in X$ consider $\Vert x+y\Vert ^2$ and $\Vert x-y\Vert ^2$ Then on adding I get $\Vert x+y\Vert ^2+\Vert x-y\Vert ^2=2(\Vert x\Vert ^2 +\Vert y\Vert ^2)$ For the second part I don’t know how can a parallelogram law prove $C[a,b]$ is not a Hilbert space, help me please.
115553
I would appreciate a hint on the following problem:     Let $V$ be a finite dimensional vector space over $F$. There are two scalar products such that: $$ \forall \ w,v \in V \ \Big(\langle v,w\rangle_1=0 \implies \langle v,w\rangle_2=0\Big) $$ Show that $$ \exists \ c \in F \ \ \forall \ w,v \in V \ \Big(\langle v,w\rangle_1=c \langle v,w\rangle_2\Big) $$   I have tried to define an orthonormal basis with respect to $\langle \cdot,\cdot\rangle_1$ hoping that the transformation matrices would be different by a constant, yet it seems to lead nowhere.
55204
Suppose $x_1, \dots, x_n$ are positive real numbers such that $x_1 \cdot \ldots \cdot x_n = 1$. Prove $x_1 + \dots + x_n \ge n$.    I don't want to use the AM-GM inequality to prove this (from which this statement would follow).   I suppose induction is the way to go. The base case is true. The case for $n=2$ is true by  $$x_1 + x+2 - 2 \sqrt{x_1 x_2} = (\sqrt{x_1} - \sqrt{x_n})^2 \ge 0.$$  Now suppose the statement holds for $n$, then $$x_1 \cdot \ldots \cdot x_n = 1 \implies x_1 + \dots + x_n \ge n.$$  So if $x_1 \cdot \ldots \cdot (x_n x_{n+1}) = 1$, then  $$ x_1 + \dots + x_n x_{n+1} \ge n. $$  Then we also have $$ x_1 + \dots + x_{n-1} + x_n + x_{n+1}  \ge n - x_{n}x_{n+1} + x_n + x_{n+1}. $$  Now I'm not sure where else to go from here.
66521
I realize that every inner product defined on a vector space can give rise to a norm on that space.  The converse, apparently is not true.  I'd like an example of a norm which no inner product can generate.
97226
If B is $n \times n$ real symmetric positive definite matrix, then $(x,y) = x^T By$ defines an inner product on $R^n$. How do you prove that $\|x\|=(x^T B x)^{1/2}$ is a norm on $R^n$?
24
Let $H$ be a vector space equipped with an inner product $(\cdot, \cdot)$ and $f:H\to H',\ f(x)=(\cdot,x)$ surjective. Now, why $H$  is a Hilbert space? The other direction is clear by Riesz' representation theorem but what about this?
57003
Let $\| \cdot \|$ be a norm on $\mathbb{R}^n$. The associated dual norm, denoted $\| \cdot \|_*$ is defined as $\| z \|_* = \sup\{ z^{t} x : \| x \| &lt; 1 \}$.   Does someone know how to prove that the dual norm of the $\mathcal l_{p}$ norm is the $\mathcal l_{q}$ norm? I read about norms and it was stated without proof in a book.
75651
Let $1\le p &lt; \infty$. Let $g \in L^p(\Bbb R^n)$, and $f \in L^1(\Bbb R^n)$.   Show: $||f*g||_p \le ||f||_1||g||_p$, where $*$ indicates the convolution defined by   $$ f*g (x) := \int f(x-t) g(t)dt.$$  Here is a relevant hint, apparently:  Let $q$ be the conjugate of $p$. Note that:  $$|f*g| \le \int [|g(t)||f(x-t)|^{1/p}]|f(x-t)|^{1/q}dt,$$ then apply Holder's inequality.  I did exactly as the hint suggested, and ended up with what looked like a meaningless inequality compared to the final answer. I even took the $p$ power of both sides.   Any help?