Inner Product, Triangle and Cauchy Schwarz Inequalities

[Lelouch]

Homework Statement

Homework Equations

I am not sure. I have not seen the triangle inequality for inner products, nor the Cauchy-Schwarz Inequality for the inner product. The only thing that my lecture notes and textbook show is the axioms for general inner products, the definition of norm (length) and distance between inner product. And some basic consequences of the inner product axioms for norm and distance, s.t. norm is greater equal to 0, norm multiplied by a constant, symmetry of distance, and distance is greater than or equal to 0.

The rest are examples of inner products i.e. on $R^n$.

The Attempt at a Solution

I have solved the definite integral for the given $p(x)$ and $q(x)$ which has the value $\frac{68}{21}$.

What I thought I need to show is that $<p, q> \leq <p> + <q>$ for the triangle inequality. But this seems nonsensical to me since the inner product is defined for two vectors and not for one.
Maybe I have to show $<p, q> \leq <p, p> + <q, q>$? But at this point I am just guessing since I do not know the triangle inequality for inner products nor the Cauchy-Schwarz inequality for inner products.

 

[FactChecker]

Maybe I have to show $<p, q> \leq <p, p> + <q, q>$? But at this point I am just guessing since I do not know the triangle inequality for inner products nor the Cauchy-Schwarz inequality for inner products.

Yes, an inner product defines an associated norm ||a|| ²= <a,a>. The triangle they are talking about is p, q, and p-q. To show that the norm satisfies the triangle inequality for that triangle, show that $√<p-q, p-q> \leq √<p, p> + √<q, q>$.

 

[Lelouch]

Yes, an inner product defines an associated norm ||a|| ²= <a,a>. The triangle they are talking about is p, q, and p-q. To show that the norm satisfies the triangle inequality for that triangle, show that $√<p-q, p-q> \leq √<p, p> + √<q, q>$.

I wanted to ask why the norm was defined as the inner product, but now right before I answered I noticed you added the $\sqrt{}$.

It seems simple I just calculate the respective inner products using the integral, then square root and check whether the inequality holds for the numerical values.
However, why do we perform $<p - q, p - q>$ instead of $<p + q, p + q>$ as shown in Theorem 1(a) in the following link?http://mathonline.wikidot.com/the-triangle-inequality-for-inner-product-spaces
http://mathonline.wikidot.com/the-triangle-inequality-for-inner-product-spaces

 

[FactChecker]

I wanted to ask why the norm was defined as the inner product, but now right before I answered I noticed you added the $\sqrt{}$.

It seems simple I just calculate the respective inner products using the integral, then square root and check whether the inequality holds for the numerical values.
However, why do we perform $<p - q, p - q>$ instead of $<p + q, p + q>$ as shown in Theorem 1(a) in the following link?
http://mathonline.wikidot.com/the-triangle-inequality-for-inner-product-spaces

Since any of the three sides of a triangle is no longer than the sum of the lengths of the other two sides, both triangle examples work.

 

[Ray Vickson]

Homework Statement

View attachment 220858

Homework Equations

I am not sure. I have not seen the triangle inequality for inner products, nor the Cauchy-Schwarz Inequality for the inner product. The only thing that my lecture notes and textbook show is the axioms for general inner products, the definition of norm (length) and distance between inner product. And some basic consequences of the inner product axioms for norm and distance, s.t. norm is greater equal to 0, norm multiplied by a constant, symmetry of distance, and distance is greater than or equal to 0.

The rest are examples of inner products i.e. on $R^n$.

The Attempt at a Solution

I have solved the definite integral for the given $p(x)$ and $q(x)$ which has the value $\frac{68}{21}$.

What I thought I need to show is that $<p, q> \leq <p> + <q>$ for the triangle inequality. But this seems nonsensical to me since the inner product is defined for two vectors and not for one.
Maybe I have to show $<p, q> \leq <p, p> + <q, q>$? But at this point I am just guessing since I do not know the triangle inequality for inner products nor the Cauchy-Schwarz inequality for inner products.

Get the terminology straight first, before attempting the question. For any two vectors $p,q$:
1. Triangle inequality: $|| p + q || \leq || p || + || q ||$.
2. Cauchy-Schwartz: $|\langle p,q \rangle | \leq || p || \cdot || q ||.$
Here, $\langle \cdot , \cdot \rangle$ is the inner product and $||w|| = \sqrt{\langle w,w \rangle }$ is the associated "norm".

In your case the easiest thing is to just do all the integrals and show that you get the above inequalities with the given $p,q$.

 

[Lelouch]

Get the terminology straight first, before attempting the question. For any two vectors $p,q$:
1. Triangle inequality: $|| p + q || \leq || p || + || q ||$.
2. Cauchy-Schwartz: $|\langle p,q \rangle | \leq || p || \cdot || q ||.$
Here, $\langle \cdot , \cdot \rangle$ is the inner product and $||w|| = \sqrt{\langle w,w \rangle }$ is the associated "norm".

In your case the easiest thing is to just do all the integrals and show that you get the above inequalities with the given $p,q$.

Thank you. I was able to solve the problem given your explanation of the terminology.