Is It An Inner Product

Question

Let $C^{1}[0,1]$ be the vector space of all continuous functions $f:[0,1]\to \mathbb{C}$ with continuous derivative on $[0,1]$ $\newcommand\inner[2]{\left\langle #1, #2 \right\rangle}$ $$\inner{f}{g}=f(0)\overline{g(0)}+f'(0)\overline{g'(0)}+f(1)\overline{g(1)}$$

1. Is it an inner product over the subspace $P_{2}=\mathrm{span}\{1,x,x^2\}$?

2. Is it an inner product over $C^{1}[0,1]$?

I have started with

a.$$\overline{\inner{g}{f}}=\overline{g(0)\overline{f(0)}+g'(0)\overline{f'(0)}+g(1)\overline{f(1)}}=f(0)\overline{g(0)}+f'(0)\overline{g'(0)}+f(1)\overline{g(1)}=\inner{f}{g}$$

b.$$\inner{\alpha f+\beta g}{h}=[\alpha f(0)+\beta g(0)]\overline{h(0)}+[\alpha f(0)+\beta g(0)]'\overline{h'(0)}+[\alpha f(1)+\beta g(1)]\overline{h(1)}=\alpha f(0)\overline{h(0)}+\alpha f'(0)\overline{h(0)}+\alpha f(1)\overline{h(1)}+\beta g(0)\overline{h(0)}+\beta g'(0)\overline{h'(0)}+\beta g(1)\overline{h(1)}=\alpha\inner{f}{h}+\beta\inner{g}{h}$$

Which hold for both $1$ and $2$

Looking at $\inner{f}{f}$ for the first case we get:

$$\inner{f}{f}=[f(0)]^2+[f'(0)]^2+[f(1)]^2$$ Which is greater than $0$ and equal to $0$ if each element is zero $f(0)=f(1)=f'(0)=0$ and it can be only if $f\equiv 0$

Looking at $\inner{f}{f}$ for the second case we get:

$$\inner{f}{f}=|f(0)|^2+|f'(0)|^2+|f(1)|^2$$ Which is greater than $0$ and equal to $0$ if each element is zero $|f(0)|=|f(1)|=|f'(0)|=0$ and it can be only if $f\equiv 0$

So it is an inner product from both $1$ and $2$? If we know that a function is an inner product on a "bigger" space let say bigger dimension vector space, can we conclude it will an inner product for all subspaces of the vector space?

Answer 1

To prove that $\newcommand\inner[2]{\langle #1, #2 \rangle}$$\inner{f}{f} = 0$ implies $f = 0$ on the space $\mathrm{span}\{1,x,x^2\}$, consider $f \in \mathrm{span}\{1,x,x^2\}$. $f$ must be of the form $f(x) = ax^2 + bx + c$. The conditions imposed on the coefficients $a, b, c$ by $\inner{f}{f} = 0$ are: $$0 = f(0) = c$$ $$0 = f(1) = a+b+c$$ $$0 = f'(0) = 2a$$

From here follows $a = b = c = 0$, which implies $f = 0$.

Answer 2

No, your second case is not correct. There are $C^1$-functions with $f(0)=f(1)=f'(0)=0$ but $f\neq 0$, e.g. $f(x)=x^3-x^2$. The rest looks good.

Answer 3

If $\newcommand\inner[2]{\langle #1, #2 \rangle}$$f \in P_2$ and if $\inner{f}{f}=0$, then $f\equiv 0$. This is true ! But you have not proved it !

Let $f(x)=x^2(x-1)$. Then $\inner{f}{f}=0$, but we do not have that $f\equiv 0$. Conclusion: $\inner{\cdot}{\cdot}$ is not an inner product on $C^1[0,1]$.

Answer 4

No, it is not an inner product in $\mathcal{C}^1\bigl([0,1]\bigr)$. Take $f(x)=x^2-x^3$. It is not the null function, but $\langle f,f\rangle=0$.

Answer 5

Your belief that $\lvert f(0) \rvert = \lvert f'(0) \rvert = \lvert f(1) \rvert = 0$ implies that $f \equiv 0$ is incorrect in general. For example, a smooth bump function centered at $1/2$ with support within $[1/4, 3/4]$ is a nonzero function whose every derivative is zero at both $0$ and $1$. This shows that the second case is not true.

This also shows that you should revisit this step in the first case. Is this true for degree $2$ polynomials? This is something that needs to be shown.