# prove that a function is an inner product

I would appreciate some assistance in answering the following problems. We are moving so quickly through our advanced linear algebra material, I can't wrap my head around the key concepts. Thank you.

Let $V$ be the space of all continuously differentiable real valued functions on $[a, b]$.

(i) Define $$\langle f,g\rangle = \int_a^bf(t)g(t) \, dt + \int_a^bf'(t)g'(t) \, dt.$$ Prove that $\langle , \rangle$ is an inner product on $V$.

(ii) Define that $||f|| = \int_a^b|f(t)| \, dt + \int_a^b|f'(t)| \, dt$. Prove that this defines a norm on V.

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You need to verify that the definitionof norm and inner products hold. Which one of the axioms are you having problems verifying ? for example, linearity for the inner product should be a simple one –  Belgi Oct 20 '12 at 16:57
I changed $<f,g>$ to $\langle f,g\rangle$. That is standard usage. –  Michael Hardy Oct 20 '12 at 17:05
You need to check (1) symmetry, ie, $\langle f,g\rangle = \langle g,f\rangle$, (2) linearity in the first argument, ie, $\langle \alpha f,g\rangle = \alpha \langle f,g\rangle$ and $\langle f_1+f_2,g\rangle = \langle f_1,g\rangle + \langle f_2,g\rangle$ and (3) positive definiteness, ie, $\langle f,f\rangle \geq 0$, with equality iff $f=0$. –  copper.hat Oct 20 '12 at 17:07
To show (ii), you need to check (1) $\|f\| \geq 0$ with equality iff $f=0$, (2) $\|\alpha f\| = |\alpha| \|f\|$, and (3) $\|f+g\| \leq \|f\| + \|g\|$. All of these properties follow from properties of the integral. –  copper.hat Oct 20 '12 at 17:20

Part (i)

If you ever want to show something is an inner product, you need to show three things for all $f, g \in V$ and $\alpha \in \mathbb{R}$:

1. Symmetry: $\newcommand{\inp}[1]{\left\langle #1 \right\rangle}$ $\inp{f, g} = \inp{g, f}$ (Or, if the field is the complex numbers, $\inp{f,g} = \overline{\inp{g,f}}$, i.e. "conjugate symmetry.)

2. Linearity: $\inp{\alpha f, g} = \alpha \inp{f, g}$. Notice this also implies $\inp{f, \alpha g} = \alpha \inp{f, g}$ ($\overline{\alpha}$ in the complex case) by symmetry, so we don't need to prove symmetry for the first case.

3. Positive-definite: $\inp{f, f} \ge 0$ with equality if and only if $f = 0$, the zero function.

The first two properties follow directly from the definition of an integral. For the third property, you have $$\int_a^b f^2 + \int_a^b (f')^2 \ge 0$$ Now when is this equal to $0$? Well, recall that if a continuous function is positive anywhere, then the integral is positive. Since $f^2$ is continuous, this means $f = 0$ everywhere. Thus $f' = 0$ everywhere also, so equality holds.

Part (ii)

If you ever want to show something is a norm, you need to show three things for all $f, g \in V$ and $\alpha \in \mathbb{R}$:

1. Scales in absolute value: $\newcommand{\norm}[1]{\left\| #1 \right\|}$ $\norm{\alpha f} = |\alpha| \norm{f}$.

2. Triangle Inequality: $\norm{f + g} \le \norm{f} + \norm{g}$.

3. Separates Points: $\norm{f} = 0$ if and only if $f = 0$.

Again the first two properties follow directly from the definition of the norm, which is an integral. For the third property, we use the same property of integrals we used before: if a continuous function is positive anywhere, its integral is positive. Since $|f|$ is continuous when $f$ is, this means $\int |f| = 0$ if and only if $f = 0$. Which in turn implies $\int |f'| = 0$.

You will notice parts (i) and (ii) seem very similar. It's almost the case that you can use part (i) for part (ii), but not quite. The problem is that with this inner product and norm, $\inp{f, f} = \norm{f^2}$, not $\norm{f}$. $\inp{f, 1}$ is closer to $\norm{f}$, but it still isn't quite there. $\inp{|f|, 1}$ doesn't work because $f$ might not be differentiable. And $\inp{\sqrt{|f|}, \sqrt{|f|}}$ doesn't work at all since the derivative of $\sqrt{|f|}$ is something really weird, even where it exists.

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All of the conditions for an inner product follow from properties of integrals. For example, $\int_a^b \alpha f(x)dx=\alpha \int_a^b f(x)dx, \forall\alpha \in \mathbb{R}$.

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Well, positive definiteness is (slightly) trickier, for that you need to use that $f$ is differentiable. Otherwise, there'd be non-zero $f$ with $\int_a^b f^2 = 0$... –  fgp Oct 20 '12 at 17:24