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Let $C[0,1]$ be the vector space of continuous real-valued functions on the interval [0,1]. The following mapping is defined: $$ \langle \bullet , \bullet \rangle : C[0,1]^2 \rightarrow \mathbb{R} : (f,g) \mapsto \int_0^1f(x)g(x)dx $$

Firstly, this map is well defined since the product of two continuous function is continuous, and a continuous function is Riemann integrable.

I can show that $\langle \bullet , \bullet \rangle$ is a scalar product:

  • Linearity follows from the linearity of the Riemann integral
  • Symmetry follows from the commutativity of functions inside the Riemann integral

I would like help showing that $\langle \bullet , \bullet \rangle$ is non-degenerate, that is: $$ If \ \ f \in C[0,1],\ \ and \ \ \langle f , g \rangle = 0 \ \ \forall \ \ g \in C[0,1], \ \ then \ \ f \equiv 0. $$ Ordinarily, if we were working over a finite-dimensional vector space, I would evaluate $g$ at each of the standard basis vectors and hope to show that the corresponding component in $f$ is zero - and so $f$ is the zero vector. However I doubt this method works for $C[0,1]$ since it is infinite dimensional.

I have also tried to derive a contradiction by assuming: $\exists \ \ c \in [0,1] s.t. f(c)\not= 0$, but to no success.

I'd appreciate any help given. Also, is there a name for this scalar product?

Reference

This was Chapter VI, Section 1, Example 2 taken from

Lang, S., Linear Algebra (2nd ed.), Addison-Wesley

Bysshed
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    Take $g=f$ — what can you tell from $\int_{[0,1]}f^2 = 0$?. – Clement C. Dec 23 '14 at 21:33
  • A followup hint to Clement's hint: suppose there is some $y$ with $|f(y)|>0$, in other words that $f$ is not identically zero. Try to argue that $f(x)^2$ is bounded away from zero on some interval centered at $y$. – Ian Dec 23 '14 at 21:36

1 Answers1

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Thanks to Clement C. for you're comment.

Take $g \equiv f$ and assume, for contradiction, $\exists \ \ c \in [0,1] s.t. f(c)\not= 0$.

Consider the case $f(c) > 0$ (the case $f(c) < 0$ being similar):

By continuity of $f$, $$ \forall \ \ \epsilon >0, \exists \ \ \delta >0, \left( \vert c-y \vert < \delta \Rightarrow \vert f(c) - f(y) \vert < \epsilon \right) $$ Taking $\epsilon = \frac{f(c)}{2} $ yields, $$ \exists \ \ \delta >0, \left( y \in (c-\delta,c+\delta) \Rightarrow f(y) \in \left(\frac{f(c)}{2},\frac{3f(c)}{2}) \right) \right) $$

Take $\delta^* = min\{c,1-c,\delta\}$ to ensure $(c-\delta^*,c+\delta^*) \subseteq (c-\delta,c+\delta) $ and $(c-\delta^*,c+\delta^*)\subset [0,1]$

Next, define the following subdivision $\Delta = (0,c-\delta^*,c+\delta^*,1)$.

Consider the lower Riemann sum: $$ \begin{align} L(f^2,\Delta) &\geq \inf\{f^2(t):t \in [c-\delta^*,c+\delta^*]\}\vert 2\delta^*\vert \\ &\geq f(c) \vert \delta^*\vert \\ &> 0 \end{align} $$

Finally, $$ 0 < L(f^2,\Delta) \leq \int_0^1f^2(t)dt =0 $$ which is a contradiction.

Bysshed
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