# Is the $L^2$ inner product really an inner product?

If $\alpha: [a,b] \rightarrow \mathbb{R}$ is an increasing function, we can define the Riemann-Stieltjes integral $$\int_a^b f d \alpha$$ Does the function $\langle f,g\rangle = \int_a^b fg d\alpha$ define an inner product on the $\mathbb{R}$-space of all Riemann-Stieltjes integrable functions on $[a,b]$ (with respect to $\alpha$)? It seems to me that the condition $\langle f,f\rangle = 0$ implies $f= 0$ is false. For example if we take the Riemann integral and take $f(x)$ to be $0$ everywhere except $1$ at $(b-a)/2$, then $\langle f,f\rangle = \int_a^b f(x)^2dx = 0$, but $f \neq 0$.

If it is not an inner product, can we still get something like the Cauchy-Schwarz inequality, i.e. $$\Biggl[\int_a^b |fg| d \alpha\Biggr]^2 \leq \Biggl(\int_a^b |f|^2 d \alpha\Biggr) \Biggl(\int_a^b |g|^2 d \alpha\Biggr)\,?$$

• One quotients the space by "null" functions and then it's an inner product. – anon Sep 28 '14 at 17:55
• You get an inner product on the space of equivalence classes of (square integrable) functions modulo the relation of being equal almost everywhere. But the Cauchy-Schwarz inequality holds for all positive semidefinite bilinear/sesquilinear forms. – Daniel Fischer Sep 28 '14 at 17:56
• Does Cauchy-Schwarz really hold for semidefinite bilinear forms? All proofs I've seen involve dividing by $<x,x>$ for $x \neq 0$, which you may not be able to do in the semidefinite case. – D_S Sep 29 '14 at 15:15

I know this has been answered in the comments but you might be interested in reading up about exactly how this is done over here. Note that in that article they talk about functions being equal almost everywhere which in the view of Riemann-Stieltjes Integral is exactly the type of issue you brought up: the functions are nearly identical (in fact they are identical almost everywhere).

Also I'll a quick overview of the process outlined by that article:

• Define a function $\lVert \cdot \rVert$ on the set of $\alpha$ square integrable functions by $$\lVert f \rVert = \sqrt{\int \lvert f \rvert^2 \, \mathrm{d} \alpha}$$
• Note that the above function is in fact a seminorm, that the space is complete under it and that the seminorm abides by the parallelogram equality: $$\lVert f \rVert^2 + \lVert g \rVert^2 = \lVert f + g \rVert^2 + \lVert f - g \rVert^2$$
• We can now "mod" out by all $0-$functions by modding out by the kernel of $\lVert \cdot \rVert$ (i.e. create equivalence classes around functions $f$ so that $\lVert f \rVert = 0$).
• Our modded out space can be checked to be a complete normed vector space that satisfies the parallelogram equality thus we can define $$\langle f, g \rangle = \frac{\lVert f + g \rVert^2 - \lVert f - g \rVert^2}{4}$$ in order to make our space a Hilbert space (and thus we have our inner product).

If you're just trying to get the Schwarz inequality, I think you can look at it like this:

If both $\int |f|^2$ and $\int |g|^2$ are non-zero, the proof which involve dividing by $\langle f, f\rangle$ or $\langle g, g\rangle$ goes through.

So say we're worried about the situation where $\int |f|^2 = 0$.
Note that for every interval $I$ in the range of integration, the infimum of the values $|f|^2$ needs to be zero; otherwise $|f|$ is bounded away from zero on that interval and the integral would be nonzero.

If $g$ is bounded, then for each interval $I$ and each $\epsilon$ you can find some $x$ in the interval so that $0 \le|f(x)g(x)| < \epsilon$.

Looking at the lower Darboux sums for $|fg|$, we see that all lower sums are zero, so that the integral $\int |fg|$ is zero.

And so the Schwarz inequality holds in this case, too.