# Inner product and distance

My question is this, is there a proof to show that suppose the distance $||h(x)- g(x)||< 4$, then $|h(x) - g(x) | <4$ for all $x\in [-\pi, \pi]$? I know from Schwarz inequality that $$|h - g| \leq ||h- g||.$$

Note that the inner product space $PC [-\pi , \pi]$ the distance between two functions is $$||h-g||^2= \int_{-\pi}^{\pi} |h(x) - g(x) |^{2} dx.$$ I hope I can just state that and there will not be anything to prove.

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What does $|h-g|$ mean? –  Gerry Myerson Dec 12 '11 at 4:23
What is $|h-g|$ and what is $\lVert h-g\rVert$? –  Arturo Magidin Dec 12 '11 at 4:27
I think what OP means is suppose that $\|h-g\| < 4$, then for all $x$ in the domain of $g$ and $h$ (which I can guess here is $[-\pi,\pi]$), then $|h(x) - g(x)| < 4$. But we'll obviously need OP to confirm that. –  Patrick Da Silva Dec 12 '11 at 4:31
Thanks but what I meant was suppose that$||h−g||<4$, then for all $x$ in the domain of g and h in ([−π,π]), then $|h(x)−g(x)|<4$. –  Beat Dec 12 '11 at 12:30
@Beat: Assuming that that comment was intended to refer to my answer: Yes, that's how I understood your question, and that's what I answered. This is false for the reasons I pointed out. –  joriki Dec 12 '11 at 14:31

No, this is not the case. Given your definition of the distance between two functions, the function values can be arbitrarily far apart if this occurs in a sufficiently small region. For instance, consider $g(x)=0$ and $h(x)$ a rectangular pulse that can be arbitrarily high as long as it is sufficiently short.