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So suppose I have a vector space, $V$ which is all continuous functions on $[0,1]$. Additionally, we have an inner product over $V$ where $\langle f,g \rangle = \int_{0}^{1}f(x)g(x)dx$.

Now suppose I have a subspace, $U \subset V$, defined to be the functions where $f(0) = 0$.

I wish to find $U^\perp$, the orthogonal complement to U.

My attempts so far are unsuccessful, just trying to use the definition of $U^\perp$ ($ =\{g \in V | \langle f,g \rangle=0 , f \in U\}$) to arrive somewhere, but I plug in the inner product and can't deduce any further.

Any help would be appreciative.

Thank you.

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Let $g \in V$ be orthogonal to $U$. Define $H_t(x) = 1 - tx$ for $0 \leq x \leq \frac{1}{t}$ and $H_t(x) = 0$ for $x \geq \frac{1}{t}$ for every $t \geq 1$ and notice $|H_t(x)| \leq 1$ for all $t$ and $x$. Then $f_t := g - g(0) H_t \in U$ for all $t \geq 1$ and so we have $\langle f_t, g \rangle = 0$, hence $$ 0 \leq |\langle g,g\rangle| = \left|\int_0^1 g(0)H_t(x)g(x) dx \right| = \left|\int_0^{1/t} g(0)H_t(x) g(x) \right| \leq \int_0^{1/t}|g(0)| |g(x)| dx \leq \frac{M}{t}$$ for some constant $M > 0$ as $g$ is continuous and $[0,1]$ is compact. As this holds for all $t \geq 1$ we conclude $g = 0$. Thus $U^\perp = \{0\}$.

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  • $\begingroup$ Thank you very much! I was wondering if there is any name for this technique you used to solve it? It is very unfamiliar to me, and I think I am at a lower level of mathematics. This conclusion seems to be drawn from topology perhaps? Thanks anyways! $\endgroup$ – anakhro Dec 1 '14 at 12:37
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    $\begingroup$ I believe there is no name for this. When I failed to think of some nontrivial element of $U^\perp$ I thought that $U^\perp$ might be trivial. Then I looked at what would happen if I had some $g \in U^\perp$ and realized that $g$ would have to be orthogonal to any function $g - g(0)f$ with $f(0) = 1$. I would like to use this to show $\langle g , g \rangle = 0$. But as we are using integrals, particular values at points do not really matter which motivated me to look for these functions $H_t$ such that $H_t(0) = 1$ and $\langle g , H_t\rangle$ becomes small for large enough $t$. $\endgroup$ – Matthias Klupsch Dec 1 '14 at 17:17
  • $\begingroup$ Oh, I see. Well thank you very much for the explanation! $\endgroup$ – anakhro Dec 1 '14 at 19:20
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    $\begingroup$ @MatthiasKlupsch Thanks for your reply. The facts $U$ dense in $V$ and $f \mapsto f(0)$ not continuous remain true though. $\endgroup$ – Dark Jun 24 '16 at 10:07
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    $\begingroup$ For any $g \in V$, $(f_t)$ converges to $g$. $\endgroup$ – Dark Jun 24 '16 at 10:42

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