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.


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 $U^\perp = \{0\}$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ For any $g \in V$, $(f_t)$ converges to $g$. $\endgroup$ – Dark Jun 24 '16 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.