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