Find $g \in f^{\perp}$ s.t. $\langle f,g \rangle =0$

Question

Let $(Y, \langle .,. \rangle )$ be an inner product space with $Y=C[0,1]$ and $$\langle f,g \rangle = \int_0^1 f(x)g(x)dx$$

In $C[0,1]$, let $f(t) =t$, and find $g \in f^{\perp}$ such that $g(0) = 5$.

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Am I right in saying $f^{\perp}= \{ g \in Y:g \perp f \}$?

So we need to find $g$ so that $ \langle f,g \rangle =0$. If we have $g(t)=0$ when $t=1$ and $g(t)=5$ when $0 \le t <1$, and $f(t)=1$ then would that work?

Because then $f(t)g(t)=0$ when $t=1$ and $f(t)g(t)=5t$ (this is the part I am unsure about) when $0 \le t <1$.

Then let the integral of $f(t)g(t)$ be $F(t)$. Then $F(t)=0$ when $t=1$ and $F(t)=5t^2/2$ when $0 \le t <1$. So then find $F(1)-F(0)=0- 5(0)^2/2=0$?

Answer 1

The candidate function $g$ you mention is not continuous, i.e., not in $C[0, 1]$.

One simple approach is the following: Choose a simple $1$-parameter family of functions $g$ satisfying the given condition $g(0) = 5$, compute $\langle f, g \rangle$, and then find a parameter value (if one exists) such that the inner product is $0$.

A natural choice for the family is the affine functions with $y$-intercept $5$, namely, $$g_a(t) := a t + 5 .$$ Now, $$\langle f , g \rangle = \int_0^1 f g \,dt = \int_0^1 t (a t + 5) \,dt .$$ Can you find a value of the parameter $a$ for which this is zero?

Answer 2

Your $g$ is not continuous. Also, you will have $\langle f,g\rangle = 5\int_0^1f(x)\,\mathrm dx\ne 0$ in general.

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If $f=0$, any $g$ will do, for example $g(t)=5$.

Otherwise, pick $a\in(0,1)$ with $f(a)\ne 0$. Wlog. $f(a)>0$. Then for some small $r>0$ we have $f(x)>0$ for all $x\in(a-r,a+r)$. Let $h(x)=\max\{0,1-\frac{|x-a|}r\}$. Then $\langle f,h\rangle >0$. Let $g_0(t)=5$. Then $g = g_0-\frac{\langle f,g_0\rangle}{\langle f,h\rangle}h$ has the desired properties. - The key point was that $h(0)=0$ and $\langle f,h\rangle \ne 0$.