Applying positive definiteness on an Inner Product Space

Question

Given that $k$ is continuous on $[0,1]$ and $$\langle k,v \rangle=\int_0^1 k(x)v(x) \,\mathrm{d}x=0, \quad \forall v\in{V}$$ where$$ V=\{v: [0,1]→\mathbb{R},\ v \text{ is continuous},\ v(0) = v(1) = 0,\\ v′\ \text{ is piecewise continuous and bounded}\}, $$ show that $ k(x)\equiv0$ for $x\in [0,1]$.

I know that for any $v\in{V}$, $\langle v,v \rangle $$\geq 0$ and $\langle v,v \rangle =0$ if and only if $v=0$.

To show that $ k(x)\equiv0$, would the best approach be to apply the FTC and solve $$\int_0^1 k(x)v(x) \,\mathrm{d}x=0 $$ to find a constant $C$ that matches the boundary conditions $v(0) = v(1) = 0$?

Answer 1

Suppose that $k(x_0) > 0$ for some $x_0\in (0,1)$. By continuity, there is an open ball $B_{\varepsilon}(x_0) = (x_0-\varepsilon,x_0+\varepsilon)$ around $x_0$ such that $k(y) > k(x_0)/2$ for all $y\in B_{\varepsilon}(x_0)$. Pick a function $v\in V$ such that $v = 1$ on $B_{\varepsilon/2}(x_0)$ and whose support is contained in $B_{\varepsilon}(x_0)$. By hypothesis, $\int_0^1 k(x)v(x)\,dx = 0$. On the other hand, \begin{align*} \int_0^1k(x)v(x)\,dx &= \int_{x_0-\varepsilon}^{x_0+\varepsilon}k(x)\cdot\color{red}{v(x)}\,dx \\ &\ge \int_{x_0-\varepsilon/2}^{x_0+\varepsilon/2}\color{green}{k(x)}\cdot\color{red}{1}\,dx \\ &\ge \int_{x_0-\varepsilon/2}^{x_0+\varepsilon/2} \color{green}{\frac{k(x_0)}{2}}\,dx = \frac{k(x_0)}{2}\cdot\varepsilon > 0, \end{align*} a contradiction. (An analogous analysis holds if $k(x_0) < 0$.) Hence $k(x_0) = 0$ for every point in $(0,1)$, but $k$ is continuous on $[0,1]$, so its values at the endpoints are $$ k(0) = \lim_{x\to 0^+}k(x) = \lim_{x\to 0^+}0 = 0=\lim_{x\to 1^-}0 = \lim_{x\to 1^{-}}k(x) = k(1), $$ so $k \equiv 0$ on $[0,1]$, as desired.

Answer 2

Here is an analysis approach.

Define$$ V_\infty = \{ φ \in C^\infty ([0, 1]) \mid φ(0) = φ(1) = 0 \}, $$ then $V_\infty$ is a subspace of $V$.

Now suppose there exists $x_0 \in [0, 1]$ such that $k(x_0) \neq 0$, without loss of generality, suppose $k(x_0) > 0$. Because $k \in C([0, 1])$, there exists $r > 0$ such that$$ k(x) > 0. \quad \forall x \in (x_0 - r_0, x_0 + r_0) \cap (0, 1) $$ Note that $(x_0 - r_0, x_0 + r_0) \cap (0, 1) $ is a non-empty open set (even if $x_0 = 0$ or $x_0 = 1$), thus there exists $x_1 \in (0, 1)$ and $r_1 > 0$ such that$$ (x_1 - 2r_1, x_1 + 2r_1) \subseteq (x_0 - r_0, x_0 + r_0) \cap (0, 1). $$

Now take$$ φ(x) = \begin{cases} \exp\left( \dfrac{1}{(x - x_1)^2 - r_1^2} \right); & x \in (x_1 - r_1, x_1 + r_1)\\ 0; & \text{otherwise} \end{cases}, $$ then $φ \in V_\infty$. However, since $k(x) > 0$ for all $x \in (x_1 - r_1, x_1 + r_1)$, then$$ \langle k, φ \rangle = \int_0^1 k(x) φ(x) \,\mathrm{d}x = \int_{x_1 - r_1}^{x_1 + r_1} k(x) φ(x) \,\mathrm{d}x > 0, $$ a contradiction. Therefore $k = 0$.

Answer 3

Without loss of generality suppose there exists some $x_0\in(0,1)$ such that $k(x_0)>0$ (the case $<0$ is analogue). By continuity of $k$ there exists some open ball $B(x_0,\varepsilon)\subset(0,1)$ such that $k(x)>0$ for all $x\in B(x_0,\varepsilon)$. Let $v^*(x)$ be a bump function with support in $B(x_0,\varepsilon)$. Then $v^*\in V$ since it is continuous and smooth (in particular $v'$ satisfies the conditions of set $V$). You can take $v^*(x)>0$ for all $x$ in the support of $v^*$ but then $$\int^{1}_0k(x)v^*(x)\,dx=\int_{B(x_0,\varepsilon)}k(x)v^*(x)\,dx>0$$ This clearly contradicts the claim
$$\int^{1}_0k(x)v(x)\,dx=0\hspace{0.2cm},\forall v\in V$$