2
$\begingroup$

Let $g:[0,1] \rightarrow \mathbb R$ be a continuous function and assume that $$ \int_{0}^{1} g(x) \phi'(x) dx = 0 $$ for all continuously differentiable functions $\phi: [0,1] \rightarrow \mathbb R$, where $\phi (0) = \phi(1) = 0$. Show that $g(x) = c$ for all $x \in [0,1]$ for some constant $c \in \mathbb R$.

My thoughts are applying integration by parts with $G(t) = \int_{0}^{t} g(x) dx $, since now we get $$ 0 = \int_{0}^{1} G(t) \phi'(t) dt + \int_{0}^{1} g(t) \phi (t) dt $$ which doesn't really go far. The second thought was applying MVT for definite integrals, which doesn't go far either as the conditions are hardly satisfied.

Any hints?

$\endgroup$
  • $\begingroup$ You're applying the formula for the integration of parts wrong, it must be $$ \int_{0}^{1} g(x) \phi'(x) \; \text{d}x = \left[g(x)\phi(x)\right]_0^1 - \int_{0}^{1} g'(x) \phi(x) \; \text{d}x = 0$$ (when you integrate $\phi$ and diff $g$). $\endgroup$ – Maximilian Gerhardt May 24 '16 at 10:35
  • $\begingroup$ I chose to define a new $G$ because we do not know if $g$ is differentiable. If it is, then I believe we can apply as you said, then use MVT and continuity to get result. (Sorry if I am misunderstanding some thing...) $\endgroup$ – Bryan Shih May 24 '16 at 11:04
2
$\begingroup$

We define $$c:=\int_{0}^{1}g\left(x\right)dx$$ and consider $$\phi\left(x\right)=\int_{0}^{x}\left(g\left(x\right)-c\right)dx.$$ We can see that $\phi\left(x\right)$ has the the required assumption. Now we can observe that $$\int_{0}^{1}g\left(x\right)\phi'\left(x\right)dx=0=c\int_{0}^{1}\phi'\left(x\right)dx $$ so we have $$\int_{0}^{1}\left(g\left(x\right)-c\right)\phi'\left(x\right)dx=\int_{0}^{1}\left(g\left(x\right)-c\right)^{2}dx=0 $$ so it follows that $$g\equiv c $$ for all $x\in\left[0,1\right].$ Note that it is possible to generalize the result for every $\left[a,b\right]$.

$\endgroup$
3
$\begingroup$

Hints: The conditions $\phi(0) = \phi(1) = 1$ imply $$ \int_{0}^{1} \phi'(x)\, dx = 0; \tag{1} $$ that is, "the continuous function $\phi'$ is orthogonal to the constants with respect to the standard inner product in $C([0, 1])$". Consequently, the condition $$ \int_{0}^{1} g(x) \phi'(x)\, dx = 0\quad\text{for all $C^{1}$ functions $\phi$ satisfying (1)} \tag{2} $$ asserts "$g$ is orthogonal to every function $\psi = \phi'$ orthogonal to the constants".

Since the inner product is non-degenerate on $C([0, 1])$, the question reduces to linear algebra.

If that's not enough, here's another hint:

Show that if $c$ is the orthogonal projection of $g$ to the constants, then there exists a $C^{1}$ function $\phi$, vanishing at the endpoints, such that $\phi' = g - c$. Then observe that $g - c$ satisfies (2), etc.

$\endgroup$
  • $\begingroup$ A very nice use of orthogonality! $\endgroup$ – Tomas May 24 '16 at 11:35
1
$\begingroup$

Suppose $f(x)$ is not constant on $[0, 1]$. Let $m$ and $M$ be the minimum and maximum value attained by $f$ on $[0, 1]$. These values are attained on $[0, 1]$ because $f$ is continuous and $[0, 1]$ is compact. By Darboux property there exists $\gamma \in (0, 1)$ such that $f(\gamma)=\overline{f}=\frac{m+M}{2}$ and a sufficiently small $\epsilon > 0$ such that $f(x)\ge \overline{f}$ for $x \in (\gamma-\epsilon, \gamma] $ and $f(x)\le \overline{f}$ for $x \in [\gamma, \gamma + \epsilon) $(Or with the inequalities reversed). Now consider the function $$\phi(x)=\exp\left(\frac{1}{x-(\gamma+\epsilon)}-\frac{1}{x-(\gamma-\epsilon)}\right)$$ when $x \in (\gamma-\epsilon, \gamma+\epsilon)$ and $0$ otherwise. Then it is not difficult to show that $\phi(x)$ is differentiable and that $\left[\phi(\gamma-\beta)\right]_{\beta\rightarrow\epsilon}=[\phi(\gamma+\beta)]_{\beta\rightarrow\epsilon}=0$. Also, one can verify $\phi'\gt 0$ for $x \in (\gamma-\epsilon, \gamma)$ and $\phi'\lt 0$ for $x \in (\gamma, \gamma+\epsilon)$ and $0$ otherwise. From here it is trivial to see that $\int_0^1(f-\overline{f})\phi'dx>0$ which is a contradiction.

$\endgroup$

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.