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Suppose $h : [a,b]\to\mathbb{R}$ is a continuous function and $$ \int_{a}^{b}h(x)g'(x)dx = 0$$ for all $g \in \mathcal{C}^{1}[a,b]$ with $g(a) = g(b) = 0$. Show that $h$ must be a constant function.

I have used integration by parts to achieve

$$ 0 = \int_{a}^{b}h(x)g'(x)dx = -\int_{a}^{b}h'(x)g(x)dx $$

but am not sure what this means or if it is even a relevant approach. I look forward to any hints or tips.

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    Density of $C^1$ function will allow you to conclude $h'(x) = 0$ which means $h$ is constant. – 2017-02-01
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    @JackyChong what is meant by density of $C^1$? I am unfamiliar with this in terms of a function space. – 2017-02-01
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    Is $h$ supposed to be differentiable or $C^1$? Otherwise you might not be allowed to write $h'$. – 2017-02-01
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    @mathbeing Hmmm, true, I only have that $h$ is continuous. – 2017-02-01
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    I'm working with your assume that $h$ is differentiable since you assume that $h'$ exists. Then you simple choose a family of $g\geq 0$ such that the total area under the curve is 1 where the $g$ spikes up at an arbitrary chosen point say $x$. (This is a standard technique usually seen in calculus of variation during the derivation of the Euler Lagrange equation). – 2017-02-01
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    @JackyChong We need not assume that $h$ is differentiable. We only need to find a $g$ judiciously and apply the Cauchy-Schwarz inequality. I've posted a solution that provides a way forward. ;-)) – 2017-02-01

2 Answers 2

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Let's choose a function $g$ such that $$g'(x) = h(x) - k$$ for some suitably chosen constant $k$ such that $g(a) =g(b) =0$. Then we can see that $$\int_{a}^{b} (h(x) - k) ^{2}\,dx=\int_{a}^{b}h(x)g'(x)\,dx - k\int_{a} ^{b} g'(x) \, dx = 0$$ It now follows from continuity of $h(x) $ that $h(x) = k$ for all $x\in[a, b] $.

The proof is now complete if we can choose a constant $k$ meeting the requirement $g(a) =g(b) =0$. Clearly we can take $$g(x) =\int_{a} ^{x} h(t) \, dt - kx + c$$ and then solve for $k, c$ using $g(a) = g(b) = 0$. We get $$ka=c, kb=c+\int_{a} ^{b} h(t) \, dt$$ so that $$k=\frac{1}{b-a}\int_{a}^{b}h(t)\,dt,c=\frac{a}{b-a}\int_{a}^{b}h(t)\,dt$$ and hence $$g(x) =\int_{a} ^{x} h(t) \, dt - \frac{x-a} {b-a} \int_{a} ^{b} h(t) \, dt$$ This explains the genesis of the function $\phi(x) $ of Dr. MV's answer.

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    Hi, I like this solution, but since we constructed $g$ in a specific way, does this violate the assumption that the condition holds for *all* $g \in \mathcal{C}^{1}?$ – 2017-02-04
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    @clocktower: since the given condition applies to all $g\in\mathcal{C} ^{1}$ with $g(a) =g(b) =0$, we have chosen such a $g$ and used this condition. – 2017-02-04
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Since $h\in C[a,b]$, then it is integrable on $[a,b]$. Let $\phi(x)$ be the function

$$\phi(x)=\int_a^x h(t)\,dt-\frac{(x-a)}{b-a}\int_a^bh(t)\,dt$$

Then, $\phi\in C^1[a,b]$ and $\phi(a)=\phi(b)=0$.

In addition, we have $\phi'(x)=h(x)-\frac{1}{b-a}\int_a^b h(t)\,dt$.

Therefore, since we have $\int_a^b h(x)g'(x)\,dx=0$ for all $g\in C^1[a,b]$ with $g(a)=g(b)=0$, then we may select $g(x)=\phi(x)$ and write

$$\begin{align} \int_a^b h(x)g'(x)\,dx&=\int_a^b h^2(x)\,dx-\frac{1}{b-a}\left(\int_a^b h(x)\,dx\right)^2\\\\ &=0 \end{align}$$

But from the Cauchy-Schwarz inequality,

$$\left(\int_a^b h(x)\,dx\right)^2\le (b-a)\int_a^b h^2(x)\,dx $$

with equality holding if and only if $h(x)$ and $1$ are linearly dependent, which implies here that $h$ is constant.

And we are done!

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    Very clever argument. +1 – 2017-02-01
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    I have tried to explain the construction of $\phi(x) $ in my answer. – 2017-02-01