Given $f\in C[0,\Lambda]$ satisfying $$\sup_{n\in \mathbb{N}} \left|\int_0^\Lambda e^{nx} f(x) dx \right| < \infty$$ Prove that $f\equiv 0$ $\,\forall x\in[0,\Lambda]$

I found a weaker proposition

If $f\in C[0,1]$ satisfies $$ \left|\int_0^1 e^{nx} f(x) dx \right| =0\,\,\,\forall n\in \mathbb{N}$$ then $f\equiv 0$ $\,\forall x\in[0,1]$

But the solution of that doesn't seem to work here.

My attempt

  1. $f(\Lambda)=0$

Suppose for contradiction that $f(\Lambda)\ne 0$. WLOG, we assume that $f(\Lambda)>0$.

Then there exists $\varepsilon > 0$ such that $f(x)>\frac{f(\Lambda)}{2}\,\,\forall x \in [\Lambda-\varepsilon,\Lambda]$.

Denote $M = \sup_{[0,\Lambda]}f$ and $c=\frac{f(\Lambda)}{2}$.

\begin{align} \int_0^\Lambda e^{nx} f(x) dx &= \int_0^{\Lambda-\varepsilon} e^{nx} f(x) dx + \int_{\Lambda-\varepsilon}^\Lambda e^{nx} f(x) dx \\ &\ge c\int_{\Lambda-\varepsilon}^\Lambda e^{nx} dx - M\int_0^{\Lambda-\varepsilon} e^{nx} dx \\ &= c\left( \frac{e^{n\Lambda}}{n} - \frac{e^{n(\Lambda-\varepsilon)}}{n} \right) - M \left( \frac{e^{n(\Lambda-\varepsilon)}}{n} - \frac{1}{n} \right) \end{align}

Thus $$ \lim_{n\to \infty} \int_0^\Lambda e^{nx} f(x) dx = +\infty $$ Contradiction.

  1. Put $X= \left\{ m : f \equiv 0 \,\, \forall x \in [m,\Lambda] \right\}$. I aim to show $\inf X = 0$.

Suppose for contradiction that $\inf X = m > 0$

If there exists $\delta>0$ such that $f(x)>0$ or $f(x)<0$ $\forall x \in ]m-\delta,m[$, using the method in $1.$ leads to a contradiction.

But how to deal with the functions like $$f(x) = (\Lambda -x) \sin \frac{1}{\Lambda - x}$$ of which we can't find such $\delta$ ?

I would highly appreciate it if you could share any thoughts on how to solve this problem. Thanks in advance!


Here is a proof. This solution completely solved the trouble I encountered. But I don't quite understand how we can figure out the lemma. I would highly appreciate it if you could give me some hints to figure it out, or post a new approach.

Proof $\ $ It suffices to show that \begin{gather} \int_{\Lambda-\lambda}^\Lambda f(x)dx=0 \quad \forall \lambda \in ]0,\Lambda] \tag{1} \end{gather} We prove $(1)$ via the following lemma, of which we attach a proof at the end.

Lemma \begin{gather} \lim_{x \uparrow \infty} \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k!} \int_0^\Lambda e^{kx(\lambda-s)} \phi(s) ds = \int_0^\lambda \phi \quad \forall \lambda \in [0,\Lambda[ \nonumber \end{gather}

Choose $\phi(s)=f(\Lambda - s)$, and then from lemma we have $\forall \lambda \in [0,\Lambda[$ $$ \lim_{x \uparrow \infty} \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k!} \int_0^\Lambda e^{kx(\lambda-s)} f(\Lambda-s) ds = \int_0^\lambda f(\Lambda-s)ds $$ $$ \lim_{x \uparrow \infty} \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k!} e^{kx(\lambda-\Lambda)} \int_0^\Lambda e^{kxu} f(u) du = \int_{\Lambda-\lambda}^\Lambda f(s)ds $$ Denote $\displaystyle\sup_{n\in \mathbb{N}} \left|\int_0^\Lambda e^{nx} f(x) dx \right| = C$. Thus \begin{align} \left|\int_{\Lambda-\lambda}^\Lambda f(s)ds\right| \nonumber &\le C \lim_{x \uparrow \infty} \left(-1 + \sum_{k=0}^\infty \frac{1}{k!} e^{kx(\lambda-\Lambda)}\right) \nonumber \\ &\le C \lim_{x \uparrow \infty} \left(-1 + \exp{\{e^{x(\lambda-\Lambda)}\}} \right) \nonumber \\ &= 0 \nonumber \end{align} Done.

Now we attach a proof of the lemma.

Proof of lemma $\ $ We aim to check \begin{align} \lim_{x \uparrow \infty} \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k!} \int_0^\Lambda e^{kx(\lambda-s)} \phi(s) ds &\overset{1}{=} \lim_{x \uparrow \infty} \int_0^\Lambda \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k!} e^{kx(\lambda-s)} \phi(s) ds \nonumber \\ &= \lim_{x \uparrow \infty} \int_0^\Lambda \left[ 1-\exp{\{ -e^{x(\lambda-s)} \}} \right] \phi(s) ds \nonumber \\ &\overset{2}{=} \int_0^\Lambda \lim_{x \uparrow \infty} \left[ 1-\exp{\{ -e^{x(\lambda-s)} \}} \right] \phi(s) ds \nonumber \\ &= \int_0^\lambda \phi \nonumber \end{align}

Denote $$ I_N = \sum_{k=1}^N \frac{(-1)^{k-1}}{k!} \int_0^\lambda e^{kx(\lambda-s)} \phi(s) ds $$ $$ J_N = \sum_{k=1}^N \frac{(-1)^{k-1}}{k!} \int_\lambda^\Lambda e^{kx(\lambda-s)} \phi(s) ds $$

Then we have \begin{align} I_N &= \int_0^\lambda \left( 1-\sum_{k=0}^\infty \frac{ (-1)^k }{k!}e^{kx(\lambda-s)} + \sum_{k=N+1}^\infty \frac{ (-1)^k }{k!}e^{kx(\lambda-s)} \right) \phi(s) ds \nonumber \\ &= \int_0^\lambda \left[ 1-\exp{\{ -e^{x(\lambda-s)} \}} \right] \phi(s) ds + \int_0^\lambda \sum_{k=N+1}^\infty \frac{ (-1)^k }{k!} e^{kx(\lambda-s)} \phi(s) ds \nonumber \\ &= : \int_0^\lambda \left[ 1-\exp{\{ -e^{x(\lambda-s)} \}} \right] \phi(s) ds + G_N \nonumber \end{align}

Note that \begin{align} |G_N| &\le ||\phi||_\infty \int_0^\lambda \sum_{k=N+1}^\infty \frac{ e^{kx(\lambda-s)} }{k!} ds \nonumber \\ &= ||\phi||_\infty \sum_{k=N+1}^\infty \int_0^\lambda \frac{ e^{kxu} }{k!} du \nonumber \\ &= ||\phi||_\infty \sum_{k=N+1}^\infty \frac{ e^{kx\lambda}-1 }{xk \cdot k!} \nonumber \end{align} which implies that $$ \lim_{N \uparrow \infty} |G_N| = 0 $$ i.e. $$ \lim_{N \uparrow \infty} I_N = \int_0^\lambda \left[ 1-\exp{\{ -e^{x(\lambda-s)} \}} \right] \phi(s) ds $$

And note that \begin{align} |J_N| &\le ||\phi||_\infty \int_\lambda^\Lambda \sum_{k=1}^\infty \frac{ e^{kx(\lambda-s)} }{k!} ds \nonumber \\ &\le ||\phi||_\infty \int_0^{\Lambda-\lambda} e^{-xu} du \nonumber \\ &= ||\phi||_\infty \frac{1-e^{x(\lambda-\Lambda)} }{x} \nonumber \end{align}

Thus \begin{align} \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k!} \int_0^\Lambda e^{kx(\lambda-s)} \phi(s) ds &= \lim_{N \uparrow \infty} \left( I_N + J_N \right) \nonumber \\ &= \int_0^\lambda \left[ 1-\exp{\{ -e^{x(\lambda-s)} \}} \right] \phi(s) ds + O(\frac{1}{x}) \nonumber \end{align} which implies that $$ \lim_{x \uparrow \infty} \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k!} \int_0^\Lambda e^{kx(\lambda-s)} \phi(s) ds = \lim_{x \uparrow \infty} \int_0^\lambda \left[ 1-\exp{\{ -e^{x(\lambda-s)} \}} \right] \phi(s) ds \nonumber \\ $$ It remains to prove that $$ R:=\lim_{x \uparrow \infty} \int_0^\lambda \exp{\{ -e^{x(\lambda-s)} \}} \phi(s) ds = 0 $$ Note that $$ |R| \le ||\phi||_\infty \lim_{x \uparrow \infty} \int_0^\lambda \exp{\{ -e^{xu} \}} du $$ and $$ \int_0^\lambda \exp{\{ -e^{xu} \}} e^{ux} du = \frac{ \frac{1}{e}-\exp{ \{ -e^{x\lambda} \} }}{x} \nonumber \ge \int_0^\lambda \exp{\{ -e^{xu} \}} du \nonumber $$ Thus we have $$ |R| \le ||\phi||_\infty \lim_{x \uparrow \infty} \int_0^\lambda \exp{\{ -e^{xu} \}} du \le ||\phi||_\infty \lim_{x \uparrow \infty} \frac{1}{ex} = 0 $$ Done.

  • $\begingroup$ I had an answer I have 2nd thoughts about. $\endgroup$ – kimchi lover May 12 '19 at 17:35
  • $\begingroup$ @kimchilover, Isn't it that your previous answer answers this question via contraposition? $\endgroup$ – Sangchul Lee May 12 '19 at 18:19
  • $\begingroup$ @SangchulLee Maybe, but I'm stupid today and don't see how. This problem involves $\limsup$ and that one $\liminf$. $\endgroup$ – kimchi lover May 12 '19 at 18:50
  • $\begingroup$ @kimchilover, My bad, never mind. I must be the real stupid one :s $\endgroup$ – Sangchul Lee May 12 '19 at 18:52
  • $\begingroup$ A very nice approach indeed! I thoroughly enjoyed the use of Lemma, and tried to simplify its proof a bit at the level of preliminary analysis. $\endgroup$ – Sangchul Lee Jul 5 '20 at 7:22

Here is a slight simplification of OP's argument. We will borrow the key lemma from OP:

Lemma. Let $f \in C([0,\Lambda])$ and write $I_n := \int_{0}^{\Lambda} e^{nx} f(x) \, \mathrm{d}x$. Then for any $0 < \lambda < \Lambda$, we have $$\lim_{N\to\infty} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k!} e^{-Nk\lambda} I_{Nk} = \int_{\lambda}^{\Lambda} f(x) \, \mathrm{d}x. $$

Assuming this lemma, the Squeezing Lemma applied to the easy bound

$$ \left| \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k!} e^{-Nk\lambda} I_{Nk} \right| \leq \Bigl(\sup_{n\geq 1} \left| I_n \right| \Bigr) \Bigl( e^{e^{-N\lambda}} - 1 \Bigr) $$

shows that $\int_{\lambda}^{\Lambda} f(x) \, \mathrm{d}x = 0$ for all $\lambda \in (0, \Lambda)$, therefore $f$ is identically zero. $\square$

So we move on to the proof of Lemma.

Remark. The proof is immediate once we can interchange the order of integration and limit operators (both $\sum_{k=1}^{\infty}$ and $\lim_{N\to\infty}$). Each interchange is easily justified if the powerful machinery called the Dominated Convergence Theorem is available. However, just in case OP is not familiar to this, we provide a more elementary approach only using uniform convergence.

Proof of Lemma. By the Weierstrass M-test, $\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k!} e^{Nk(x-\lambda)} f(x)$ converges uniformly on $[0, \Lambda]$. So we can interchange the order of summation and integration to obtain

\begin{align*} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k!} e^{-Nk\lambda} I_{Nk} = \int_{0}^{\Lambda} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k!} e^{Nk(x-\lambda)} f(x) \, \mathrm{d}x = \int_{0}^{\Lambda} g_N(x) f(x) \, \mathrm{d}x, \end{align*}

where $g_N$ is defined by

$$g_N(x) := 1 - e^{-e^{N(x-\lambda)}}. $$

Then each $g_N$ is continuous and satisfies $0 \leq g_N(x) \leq 1$. Also, fix $\alpha, \beta$ such that $0 < \alpha < \lambda < \beta < \Lambda$. Then

  • On $[0, \alpha]$, we have $g_1(x) \geq g_2(x) \geq g_3(x) \geq \dots $ and $g_N(x) \to 0$.

  • On $[\beta, \Lambda]$, we have $g_1(x) \leq g_2(x) \leq g_3(x) \leq \dots $ and $g_N(x) \to 1$.

So by the Dini's Theorem, the convergence is uniform on each of the intervals $[0, \alpha]$ and $[\beta, \Lambda]$. Since $f$ is bounded, this shows that $g_N(x)f(x) \to 0$ uniformly on $[0, \alpha]$ and $g_N(x)f(x) \to f(x)$ uniformly on $[\beta, \Lambda]$. So again, we can interchange the order of limit and integration to get

$$ \lim_{N\to\infty} \int_{0}^{\alpha} g_N(x) f(x) \, \mathrm{d}x = 0 \qquad \text{and} \qquad \lim_{N\to\infty} \int_{\beta}^{\Lambda} g_N(x) f(x) \, \mathrm{d}x = \int_{\beta}^{\Lambda} f(x) \, \mathrm{d}x. $$

Then by these and the triangle inequality,

\begin{align*} L &:=\limsup_{N\to\infty} \left| \int_{0}^{\Lambda} g_N(x)f(x) \, \mathrm{d}x - \int_{\lambda}^{\Lambda} f(x) \, \mathrm{d}x \right| \\ &\leq \limsup_{N\to\infty} \left| \int_{\alpha}^{\beta} g_N(x)f(x) \, \mathrm{d}x - \int_{\lambda}^{\beta} f(x) \, \mathrm{d}x \right| \\ &\leq 2 \int_{\alpha}^{\beta} \left| f(x) \right| \, \mathrm{d}x. \end{align*}

Since $L$ does not depend on $\alpha$ and $\beta$, letting $\alpha \uparrow \lambda$ and $\beta \downarrow \lambda$ proves that the $L = 0$, which in turn implies the desired conclusion. $\square$


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