**Claim:** There exists no \(\displaystyle f \in C^4[-1, \infty)\) of the form \(\displaystyle f(x) = c_1 e^{x \sqrt{2}} + c_2 e^{-x \sqrt{2}} + c_3 \cos(x \sqrt{2}) + c_4 \sin(x \sqrt{2}), c-1, c_2, c_3, c_4 \in \mathbb{R}\) satisfying \(\displaystyle f(0) = 1, f(1) = 0, \lim_{x \to \infty} f(x) = 0\).

Note 1: The satisfaction of y'''' = y at any point implies that y is 4 times differentiable there, and so its third derivative there is continuous. Also, y is continuous there, since its first derivative exists there. But then y'''' = y says that y'''' is also continuous there... thus the \(\displaystyle C^4\) statement. Moreover, given its form, f is obviously \(\displaystyle C^{\infty}\) on (the interior of) its domain.

Note 2: The choice of a single interval, \(\displaystyle [-1, \infty)\), was based on the initial conditions. If you permit a break in the domain, so that the domain isn't that entire interval, then it becomes possible to satisfy that differential equation with those initial conditions. That's because a break in the domain means that the restrictions imposed by the initial conditions at 0, 1, and -1 no longer will carry over and influence the function on the disconnected interval where it's going to infinity. If you permitted a break in the domain, you'd in effect have different initial value problems on the different intervals, and there would be no conflict harmonizing into a single function, because the intervals would be disconnected.

Note 3: The -1 left endpoint of the domain was chosen because of the conditions, but its role in proving that claim is to have x = 0 in the interior of the function's domain. That's useful because the claim hinges on f(0) = 1, and you'd want to avoid any complications arising from derivatives and differential equations at closed interval endpoints, which might've posed a concern had the stated domain been \(\displaystyle [0, \infty)\).

**Proof:** Assume there is such a function \(\displaystyle f\) with all of those properties.

**Define:** \(\displaystyle g(x) = c_2 e^{-x \sqrt{2}} + c_3 \cos(x \sqrt{2}) + c_4 \sin(x \sqrt{2})\). In other words \(\displaystyle g(x) = f(x) - c_1 e^{x \sqrt{2}}\).

**Claim:** The function \(\displaystyle g\) is bounded on the positive reals.

Since for \(\displaystyle x> 0\) you have that \(\displaystyle 0 < e^{-x \sqrt{2}} < 1\), you get, for \(\displaystyle x> 0\), that:

\(\displaystyle | c_2 e^{-x \sqrt{2}} + c_3 \cos(x \sqrt{2}) + c_4 \sin(x \sqrt{2}) |\)

\(\displaystyle \le | c_2 | \ | e^{-x \sqrt{2}} | + | c_3 | \ | \cos(x \sqrt{2}) | + | c_4 | \ | \sin(x \sqrt{2}) |\)

\(\displaystyle \le | c_2 | + | c_3 | + | c_4 |\).

Thus the function \(\displaystyle g(x) = c_2 e^{-x \sqrt{2}} + c_3 \cos(x \sqrt{2}) + c_4 \sin(x \sqrt{2})\) is bounded (by \(\displaystyle |c_2|+|c_3|+|c_4|\)) on the positive reals.

**Claim:** \(\displaystyle c_1 = 0\) (and so \(\displaystyle f = g\)).

Assume \(\displaystyle c_1 \ne 0\). Then since \(\displaystyle \lim_{x \to \infty} e^{x \sqrt{2}} = \infty\), you'd have that \(\displaystyle \lim_{x \to \infty} c_1 e^{x \sqrt{2}} = \pm \infty\).

But then \(\displaystyle 0 = \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left( c_1 e^{x \sqrt{2}} + g(x) \right)\) would require that \(\displaystyle \lim_{x \to \infty} |g(x)| = \infty\),

since \(\displaystyle g\) would have to "blow up in the opposite way" to counter \(\displaystyle \lim_{x \to \infty} c_1 e^{x \sqrt{2}} = \pm \infty\) in order for the sum \(\displaystyle c_1 e^{x \sqrt{2}} + g(x)\) to go to 0 as x went to infinity.

But since \(\displaystyle g\) is bounded on the positive reals, that can't hold. Thus \(\displaystyle c_1 \ne 0\) produces a contradiction, so \(\displaystyle c_1 = 0\).

(Technically, if \(\displaystyle g\) is bounded by \(\displaystyle B\) on the positive reals, and \(\displaystyle c_1 >0\), use that \(\displaystyle c_1 e^{x \sqrt{2}}\) is eventually always greater than \(\displaystyle 1 + B\) to conclude that \(\displaystyle f(x) = c_1 e^{x \sqrt{2}} + g(x)\) is eventually always greater than 1,

and so it's limit to infinity can't be 0. Likewise, if \(\displaystyle c_1 < 0\), then \(\displaystyle c_1 e^{x \sqrt{2}}\) is eventually always less than \(\displaystyle -1 - B\), so that \(\displaystyle f(x) = c_1 e^{x \sqrt{2}} + g(x)\) is eventually always less than -1, and so it's limit to infinity can't be 0.)

**Claim:** \(\displaystyle c_3 = c_4 = 0\).

Since \(\displaystyle \lim_{x \to \infty} f(x) = 0\), have that \(\displaystyle \lim_{n \to \infty} f(x_n) = 0\) for any sequence \(\displaystyle \{x_n\}_{n = 1}^{\infty}\) such that \(\displaystyle x_n \to \infty\).

Consider the sequences \(\displaystyle s_n = \frac{2 \pi n}{\sqrt{2}}\) and \(\displaystyle t_n = \frac{(\pi/2) + 2 \pi n}{\sqrt{2}}\). Clearly both go to infinity.

But \(\displaystyle f(s_n) = f\left( \frac{2 \pi n}{\sqrt{2}} \right)\)

\(\displaystyle = c_2 e^{-\left( \frac{2 \pi n}{\sqrt{2}} \right) \sqrt{2}} + c_3 \cos\left(\sqrt{2} \left( \frac{2 \pi n}{\sqrt{2}} \right)\right) + c_4 \sin\left(\sqrt{2}\left( \frac{2 \pi n}{\sqrt{2}} \right)\right)\)

\(\displaystyle = c_2 e^{-2 \pi n} + c_3 \cos( 2 \pi n ) + c_4 \sin(2 \pi n) = c_2 e^{-2 \pi n} + c_3\).

Thus \(\displaystyle 0 = \lim_{n \to \infty} f(s_n) = \lim_{n \to \infty} \left( c_2 e^{-2 \pi n} + c_3\right) = c_3\).

Likewise \(\displaystyle f(t_n) = f\left( \frac{(\pi/2) + 2 \pi n}{\sqrt{2}} \right)\)

\(\displaystyle = c_2 e^{-(\pi/2) -2 \pi n} + c_3 \cos\left( (\pi/2) + 2 \pi n \right) + c_4 \sin\left((\pi/2) + 2 \pi n\right)\)

\(\displaystyle = c_2 e^{-(\pi/2)}e^{ -2 \pi n} + c_4\),

and so \(\displaystyle 0 = \lim_{n \to \infty} f(t_n) = \lim_{n \to \infty} \left(c_2 e^{-(\pi/2)}e^{ -2 \pi n} + c_4 \right) = c_4\).

**Completion of proof**

Have shown that \(\displaystyle \lim_{x \to \infty} f(x) = 0\) implies that \(\displaystyle c_1 = c_3 = c_4 = 0\), and so have that \(\displaystyle f(x) = c_2 e^{-x \sqrt{2}}\).

But then using that f(1) = 0 gives: \(\displaystyle f(1) = 0 \Rightarrow c_2 e^{-\sqrt{2}} = 0 \Rightarrow c_2 = 0\). Thus \(\displaystyle f\) is the constant function 0.

But that contradicts f(0) = 1, and so the original assumption - that there exists some f meeting all those stated criteria - has produced a contradiction. Therefore no such f exists. QED.