Let $f$ be a real-valued continuous function on $[0,1]$ which is twice continu-ously differentiable on $(0,1)$. Suppose that $f(0) = f(1) = 0$

Let $f$ be a real-valued continuous function on $[0,1]$ which is twice continu-ously differentiable on $(0,1)$. Suppose that $f(0) = f(1) = 0$ and $f$ satisfies the following equation:

$$x^2f''(x) + x^4f'(x) - f(x) = 0$$

(a) If $f$ attains its maximum $M$ at some point $x_0$ in the open interval $(0,1)$, then prove that $M= 0$.

(b)Prove that $f$ is identically zero on$[0,1]$.

My idea: At $x_0$, $f'(x_0) = 0$ and $f''(x_0)$ is negative as it is given that the function is twice differentiable and attains maximum at $x_0$. The given equation at $x_0$ reduces to $$f''(x_0) = \frac{f(x_0)}{x_0^2}$$

$f''(x_0) < 0$ implies $f(x_0) < 0$. That is the maximum value attained by the function $f$ is negative on $[0,1]$. Which is a contradiction as $f$ attains $0$ at the point $x=0$. (It is given $f$ attains maximum at $x_0$)

Hence the only way this can be satisfied is if $f(x_0) = M = 0$. Similarly we can prove that the minimum attained by the function $m = 0$.

since this is a continous value function on $[0,1]$, $m \le f \le M$.

Hence $f$ is identically equal to $0$.

• It sounds like you've got it to me. – John Martin Apr 26 '16 at 21:27

If $f(x_0) \neq 0$, from the differential equation $x_0^2f''(x_0)-f(x_0)=0$ and from the condition that $x_0$ is a maximum we simply get that $f(x_0)<0$.
Finally, from continuity arguments, one can say that from Weierstrass theorem that the inequality $f(x_0)\geq f(x)$ is satisfied over all points of the interval $[0,1]$ so that $f(x_0)\geq 0$, since $f(0)=f(1)=0$.