Question

Let f be differentiable function from R to R such that $\left| f\left( x \right)-f\left( y \right) \right|\le 2{{\left| x-y \right|}^{\dfrac{3}{2}}}$ , for all $x,y\in R$. If $f\left( 0 \right)=1$ then find $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}$

Answer 1

Hint: We start solving this problem by dividing both sides of the given inequality by \[\left| x-y \right|\] . After dividing the both sides of the inequality by \[\left| x-y \right|\], we get a new inequality. Then we apply the formula $$\displaystyle \lim_{h \to 0}$$ $\dfrac{f\left( x+h \right)-f\left( x \right)}{\left( x+h \right)-x}={f}'\left( x \right)$ . Then we find the function $f\left( x \right)$ . By squaring it, we get ${{f}^{2}}\left( x \right)$ , then we find the value of $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}$.

Complete step-by-step answer:
Let us consider the given inequality, $\left| f\left( x \right)-f\left( y \right) \right|\le 2{{\left| x-y \right|}^{\dfrac{3}{2}}}$
Now we divide both the sides of the above inequality by \[\left| x-y \right|\], we get,
$\dfrac{\left| f\left( x \right)-f\left( y \right) \right|}{\left| x-y \right|}\le 2{{\left| x-y \right|}^{\dfrac{3}{2}}}$
Let us apply limits on both the sides of the above obtained inequality, we get,
$$\displaystyle \lim_{x \to y}$$ \[\left| \dfrac{f\left( x \right)-f\left( y \right)}{x-y} \right|\le \displaystyle \lim_{x \to y},2{{\left| x-y \right|}^{\dfrac{3}{2}}}..................\left( 1 \right)\]
Now, let us consider the formula $$\displaystyle \lim_{h \to 0}$$ $\dfrac{f\left( x+h \right)-f\left( x \right)}{\left( x+h \right)-x}={f}'\left( x \right)$
Applying the above formula to equation (1), we get,

\[\begin{align}
  & \underset{x\to y}{\mathop{\lim }}\,\left| \dfrac{f\left( x \right)-f\left( y \right)}{x-y} \right|\le \underset{x\to y}{\mathop{\lim }}\,2{{\left| x-y \right|}^{\dfrac{3}{2}}} \\
 & \\
 & \Rightarrow \left| {f}'\left( y \right) \right|\le 2{{\left| y-y \right|}^{\dfrac{3}{2}}} \\
 & \\
 & \Rightarrow \left| {f}'\left( y \right) \right|\le 2{{\left| 0 \right|}^{\dfrac{3}{2}}} \\
 & \\
 & \Rightarrow \left| {f}'\left( y \right) \right|\le 0 \\
\end{align}\]
As \[\left| {f}'\left( y \right) \right|\] is always non-negative, because a modulus of any number is positive, we get\[{f}'\left( y \right)=0\] for any real number y.
Let us consider the known theorem that if $f\left( x \right)$ is a constant function, then ${f}'\left( x \right)=0$ and vice-versa.
By using the above theorem, we get,
$f\left( y \right)=K$ , where K is some constant.
So, we get,
$f\left( x \right)=K$
We were given that $f\left( 0 \right)=1$.
So, we get,
$f\left( x \right)=1$ as $f\left( x \right)$ is a constant function.
Now, we find ${{f}^{2}}\left( x \right)$ by squaring the function $f\left( x \right)=1$ on both sides, we get
$\begin{align}
  & {{f}^{2}}\left( x \right)={{1}^{2}} \\
 & \Rightarrow {{f}^{2}}\left( x \right)=1 \\
\end{align}$
Now, we were asked to find the value of $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}$
So, let us consider $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}$,
$\begin{align}
  & \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=\int\limits_{0}^{1}{\left( 1 \right)dx} \\
 & \\
 & \Rightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=\left[ x+C \right]_{0}^{1} \\
 & \\
 & \Rightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=\left[ \left( 1+C \right)-\left( 0+C \right) \right] \\
 & \\
 & \Rightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=1+C-0-C \\
 & \\
 & \Rightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=1 \\
\end{align}$
Therefore, $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=1$
Hence, the answer is 1.

Note: The possibility of making mistakes in this problem is one may confuse what limits to be taken while applying limits to the given inequality. One may also make a mistake by considering ${{f}^{2}}\left( x \right)$ as the second derivative of $f\left( x \right)$ . So, one must know the difference between the terms ${{f}^{2}}\left( x \right)$ and ${f}''\left( x \right)$ .