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

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,

& \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{aligned}$$ As $$\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 $\begin{aligned} & {{f}^{2}}\left( x \right)={{1}^{2}} \\\ & \Rightarrow {{f}^{2}}\left( x \right)=1 \\\ \end{aligned}$ Now, we were asked to find the value of $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}$ So, let us consider $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}$, $\begin{aligned} & \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=\int\limits_{0}^{1}{\left( 1 \right)dx} \\\ & \\\ & \Rightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=\left[ x+C \right]_{0}^{1} \\\ & \\\ & \Rightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=\left[ \left( 1+C \right)-\left( 0+C \right) \right] \\\ & \\\ & \Rightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=1+C-0-C \\\ & \\\ & \Rightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=1 \\\ \end{aligned}$ 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 ${f}''\left( x \right)$ .