 TIFR 2014 Problem 6 Solution is a part of TIFR entrance preparation series offered by Cheenta. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects. Generally, the exams scheduled in December.
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## Problem

Let $$f:[0,1] \to \mathbb{R}$$ be a continuous function. Which of the following statement is always true?

A. $$\int_{0}^{1} f^2(x)dx = (\int_{0}^{1}f(x)dx)^2$$

B.  $$\int_{0}^{1} f^2(x)dx \le (\int_{0}^{1}|f(x)|dx)^2$$

C.  $$\int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2$$

D.  $$\int_{0}^{1} f^2(x)dx < (\int_{0}^{1}f(x)dx)^2$$

## Discussion:

We first recall the Cauchy-Schwartz inequality for an inner product space $$V$$ and two vectors $$a,b\in V$$

$$<a,b> \le ||a||||b||$$.

Here, we also remember the fact that $$C[0,1]$$ (the set of all continuous real (/complex) valued functions on [0,1] ) forms an inner product space with respect to the inner product

$$<f,g>= \int_{0}^{1} f(x)g(x)dx$$ (We are only taking real valued functions so we neglect the conjugation…)

We want to apply this inequality to suitable functions so that we get some inequality from the options above.

Let’s try $$|f|$$ and $$1$$ (the constant function).

We get: $$\int_{0}^{1} |f(x)|1dx \le (\int_{0}^{1} |f(x)|^2dx)^{1/2} (\int_{0}^{1} 1^2dx)^{1/2}$$

Squaring, we get

$$\int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2$$.

So option C $$\int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2$$ is correct.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Inner Product Space, Cauchy – Schwartz inequality,continuous real (/complex) valued functions
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert