# Bounding $\int_0^1 f(x) dx$, given $\int_0^1 f'(x)^2 dx \leq 1$ and $f(0) = 0$.

Let $S$ be the set of all differentiable function $f \colon [0,1] \rightarrow \mathbb{R}$ such that $\int_0^1 f'(x)^2 dx \leq 1$ and $f(0) = 0$. Define $J(f) := \int_0^1 f(x) dx$. Show that $J$ is bounded on $S$, find its supremum and see if there is a function $f_0$ in $S$ at which $J$ attains its maximum value.

-

Looks alot like a calculus of variations minimization problem. – toypajme May 27 '12 at 19:14

## 1 Answer

To get you started:

1. Integration by parts yields $J(f)=\int_0^1 (1-x)f\,'(x)\,dx$.

2. Cauchy-Schwarz will appear sooner or later.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8900848627090454, "perplexity_flag": "head"}