# version of the fundamental lemma of calculus of variations

Lemma. If a real function $f$ is continuous^{} on the interval $[a,b]$ and if

$${\int}_{a}^{b}f(x)\phi (x)\mathit{d}x=\mathrm{\hspace{0.33em}0}$$ |

for all functions $\phi $ continuously differentiable on the interval and vanishing at its end points, then
$f(x)\equiv \mathrm{\hspace{0.17em}0}$ on the whole interval.

*Proof.* We make the antithesis that $f$ does not vanish identically. Then there exists a point ${x}_{0}$ of the open interval $(a,b)$ such that $f({x}_{0})\ne 0$; for example $f({x}_{0})>0$. The continuity of $f$ implies that there are the numbers $\alpha $ and $\beta $ such that $$ and $f(x)>0$ for all
$x\in [\alpha ,\beta ]$. Now the function ${\phi}_{0}$ defined by

${\phi}_{0}(x):=\{\begin{array}{cc}{(x-\alpha )}^{2}{(x-\beta )}^{2}\hspace{1em}\text{for}\alpha \leqq x\leqq \beta ,\hfill & \\ 0\hspace{1em}\hspace{1em}\text{otherwise}\hfill & \end{array}$ |

fulfils the requirements for the functions $\phi $. Since both $f$ and ${\phi}_{0}$ are positive on the open interval $(\alpha ,\beta )$, we however have

$${\int}_{a}^{b}f(x){\phi}_{0}(x)\mathit{d}x={\int}_{\alpha}^{\beta}f(x){\phi}_{0}(x)\mathit{d}x>\mathrm{\hspace{0.33em}0}.$$ |

Thus the antithesis causes a contradiction^{}. Consequently, we must have $f(x)\equiv \mathrm{\hspace{0.17em}0}$.

Title | version of the fundamental lemma of calculus of variations^{} |
---|---|

Canonical name | VersionOfTheFundamentalLemmaOfCalculusOfVariations |

Date of creation | 2013-03-22 19:12:04 |

Last modified on | 2013-03-22 19:12:04 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 4 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 26A15 |

Classification | msc 26A42 |

Synonym | fundamental lemma of calculus of variations |

Related topic | EulerLagrangeDifferentialEquation |