How do I prove $\int_a^b |f(x)|^2 dx \leq \frac{(b-a)^2}{2} \int_a^b |f'(x)|^2 dx$.

Question

Let $f$ be $C^1[a,b]$ continuously differentiable) with $f(a)=0$. [This may be generalized to $f(b)=0$ or $f(a)f(b)=0$]

I want to show

$$\int_a^b |f(x)|^2\ \mathrm{d}x\ \leq\ \frac{(b-a)^2}{2} \int_a^b |f'(x)|^2\ \mathrm{d}x.$$

Since $f$ is differentiable, it is continuous so it is integrable. And since $f'$ is continuous so it is integrable. So, L.H.S and R.H.S are well defined.

My first trial was the usage of integration by parts, but since I am dealing with $|\cdot |$, the absolute value of a function, it seems it is not a good direction.

Does this answer your question? https://math.stackexchange.com/questions/3646679/integral-inequality-with-cauchy-schwarz-inequality?noredirect=1 or this https://artofproblemsolving.com/community/c7h546384p3162147 — Sumanta, Jun 07 '21 at 08:49
@Sumanta, WOW Thanks! It was Wirtinger's inequality! The second link really helps me a lot — phy_math, Jun 07 '21 at 09:08
@phy_math I guess you are going through this proof: geodesic is locally(in a chart) length minimizing curve among all smooth curves in the same chart having same endpoints. — Sumanta, Jun 07 '21 at 11:15

score 2 · Accepted Answer · edited Jun 07 '21 at 11:09

Following @Sumanta comments, Based on the second link, I just make up a solution.

The first link Integral inequality with Cauchy-Schwarz-inequality, use Holder inequality and Cauchy-Schwarz inequality with proper and do integral with proper assumption, and the second link use only Cauchy-Schwarz inequality.

Frankly speaking, it is nothing but a re-arranging(=copy paste) the proof in the second link

Let $\displaystyle f(x) = \int_a^x f'(t)\ \mathrm{d}t$. Now from Cauchy-Schwartz one have \begin{align} \left( \int_a^x |f'(t)|\ \mathrm{d}t \right)^2 \leq \left( \int_a^x |f'(t) |^2\ \mathrm{d}t \right) \left(\int_a^x 1^2\ \mathrm{d}t \right) \end{align} Since $\displaystyle f(x)^2 = \left( \int_a^x f'(t)\ \mathrm{d}t \right)^2 \leq \left( \int_a^x |f'(t)|\ \mathrm{d}t \right)^2$ we have \begin{align} &f^2 (x) = \left( \int_a^x f'(t)\ \mathrm{d}t \right)^2 \leq \left( \int_a^b |f'(t)|^2\ \mathrm{d}t \right) (x-a) \\ &\quad \implies \quad \int_a^b f(x)^2\ \mathrm{d}x \leq \left(\int_a^b |f'(t)|^2\ \mathrm{d}t \right) \int_a^b (t-a)\ \mathrm{d}t \\ & \quad \implies \quad \int_a^b |f(t)|^2\ \mathrm{d}t \leq \frac{(b-a)^2}{2} \left( \int_a^b |f'(t) |^2\ \mathrm{d}t \right) \end{align} This inequality is called "Wirtinger's inequality."

How do I prove $\int_a^b |f(x)|^2 dx \leq \frac{(b-a)^2}{2} \int_a^b |f'(x)|^2 dx$.

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