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Let $a$ and $b$ be a real numbers and $t,L$ are positive constants and $f$ a function that satisfies $f(t,0)=f(t,L)=0$, is there any conditions on $a$ and $b$ independent of $t$ to make this inequality hold?

$$a\int_{0}^{L}{f_{t}}^{2}(t,x)dx-b\int_{0}^{L}\left(\int_{0}^{t}f_{x}(y,x)dy\right)^{2}dx\geqslant0$$ Where $$f_t = \frac{\partial f}{\partial t} \hspace{1cm} f_x = \frac{\partial f}{\partial x}$$

I have tried to use the Poincaré's inequality and the Cauchy-Scwartz's inequality but it didn't work. Thank you.

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  • $\begingroup$ Seems like it could be an interesting question but what does the notation $f_t$ and $f_x$ mean? $\endgroup$ – mathreadler Aug 25 '17 at 8:54
  • $\begingroup$ It is mean the derivative with respect to t and x. $\endgroup$ – Gustave Aug 25 '17 at 9:08
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Let $f(t,x)=x(L-x)$. Then $f_t=0$, so the first integral is zero. On the other hand, $f_x=L-2x$, so $$\int_{0}^{t}f_{x}(y,x)dy=\int_{0}^{t}(L-2x)dy=(L-2x)t$$ and $$\int_{0}^{L} ((L-2x)t)^2dx=\frac{L^3t^2}3.$$

Thus $b\le 0$.

So remains the only non-trivial case $a<0$. Let $f(t,x)=g(t)x(L-x)$ for some function $g(t)$. Then $f_t=g_t x(L-x)$, so the first integral is $$\int_0^L (g_t(t)x(L-x))^2dx=g_t(t)^2\int_0^L (x(L-x))^2dx=\frac{g_t(t)^2L^5}{30}.$$

On the other hand, $f_x(y,x)=g(y)(L-2x)$, so $$\int_{0}^{t}f_{x}(y,x)dy=\int_{0}^{t}g(y)(L-2x)dy=(L-2x)\int_0^t g(y)dy$$ and $$\int_{0}^{L} ((L-2x)\int_0^t g(y)dy)^2dx=\frac{L^3(\int_0^t g(y)dy)^2}3.$$

Thus we have

$$|a|\frac{g_t(t)^2L^5}{30}\le |b|\frac{L^3(\int_0^t g(y)dy)^3}2$$

$$|a|g_t(t)^2L^2\le 10|b|\left(\int_0^t g(y)dy\right)^2.$$

Now put $g(t)=e^{Nt}$ for $N>0$. Then $g_t(t)=Ne^{Nt}$ and $\int_0^t g(y)dy=\frac{e^{Nt}-1}{N}$, so we have

$$|a|N^2e^{2Nt}L^2\le 10|b|\left(\frac{e^{Nt}-1}{N}\right)^2\le 10|b|\left(\frac{e^{Nt}}{N}\right)^2$$

$$|a|N^4L^2\le 10|b|.$$

Since $N$ can be arbitrary big, we obtain $|a|=0$.

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    $\begingroup$ thank you brother. sorry for the delay. $\endgroup$ – Gustave Sep 3 '17 at 18:24

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