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Use the inequality \alpha \beta \leq {\frac  {\alpha ^{p}}{p}}+{\frac  {\beta ^{q}}{q}}\, for \alpha ,\beta \geq 0,{\frac  {1}{p}}+{\frac  {1}{q}}=1\, to prove that

\int _{a}^{b}\left|f(t)g(t)\right|dt\leq \left(\int _{a}^{b}\left|f(t)\right|^{p}\right)^{{1/p}}\left(\int _{a}^{b}\left|g(t)\right|^{q}\right)^{{1/q}}\,


for p>1\, and {\frac  {1}{p}}+{\frac  {1}{q}}=1\,, and then use this to prove the triangular inequality:


\left(\int _{a}^{b}\left|f(t)+g(t)\right|^{p}dt\right)^{{1/p}}\leq \left(\int _{a}^{b}\left|f(t)\right|^{p}\right)^{{1/p}}+\left(\int _{a}^{b}\left|g(t)\right|^{p}\right)^{{1/p}}\,


First Part

Let {\hat  {f}}\, and {\hat  {g}}\, be such that \int _{a}^{b}|{\hat  {f}}|^{p}dt=\int _{a}^{b}|{\hat  {g}}|dt=1\,.


Let \alpha =|{\hat  {f}}|\, and \beta =|{\hat  {g}}|\,.


Using the inequality, |{\hat  {f}}{\hat  {g}}|\leq |{\hat  {f}}||{\hat  {g}}|\leq {\frac  {|{\hat  {f}}|^{p}}{p}}+{\frac  {|{\hat  {g}}|^{q}}{q}}\,.


Now integrate from a\, to b\,.


\int _{a}^{b}|{\hat  {f}}{\hat  {g}}|dt\leq \int _{a}^{b}{\frac  {|{\hat  {f}}|^{p}}{p}}dt+\int _{a}^{b}{\frac  {|{\hat  {g}}|^{q}}{q}}dt={\frac  {1}{p}}+{\frac  {1}{q}}=1\,


Set {\hat  {f}}={\frac  {f}{\left(\int _{a}^{b}|f|^{p}dt\right)^{{1/p}}}}\, and {\hat  {g}}={\frac  {g}{\left(\int _{a}^{b}|g|^{q}dt\right)^{{1/q}}}}\,.


Plug these into the previous equation and multiply by the denominator:


\int _{a}^{b}|fg|dt\leq \left(\int _{a}^{b}|f|^{p}dt\right)^{{1/p}}\left(\int _{a}^{b}|g|^{q}dt\right)^{{1/q}}\,


Second Part

|f+g|^{p}=|f+g||f+g|^{{p-1}}\,

\leq (|f|+|g|)|f+g|^{{p-1}}\,

|f||f+g|^{{p-1}}+|g||f+g|^{{p-1}}\,

Therefore

\int _{a}^{b}|f+g|^{p}dt\leq \int _{a}^{b}|f||f+g|^{{p+1}}dt+\int _{a}^{b}|g||f+g|^{{(p-1)q}}dt\,


From the first part,


\int _{a}^{b}|f||f+g|^{{p-1}}dt\leq \left(\int _{a}^{b}|f|^{p}dt\right)^{{1/p}}\left(\int _{a}^{b}|f+g|^{{(p-1)q}}\right)^{{1/q}}\,


\int _{a}^{b}|g||f+g|^{{p-1}}dt\leq \left(\int _{a}^{b}|g|^{p}dt\right)^{{1/p}}\left(\int _{a}^{b}|f+g|^{{(p-1)q}}\right)^{{1/q}}\,


{\frac  {1}{p}}+{\frac  {1}{q}}=1\implies q+p=pq\implies (p-1)q=p\,


Adding these two integrals,


\int _{a}^{b}|f+g|^{p}dt\leq \left[\left(\int _{a}^{b}|f|^{p}dt\right)^{{1/p}}+\left(\int _{a}^{b}|g|^{p}dt\right)^{{1/p}}\right]\left(\int _{a}^{b}|f+g|^{p}dt\right)^{{1/q}}\,


Multiply both sides by the inverse of the last term and use 1-{\frac  {1}{q}}={\frac  {1}{p}}\,


\left(\int _{a}^{b}|f+g|^{p}dt\right)^{{1/p}}\leq \left(\int _{a}^{b}|f|^{p}dt\right)^{{1/p}}+\left(\int _{a}^{b}|g|^{p}dt\right)^{{1/p}}\,



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